# Linear Equations of the First Kind with Constant Limits of Integration

Authored by: Andrei D. Polyanin , Alexander V. Manzhirov

# Handbook of Integral Equations

Print publication date:  February  2008
Online publication date:  February  2008

Print ISBN: 9781584885078
eBook ISBN: 9780203881057

10.1201/9781420010558.ch3

#### Abstract

► Notation: f = f (x), g = g(x), h = h(x), K = K (x), and M = M (x) are arbitrary functions (these may be composite functions of the argument depending on two variables x and t); A, B, C, a, b, c, k, α, β, γ, λ, and μ are free parameters; and n is a nonnegative integer.

#### Linear Equations of the First Kind with Constant Limits of Integration

► Notation: f = f (x), g = g(x), h = h(x), K = K (x), and M = M (x) are arbitrary functions (these may be composite functions of the argument depending on two variables x and t); A, B, C, a, b, c, k, α, β, γ, λ, and μ are free parameters; and n is a nonnegative integer.

#### 3.1  Equations Whose Kernels Contain Power-Law Functions

In Section 3.1, we mean that kernels of the integral equations discussed may contain power-law functions or modulus of power-law functions.

#### 3.1-1  Kernels Linear in the Arguments x and t.

1. $∫ 0 1 | x − t | y ( t ) d t = f ( x )$

.

1°.Let us remove the modulus in the integrand:

1 $∫ 0 x ( x − t ) y ( t ) d t + ∫ x 1 ( t − x ) y ( t ) d t = d f ( x ) .$

Differentiating (1) with respect to x yields

2 $∫ 0 x y ( t ) d t − ∫ x 1 y ( t ) d t = f ′ x ( x ) .$

Differentiating (2) yields the solution

3 $y ( x ) = 1 2 f ″ x x ( x ) .$

2°.Let us demonstrate that the right-hand side f(x)of the integral equation must satisfy certain relations. By setting x = 0 and x =1in(1), we obtain two corollaries $∫ 0 1 t y ( t ) d t = f ( 0 )$

and $∫ 0 1 ( 1 − t ) y ( t ) d t = f ( 1 )$ , which can be rewritten in the form

4 $∫ 0 1 t y ( t ) d t = f ( 0 ) , ∫ 0 1 y ( t ) d t = f ( 0 ) + f ( 1 ) .$

Substitute y(x)of (3) into (4). Integration by parts yields $f x ′ ( 1 ) = f ( 1 ) + f ( 0 )$

and $f x ′ ( 1 ) − f x ′ ( 0 ) =$ 2f (1) + 2f (0). Hence, we obtain the desired constraints for f(x):

5 $f ′ x ( 1 ) = f ( 0 ) + f ( 1 ) , f ′ x ( 0 ) + f ′ x ( 1 ) = ( 0 ) .$

Conditions (5) make it possible to find the admissible general form of the right-hand side of the integral equation:

$f ( x ) = F ( x ) + A x + B , A = − 1 2 [ F ′ x ( 1 ) + F ′ x ( 0 ) ] , B = 1 2 [ F ′ x ( 1 ) − F ( 1 ) − F ( 0 ) ] ,$

where F(x)is anarbitrary bounded twice differentiable function with bounded first derivative.

2. $∫ a b | x − t | y ( t ) d t = f ( x ) , 0 ≤ a ≤ b < ∞$

.

This is a special case of equation 3.8.3 with g(x) = x.

Solution:

$y ( x ) = 1 2 f ″ x x ( x ) .$

The right-hand side f(x) of the integral equation must satisfy certain relations. The general form of f(x)isasfollows:

$f ( x ) = F ( x ) + A x + B , A = − 1 2 [ F ′ x ( a ) + F ′ x ( b ) ] , B = 1 2 [ a F ′ x ( a ) + b F ′ x ( b ) − F ( a ) − F ( b ) ] ,$

where F(x) is an arbitrary bounded twice differentiable function (with bounded first derivative).

3. $∫ 0 a | λ x − t | y ( t ) d t = f ( x ) , λ > 0$

.

Here 0 ≤ xa and 0 ≤ ta.

1°.Let us remove the modulus in the integrand:

1 $∫ 0 λ x ( λ x − t ) y ( t ) d t + ∫ λ x a ( t − λ x ) y ( t ) d t = f ( x ) .$

Differentiating (1) with respect to x,we find that

2 $λ ∫ 0 λ x y ( t ) d t − λ ∫ λ x a y ( t ) d t = f ′ x ( x ) .$

Differentiating (2) yields $2 λ 2 y ( λ x ) = f x x ″ ( x )$

. Hence, we obtain the solution

3 $y ( x ) = 1 2 λ 2 f ″ x x ( x λ ) .$

2°.Letusdemonstrate that the right-hand side f(x) of the integral equation must satisfy certain relations. By setting x =0in(1)and(2), we obtain two corollaries

4 $∫ 0 a t y ( t ) d t = f ( 0 ) , λ ∫ 0 a y ( t ) d t = − f ′ x ( 0 ) .$

Substitute y(x) from (3) into (4). Integrating by parts yields the desired constraints for f(x):

5 $( a / λ ) f ′ x ( a / λ ) = f ( 0 ) + f ( a / λ ) , f ′ x ( 0 ) + f ′ x ( a / λ ) = 0.$

Conditions (5) make it possible to establish the admissible general form of the right-hand side of the integral equation:

$f ( x ) = F ( z ) + A z + B , z = λ x ; A = − 1 2 [ F ′ z ( a ) + F ′ z ( 0 ) ] , B = 1 2 [ a F ′ z ( a ) − F ( a ) − F ( 0 ) ] ,$

where F(x) is an arbitrary bounded twice differentiable function (with bounded first derivative).

4. $∫ 0 a | x − λ t | y ( t ) d t = f ( x ) , λ > 0$

Here 0 ≤ xa and 0 ≤ ta.

Solution:

$y ( x ) = 1 2 λ f ″ x x ( λ x ) .$

The right-hand side f(x) of the integral equation must satisfy the relations

$a λ f ′ x ( a λ ) = f ( 0 ) + f ( a λ ) , f ′ x ( 0 ) + f ′ x ( a λ ) = 0.$

Hence, it follows the general form of the right-hand side:

$f ( x ) = F ( x ) + A x + B , A = − 1 2 [ F ′ x ( λ a ) + F ′ x ( 0 ) ] , B = 1 2 [ a λ F ′ x ( a λ ) − F ( λ a ) − F ( 0 ) ] ,$

where F(x)is anarbitrary bounded twice differentiable function (with bounded first derivative).

#### 3.1-2  Kernels Quadratic in the Arguments x and t.

5. $∫ 0 a | A x + B x 2 − t | y ( t ) d t = f ( x ) , A > 0 , B > 0$

.

This is a special case of equation 3.8.5 with g(x) = Ax + Bx 2.

6. $∫ 0 a | x − A t − B t 2 | y ( t ) d t = f ( x ) , A > 0 , B > 0$

.

This is a special case of equation 3.8.6 with g(x) = At + Bt 2.

7. $∫ a b | x t − t 2 | y ( t ) d t = f ( x ) , 0 ≤ a < b < ∞$

.

The substitution w(t)= ty(t)leads to an equation of the form 3.1.2:

$∫ a b | x − t | w ( t ) d t = f ( x ) .$

8. $∫ a b | x 2 − t 2 | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x) = x 2.

Solution: $y ( x ) = d d x [ f x ′ ( x ) 4 x ]$

. The right-hand side f(x) of the equation must satisfy certain constraints, given in 3.8.3.

9. $∫ 0 a | x 2 − β t 2 | y ( t ) d t = f ( x ) , β > 0$

.

This is a special case of equation 3.8.4 with g(x) = x 2 and β = λ2.

10. $∫ 0 a | A x + B x 2 − A λ t − B λ 2 t 2 | y ( t ) d t = f ( x ) , λ > 0$

.

This is a special case of equation 3.8.4 with g(x)=Ax + Bx 2.

#### 3.1-3  Kernels Containing Integer Powers of x and t or Rational Functions.

11. $∫ a b | x − t | 3 y ( t ) d t = f ( x )$

.

Let us remove the modulus in the integrand:

1 $∫ a x ( x − t ) 3 y ( t ) d t + ∫ x b ( t − x ) 3 y ( t ) d t = f ( x ) .$

Differentiating (1) twice yields

$6 ∫ a x ( x − t ) y ( t ) d t + 6 ∫ x b ( t − x ) y ( t ) d t = f ″ x x ( x ) .$

This equation can be rewritten in the form 3.1.2:

2 $∫ a b | x − t |y ( t ) d t = 1 6 f ″ x x ( x ) .$

Therefore the solution of the integral equation is given by

3 $y ( x ) = 1 12 y ″ ″ x x x x ( x ) .$

The right-hand side f(x) of the equation must satisfy certain conditions. To obtain these conditions, one must substitute solution (3) into (1) with x = a and x = b and into(2) with x = a and x = b,and then integrate the four resulting relations by parts.

12. $∫ a b | x 3 − t 3 | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x) = x 3.

13. $∫ a b | x t 2 − t 3 | y ( t ) d t = f ( x ) 0 ≤ a < b < ∞$

.

The substitution w(t)= t 2 y ( t )leads to an equation of the form 3.1.2:

$∫ a b | x − t | w ( t ) d t = f ( x ) .$

14. $∫ a b | x 2 t − t 3 | y ( t ) d t = f ( x )$

.

The substitution w(t) = |t| y(t)leads to an equation of the form 3.1.8:

$∫ a b | x 2 − t 2 | w ( t ) d t = f ( x ) .$

15. $∫ 0 a | x 3 − β t 3 | y ( t ) d t = f ( x ) , β > 0$

.

This is a special case of equation 3.8.4 with g ( x ) = x 3 and β = λ 3 .

16. $∫ a b | x − t | 2 n + 1 y ( t ) d t = f ( x ) , n = 0 , 1 , 2 , …$

Solution:

1 $y ( x ) = 1 2 ( 2 n + 1 ) ! f x ( 2 n + 2 ) ( x ) .$

The right-hand side f(x) of the equation must satisfy certain conditions. To obtain these conditions, one must substitute solution (1) into the relations

$∫ a b ( t − a ) 2 n + 1 y ( t ) d t = f ( a ) , ∫ a b ( t − a ) 2 n − k y ( t ) d t = ( − 1 ) k + 1 A k f x ( k + 1 ) ( a ) , A k = ( 2 n + 1 ) ( 2 n ) … ( 2 n + 1 − k ) ; k = 0 , 1 , … , 2 n ,$

and then integrate the resulting equations by parts.

17. $∫ 0 ∞ y ( t ) d t x + t = f ( x )$

The left-hand side of this equation is the Stieltjes transform.

1 °. By setting

$x = e z , t = e τ y ( t ) = e − τ / 2 ω ( τ ) , f ( x ) = e − z / 2 g ( z ) ,$

we obtain an integral equation with difference kernel of the form 3.8.15:

$∫ − ∞ ∞ ω ( τ ) d τ 2 cosh [ 1 2 ( z − τ ) ] = g ( z ) ,$

whose solution is given by

$w ( z ) = 1 2 π 3 ∫ − ∞ ∞ cosh ( π u ) g ˜ ( u ) e i u x d u , g ˜ ( u ) = 1 2 π ∫ − ∞ ∞ g ( z ) e − i u x d z , i 2 = − 1.$

2°.Solution:

$y ( x ) = 1 2 π i lim ε → + 0 [ f ( − x − i ε ) − f ( − x + i ε ) ] = 1 π x ∑ k = 0 ∞ ( − 1 ) k ( 2 k ) ! ( π x d d x ) 2 k [ x f ( x ) ] .$

3°.Under some assumptions, the solution of the original equation can be represented in the form

1 $y ( x ) = lim n → ∞ ( − 1 ) n ( n + 1 ) ! ( n − 1 ) [ x 2 n + 1 f x ( n ) ( x ) ] x ( n + 1 ) ,$

which is the real inversion of the Stieltjes transform.

An alternative form of the solution is

2 $y ( x ) = lim n → ∞ ( − 1 ) n 2 π ( e n ) 2 n [ x 2 n f x ( n ) ( x ) ] x ( n ) .$

To obtain an approximate solution of the integral equation, one restricts oneself to a specific value of n in (1) or (2) instead of taking the limit.

#### 3.1-4  Kernels Containing Square Roots.

18. $∫ 0 a | x − t | y ( t ) d t = f ( x ) , 0 < a < ∞$

.

This is a special case of equation 3.8.3 with g(x) $x$

Solution:

$y ( x ) = d d x [ x f ′ x ( x ) ] .$

The right-hand side f(x) of the equation must satisfy certain conditions. The general form of the right-hand side is

$f ( x ) = F ( x ) + A x + B , A = − F ′ x ( a ) , B = 1 2 [ a F ′ x ( a ) − F ( a ) − F ( 0 ) ] ,$

where F( x ) is an arbitrary bounded twice differentiable function (with bounded first derivative).

19. $∫ 0 a | x − β t | y ( t ) d t = f ( x ) , β > 0$

.

This is a special case of equation 3.8.4 with g(x) = $x$

and β =

20. $∫ 0 a | x − t | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.5 with g(x) = $x$

(see item 3° of 3.8.5).

21. $∫ 0 a | x − t | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.6 with g(t) = $t$

(see item 3° of 3.8.6).

22. $∫ 0 a y ( t ) | x − t | d t = f ( x ) , 0 < a ≤ ∞$

.

This is a special case of equation 3.1.30 with k = $1 2$

.

Solution:

$y ( x ) = − A x 1 / 4 d d x [ ∫ x a d t ( t − x ) 1 / 4 ∫ 0 t f ( s ) d s s 1 / 4 ( t − s ) 1 / 4 ] , A = 1 8 π Γ 2 ( 3 / 4 ) .$

23. $∫ − ∞ ∞ y ( t ) | x − t | d t = f ( x )$

.

This is a special case of equation 3.1.35 with λ = $1 2$

Solution:

$y ( x ) = 1 4 π ∫ − ∞ ∞ f ( x ) − f ( t ) | x − t | 3/2 d t .$

24. $∫ − 1 1 y ( t ) d t 1 + x 2 − 2 x t = f ( x )$

.

Solution:

$y ( x ) = 1 2 ∑ n = 0 ∞ 2 n + 1 n ! f x ( n ) ( 0 ) P n ( x ) ,$

where P n ( x ) are the Legendre polynomials (see Supplement 11.11-1)

$P n ( x ) = 1 n ! 2 n d n d x n ( x 2 − 1 ) n .$

#### 3.1-5  Kernels Containing Arbitrary Powers.

25. $∫ 0 a | x k − t k | y ( t ) d t = f ( x ) , 0 < k < 1 , 0 < a < ∞$

.

1°.Let us remove the modulus in the integrand:

1 $∫ 0 x ( x k − t k ) y ( t ) d t + ∫ x a ( t k − x k ) y ( t ) d t = f ( x ) .$

Differentiating (1) with respect to x yields

2 $k x k − 1 ∫ 0 x y ( t ) d t − k x k − 1 ∫ x a y ( t ) d t = f ′ x ( x ) .$

Let us divide both sides of (2) by kx k–1 and differentiate the resulting equation. As a result, we obtain the solution

3 $y ( x ) = 1 2 k d d x [ x 1 − k f ′ x ( x ) ] .$

2°.Let us demonstrate that the right-hand side f ( x ) of the integral equation must satisfy certainrelations. By setting x=0 and x=a, in (1), we obtain two corollaries $∫ 0 a t k y ( t ) d t = f ( 0 )$

and $∫ 0 a ( a k − t k ) y ( t ) d t = f ( a )$ which can be rewritten in the form

4 $∫ 0 a t k y ( t ) d t = f ( 0 ) , a k ∫ 0 a y ( t ) d t = f ( 0 ) + f ( a ) .$

Substitute y ( x ) of (3) into (4). Integrating by parts yields the relations $a f x ′ ( a ) = k f ( a ) + k f ( 0 )$

and $a f x ′ ( a ) = 2 k f ( a ) + 2 k f ( 0 )$ . Hence, the desired constraints for f ( x ) have the form

5 $f ( 0 ) + f ( a ) = 0 , f ′ x ( a ) = 0.$

Conditions (5) make it possible to find the admissible general form of the right-hand side of the integral equation:

$f ( x ) = F ( x ) + A x + B , A = − F ′ x ( a ) , B = 1 2 [ a F ′ x ( a ) − F ( a ) − F ( 0 ) ] ,$

where F( x ) is an arbitrary bounded twice differentiable function with bounded first derivative. The first derivative may be unbounded at x = 0, in which case the conditions $[ x 1 − k F x ′ ] x = 0 = 0$

must hold.

26. $∫ 0 a | x k − β t k | y ( t ) d t = f ( x ) , 0 < k < 1 , β > 0$

.

This is a special case of equation 3.8.4 with g(x)=x k and β = λ k .

27. $∫ 0 a | x k t m − t k + m | y ( t ) d t = f ( x ) , 0 < k < 1 , 0 < a < ∞$

.

The substitution w(t)= t m y ( t ) leads to an equation of the form 3.1.25:

$∫ 0 a | x k − t k | w ( t ) d t = f ( x ) .$

28. $∫ 0 1 | x k − t m | y ( t ) d t = f ( x ) , k > 0 , m > 0$

.

The transformation

$z = x k , τ = t m , w ( τ ) = τ 1 − m m y ( t )$

leads to anequation of the form 3.1.1:

$∫ 0 1 | z − τ | w ( τ ) d τ = F ( z ) , F ( z ) = m f ( z 1 / k ) .$

29. $∫ a b | x − t | 1 + λ y ( t ) d t = f ( x ) , 0 ≤ λ < 1$

.

For λ =0,seeequation 3.1.2. Assume that 0 < λ <1.

1 °. Let us remove the modulus in the integrand:

1 $∫ a x ( x − t ) 1 + λ y ( t ) d t + ∫ x b ( t − x ) 1 + λ y ( t ) d t = f ( x ) .$

Let us differentiate (1) with respect to x twice and then divide both the sides by λ(λ +1). As aresult, we obtain

2 $∫ a x ( x − t ) λ − 1 y ( t ) d t + ∫ x b ( t − x ) λ − 1 y ( t ) d t = 1 λ ( λ+1 ) f ″ x x ( x ) .$

Rewrite equation (2) in the form

3 $∫ a b y ( t ) d t | x − t | k = 1 λ ( λ + 1 ) f ″ x x ( x ) , k = 1 − λ .$

See 3.1.30 and 3.1.31 for the solutions of equation (3) for various a and b.

2°.Theright-handside f(x) of the integral equation must satisfy certain relations. By setting x = a and x = b in (1), we obtain two corollaries

4 $∫ a b ( t − a ) 1 + λ y ( t ) d t = f ( a ) , ∫ a b ( b − t ) 1+λ y ( t ) d t = f ( b ) .$

On substituting the solution y(x) of (3) into (4) and then integrating by parts, we obtain the desired constraints for f(x).

30. $∫ 0 a y ( t ) | x − t | k d t = f ( x ) , 0 ≤ k < 1 , 0 < a ≤ ∞$

.

1 °. Solution:

$y ( x ) = − A x k − 1 d x d d x ⌈ ∫ x a t 1 − 2 k 2 d t ( t − x ) 1 − k 2 ∫ 0 t f ( s ) d s s 1 − k 2 ( t − s ) 1 − k 2 ⌉ , A = 1 2 π cos ( π k 2 ) Γ ( k ) [ Γ ( 1 + k 2 ) ] − 2 ,$

where Γ(k) is the gamma function.

2°.The transformation $x = z 2 , t = ξ 2 , w ( ξ ) = 2 ξ y ( t )$

leads to an equation of the form 3.1.32:

$∫ 0 a w ( ξ ) | z 2 − ξ 2 | k d ξ = f ( z 2 ) .$

31. $∫ a b y ( t ) | x − t | k d t = f ( x ) , 0 < k < 1$

.

It is assumed that \a\ + \b\ < ∞.Solution:

$y ( x ) = 1 2 π cot ( 1 2 π k ) d d x ∫ a x f ( t ) d t ( x − t ) 1 − k − 1 π 2 cos 2 ( 1 2 π k ) ∫ a x Z ( t ) F ( t ) ( x − t ) 1 − k d t ,$

where

$Z ( t ) = ( t − a ) 1 + k 2 ( b − t ) 1 − k 2 , F ( t ) = d d t [ ∫ a t d τ ( t − τ ) k ∫ τ b f ( s ) d s Z ( s ) ( s − τ ) 1 − k ] .$

⊙ Reference: F. D. Gakhov (1977).

32. $∫ 0 a y ( t ) | x 2 − t 2 | k d t = f ( x ) , 0 < k < 1 , 0 < a ≤ ∞$

.

Solution:

$y ( x ) = − 2 Γ ( k ) cos ( 1 2 π k ) π [ Γ ( 1 + k 2 ) ] 2 x k − 1 d d x ∫ x a t 2 − 2 k F ( t ) d t ( t 2 − x 2 ) 1 − k 2 , F ( t ) = ∫ 0 t s k f ( s ) d s ( t 2 − s 2 ) 1 − k 2 .$

⊙ Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975).

33. $∫ a b y ( t ) | x λ − t λ | k d t = f ( x ) , 0 < k < 1 , λ > 0$

.

1 °. The trans formation

$z = x λ , τ = t λ , w ( τ ) = τ 1 − λ λ y ( t )$

leads to an equation of the form3.1.31:

$∫ A B w ( τ ) | z − τ | k d τ = F ( z ) ,$

where $A = a λ , B = b λ , F ( z ) = λ f ( z 1 / λ )$

2°.Solution with a = 0:

$y ( x ) = − A x λ ( k − 1 ) 2 d d x ⌈ ∫ x b t λ ( 3 − 2 k ) − 2 2 d t ( t λ − x λ ) 1 − k 2 ∫ 0 t s λ ( k + 1 ) − 2 2 f ( s ) d s ( t λ − s λ ) 1 − k 2 ⌉ , A = λ 2 2 π cos ( π k 2 ) Γ ( k ) [ Γ ( 1 + k 2 ) ] − 2 ,$

where Γ(k) is the gamma function.

34. $∫ 0 1 y ( t ) | x λ − t m | k d t = f ( x ) , 0 < k < 1 , λ > 0 , m > 0$

.

The transformation

$z = x λ , τ = t m , w ( τ ) = τ 1 − m m y ( t )$

leads to anequation of the form 3.1.31:

$∫ 0 1 w ( τ ) | z − τ | k d τ = F ( z ) , F ( z ) = m f ( z 1 / λ ) .$

35. $∫ − ∞ ∞ y ( t ) | x − t | 1 − λ d t = f ( x ) , 0 < R e λ < 1$

.

Solution:

$y ( x ) = λ 2 π tan ( π λ 2 ) ∫ − ∞ ∞ f ( x ) − f ( t ) | x − t | 1+λ d t = λ 2 π tan ( π λ 2 ) ∫ 0 ∞ 2 f ( x ) − f ( x + t ) − f ( x − t ) t 1 + λ d t .$

It is assumed that the condition $∫ − ∞ ∞ | f ( x ) | p d x < ∞$

is satisfied for some p,1 < p <1/λ.

The integral equation and its solution form the Riesz transform pair (the Riesz potential).

36. $∫ − ∞ ∞ y ( t ) | x 3 − t | 1 − λ d t = f ( x ) , 0 < λ < 1$

.

The substitution z = x 3 leads to an equation of the form 3.1.35:

$∫ − ∞ ∞ y ( t ) | z − t | 1 − λ d t = f ( z 1 / 3 ) .$

37. $∫ − ∞ ∞ y ( t ) | x 3 − t 3 | 1 − λ d t = f ( x ) , 0 < λ < 1$

.

The transformation

$z = x 3 , τ = t 3 , w ( τ ) = τ − 2 / 3 y ( t )$

leads to anequation of the form 3.1.35:

$∫ − ∞ ∞ w ( τ ) | z − τ | 1 − λ d τ = F ( z ) , F ( z ) = 3 f ( z 1 / 3 ) .$

38. $∫ − ∞ ∞ s i g n ( x − t ) | x − t | 1 − λ y ( t ) d t = f ( x ) , 0 < R e λ < 1$

.

Solution:

$y ( x ) = λ 2 π cot ( π λ 2 ) ∫ − ∞ ∞ f ( x ) − f ( x ) | x − t | 1+λ sign ( x − t ) d t = λ 2 π cot ( π λ 2 ) ∫ 0 ∞ f ( x + t ) − f ( x − t ) t 1+λ d t = λ 2 π cot ( π λ 2 ) d d x ∫ − ∞ ∞ f ( t ) | x − t | λ d t .$

The integral equation and its solution form the Feller transform pair (the Feller potential).

39. $∫ − ∞ ∞ a + b s i g n ( x − t ) | x − t | 1 − λ y ( t ) d t = f ( x ) , 0 < R e λ < 1$

.

Solution:

$y ( x ) = C λ ∫ − ∞ ∞ a + b sign ( x − t ) | x − t | 1+ λ [ f ( x ) − f ( t ) ] d t = C λ ∫ 0 ∞ t − 1 − λ [ 2 a f ( x ) − ( a + b ) f ( x − t ) − ( a − b ) f ( x + t ) ] d t = C d d x ∫ − ∞ ∞ b + a sign ( x − t ) | x − t | λ f ( t ) d t ,$

where

$C = sin ( π λ ) 4 π [ a 2 cos 2 ( 1 2 π λ ) + b 2 sin 2 ( 1 2 π λ ) ] .$

40. $∫ 0 ∞ y ( t ) d t ( a x + b t ) k = f ( x ) , a > 0 , b > 0 , k > 0$

.

By setting

$x = 1 2 a e 2 z , t = 1 2 b e 2 τ , y ( t ) = b e ( k − 2 ) τ w ( τ ) , f ( x ) = e − k z g ( z ) ,$

we obtain an integral equation with the difference kernel of the form 3.8.15:

$∫ − ∞ ∞ w ( τ ) d τ cosh k ( z − τ ) = g ( z ) .$

41. $∫ 0 ∞ t z − 1 y ( t ) d t = f ( z )$

.

The left-hand side of this equation is the Mellin transform of y(t)(z is treated as a complex variable).

Solution:

$y ( t ) = 1 2 π i ∫ c − i ∞ c + i ∞ t − z f ( z ) d z , i 2 = − 1.$

For specific f (z), one can use tables of Mellin and Laplace integral transforms to calculate the integral.

⊙ References: H. Bateman and A. Erdelyi (vol. 2, 1954), V. A. Ditkin and A. P. Prudnikov (1965).

#### 3.1-6  Equations Containing the Unknown Function of a Complicated Argument.

42. $∫ 0 1 y ( x t ) d t = f ( x )$

.

Solution:

$y ( x ) = x f ′ x ( x ) + f ( x ) .$

The function f(x) is assumed to satisfy the condition xf(x) x=0 =0.

43. $∫ 0 1 t λ y ( x t ) d t = f ( x )$

.

The substitution ξ = xt leads to equation $∫ 0 x ξ λ y ( ξ ) d ξ = x λ + 1 f ( x )$

. Differentiating with respect to x yields the solution

$y ( x ) = x f ′ x ( x ) + ( λ + 1 ) f ( x ) .$

The function f(x) is assumed to satisfy the condition $[ x λ + 1 f ( x ) ] x = 0 = 0$

44. $∫ 0 1 ( A x k + B t m ) y ( x t ) d t = f ( x )$

.

The substitution ξ = xt leads to an equation of the form 1.1.51:

$∫ 0 x ( A x k + m + B ξ m ) y ( ξ ) d ξ = x m + 1 f ( x ) .$

45. $∫ 0 1 y ( x t ) d t 1 − t = f ( x )$

.

The substitution ξ = xt leads to Abel’s equation 1.1.36:

$∫ 0 x y ( ξ ) d ξ x − ξ = x f ( x ) .$

46. $∫ 0 1 y ( x t ) d t ( 1 − t ) λ = f ( x ) , 0 < λ < 1$

.

The substitution ξ = xt leads to the generalized Abel equation 1.1.47:

$∫ 0 x y ( ξ ) d ξ ( x − ξ ) λ = x 1 − λ f ( x ) .$

47. $∫ 0 1 t μ y ( x t ) ( 1 − t ) λ d t = f ( x ) , 0 < λ < 1$

.

The transformation ξ = xt, ω(ξ) = ξ μ y(ξ) leads to the generalized Abel equation 1.1.47:

$∫ 0 x w ( ξ ) d ξ ( x − ξ ) λ = x 1 + μ − λ f ( x ) .$

48. $∫ 0 ∞ y ( x + t ) − y ( x − t ) t d t = f ( x )$

.

Solution:

$y ( x ) = − 1 π 2 ∫ 0 ∞ f ( x + t ) − f ( x − t ) t d t .$

⊙ References: V. A. Ditkin and A. P. Prudnikov (1965), A.P.Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 427).

#### 3.1-7  Singular Equations.

In this subsection, all singular integrals are understood in the sense of the Cauchy principal value.

49. $∫ − ∞ ∞ y ( t ) d t t − x = f ( x )$

.

Solution:

$y ( x ) = − 1 π 2 ∫ − ∞ ∞ f ( t ) d t t − x .$

The integral equation and its solution form a Hilbert transform pair (in the asymmetric form).

⊙ References: V. A. Ditkin and A. P. Prudnikov (1965), A.P.Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 427).

50. $∫ 0 ∞ y ( t ) d t t − x = f ( x )$

.

Solution:

$y ( x ) = − x π 2 ∫ 0 ∞ f ( t ) t ( t − x ) d t .$

The integral equation and its solution form a Hilbert transform pair on the semiaxis (in the asymmetric form).

⊙ References: D. Hilbert (1953), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 427), I. K. Lifanov, L. N. Poltavskii, and G. M. Vainikko (2004, p. 8).

51. $∫ a b y ( t ) d t t − x = f ( x )$

.

This equation is encountered in hydrodynamics in solving the problem on the flow of an ideal inviscid fluid around a thin profile (a ≤ xb). It is assumed that \a\ + \b\ < ∞.

1 °.The solution bounded at the endpoints is

$y ( x ) = − 1 π 2 ( x − a ) ( b − x ) ∫ a b f ( t ) ( t − a ) ( b − t ) d t t − x ,$

provided that

$∫ a b f ( t ) d t ( t − a ) ( b − t ) = 0.$

2°. The solution bounded at the endpoint x = a and unbounded at the endpoint x = b is

$y ( x ) = − 1 π 2 x − a b − x ∫ a b b − t t − a f ( t ) t − x d t .$

3°.The solution unbounded at the endpoints is

$y ( x ) = − 1 π 2 ( x − a ) ( b − x ) [ ∫ a b ( t − a ) ( b − t ) t − x f ( t ) d t + C ] ,$

where C is an arbitrary constant. The formula $∫ a b y ( t ) d t = C / π$

holds.

Solutions that have a singularity point x = s inside the interval [a, b] can be found in Subsection 14.4-3.

⊙ Reference: F. D. Gakhov (1977).

52. $∫ − 1 1 ( 1 t − x + 1 x + t + 2 ) y ( t ) d t = f ( x ) , − 1 < x < 1$

.

Solution for f ( x ) = πq = const:

$y ( t ) = q 1 + t ( 1 − t ) ( 3 + t ) .$

⊙ Reference: H. F. Bueckner (1966).

53. $∫ 0 1 ( 1 t − x + λ t + x ) y ( t ) d t = f ( x ) , 0 < x < 1$

.

Solution for f (x) = πq = const:

$y ( x ) = q 2 sin ⁡ ( 1 2 π β ) [ ( x 1 + 1 − x 2 ) β ( β 1 − x 2 + 1 ) + ( x 1 + 1 − x 2 ) − β ( β 1 − x 2 − 1 ) ] ,$

where β is given by

$cos ⁡ ( π β ) = − λ , 0 < β < 1.$

We assume that the following necessary condition holds

$∫ 0 1 y ( t ) d t = 0.$

⊙ References: H. F. Bueckner (1966), P. S. Theocaric and N. I. Ioakimidis (1977).

54. $1 π i ∫ − a a ( 1 t − x − λ x x t − a 2 ) y ( t ) d t = f ( x ) , − a < x < a ( i 2 = − 1 )$

.

1°.Solution:

$y ( x ) = ( a − x − a − x ) β 1 2 π i ∫ − a a ( a − t − a − t ) − β ( 1 t − x − x x t − a 2 ) f ( t ) d t + ( a − x − a − x ) − β 1 2 π i ∫ − a a ( a − t − a − t ) β ( 1 t − x − x x t − a 2 ) f ( t ) d t ,$

where λ = cos θ and $β = 1 − θ π$

We assume that the following necessary condition holds

$1 2 π i ∫ − a a [ e − π i β ( a − t − a − t ) β − e π i β ( a − t − a − t ) − β ] f ( t ) t d t = 0.$

2°.Solution for f (x) = 0:

$y ( x ) = C 1 Λ 1 ( x ) + C 2 Λ 2 ( x ) + C 3 Λ 3 ( x ) ,$

where C 1 , C 2, and C 3 are arbitrary constants, and

$Λ 1 ( x ) = ( 1 + λ ) e i π β ( a − t − a − t ) 1 − β + ( 1 − λ ) e − i π β ( a − t − a − t ) β , Λ 2 ( x ) = ( 1 + λ ) e − i π β ( a − t − a − t ) − 1 − β + ( 1 − λ ) e i π β ( a − t − a − t ) − β , Λ 3 ( x ) = e i π β ( a − t − a − t ) 1 − β + e − i π β ( a − t − a − t ) − 1 + β .$

⊙Reference: D. I. Sherman (1969).

55. $∫ a b y ( t ) ( x − t ) 2 d t = f ( x ) , a ≤ x ≤ b$

.

The simple hypersingular equation of the first kind with Cauchy-type kernel. This equation governs circulation-free flow of an ideal incompressible fluid past the segment [a, b].

Let the conditions y ( a ) = y ( b ) = 0 be satisfied. Then the solution is

$y ( x ) = 1 π 2 ∫ a b ln | ( b − t ) ( x − a ) − ( b − x ) ( t − a ) ( b − t ) ( x − a ) + ( b − x ) ( t − a ) | f ′ t ( t ) d t .$

This equation is discussed in Subsection 14.6-3 in detail.

⊙ Reference: I. K. Lifanov, L. N. Poltavskii, and G. M. Vainikko (2004, p. 7).

56. $1 π 2 ∫ − 1 1 ∫ − 1 1 u ( x , y ) d x d y ( x 0 − x ) ( y 0 − y ) = f ( x 0 , y 0 )$

.

A two-dimensional singular equation.

A solution, which is bounded on the lines x = ±1 and y = ±1 but which is unbounded on the line x = q (–1 < q <1), is given by the formula

$u ( x 0 , y 0 ) = ( 1 − x 0 2 ) ( 1 − y 0 2 ) π 2 ∫ − 1 1 ∫ − 1 1 f ( x , y ) d x d y ( 1 − x 2 ) ( 1 − y 2 ) ( x − x 0 ) ( y − y 0 ) − ( 1 − x 0 2 ) ( 1 − y 0 2 ) π 2 ( q − x 0 ) ∫ − 1 1 d x 1 − x 2 ( 1 π 2 ∫ − 1 1 f ( x , y ) d y 1 − y 2 ( y − y 0 ) ) ,$

provided that

$∫ − 1 1 f ( x 0 , y ) d y 1 − y 2 = 0 , − 1 ≤ x 0 ≤ 1.$

⊙ Reference: I. K. Lifanov, L. N. Poltavskii, and G. M. Vainikko (2004, pp. 16–20).

#### 3.2-1  Kernels Containing Exponential Functions of the Form e λ | x − t |

1. $∫ − ∞ ∞ e − λ | x − t | y ( t ) d t = f ( x ) , f ( ± ∞ ) = 0$

.

Solution:

$y ( x ) = 1 2 λ [ λ 2 f ( x ) − f ″ x x ( x ) ] .$

⊙ References: 1.1. Hirschman and D. V. Widder (1955), F. D.Gakhov and Yu. I. Cherskii (1978), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 433).

2. $∫ 0 ∞ e − λ | x − t | y ( t ) d t = f ( x ) , f ( ∞ ) = 0$

.

1 °. Solution:

$y ( x ) = 1 2 λ e − λ x d d x e 2 λ x d d x e − λ x f ( x ) .$

$If f x ′ ( 0 ) − λ f ( 0 ) = 0$

then

$y ( x ) = 1 2 λ [ λ 2 f ( x ) − f ″ x x ( x ) ] .$

3. $∫ a b e λ | x − t | y ( t ) d t = f ( x ) , − ∞ < a < b < ∞$

.

1 °. Let us remove the modulus in the integrand:

1 $∫ a x e λ ( x − t ) y ( t ) d t + ∫ x b e λ ( t − x ) y ( t ) d t = f ( x ) .$

Differentiating (1) with respect to x twice yields

2 $2 λ y ( x ) + λ 2 ∫ a x e λ ( x − t ) y ( t ) d t + λ 2 ∫ x b e λ ( t − x ) y ( t ) d t = f ″ x x ( x ) .$

By eliminating the integral terms from (1) and (2), we obtain the solution

3 $y ( x ) = 1 2 λ [ f ″ x x ( x ) − λ 2 f ( x ) ] .$

2°.The right-hand side f(x) of the integral equation must satisfy certain relations. By setting x = a and x = b in (1), we obtain two corollaries

4 $∫ a b e λ t y ( t ) d t = e λ a f ( a ) , ∫ a b e − λ t y ( t ) d t = e − λ b f ( b ) .$

On substituting the solution y(x) of (3) into (4) and then integrating by parts, we see that

$e λ b f ′ x ( b ) − e λ a f ′ x ( a ) = λ e λ a f ( a ) + λ e λ b f ( b ) , e − λ b f ′ x ( b ) − e − λ a f ′ x ( a ) = λ e − λ a f ( a ) + λ e − λ b f ( b ) .$

Hence, we obtainthedesired constraints for f(x):

5 $f ′ x ( a ) + λ f ( a ) = 0 , f ′ x ( b ) − λ f ( b ) = 0.$

The general form of the right-hand side satisfying conditions (5) is given by

$f ( x ) = F ( x ) + A x + B , A = 1 b λ − a λ − 2 [ F ′ x ( a ) + F ′ x ( b ) + λ F ( a ) − λ F ( b ) ] , B = − 1 λ [ F ′ x ( a ) + λ F ( a ) + A a λ + A ] ,$

where F(x) is an arbitrary bounded, twice differentiable function.

4. $∫ a b ( A e λ | x − t | + B e μ | x − t | ) y ( t ) d t = f ( x ) , − ∞ < a < b < ∞$

.

Let us remove the modulus in the integrand anddifferentiate the resulting equation with respect to x twice to obtain

1 $2 ( A λ + B μ ) y ( x ) + ∫ a b ( A λ 2 e λ | x − t | + B μ 2 e μ | x − t | ) y ( t ) d t = f ″ x x ( x ) .$

Eliminating the integral term with $e μ | x − t |$

from (1) with the aid of the original integral equation, we find that

2 $2 ( A λ + B μ ) y ( x ) + A ( λ 2 − μ 2 ) ∫ a b e λ | x − t | y ( t ) d t = f ″ x x ( x ) − μ 2 f ( x ) .$

For + = 0, this is anequation of the form 3.2.3, and for + ≠ 0, this is an equation of the form 4.2.15.

The right-hand side f(x) must satisfy certain relations, which can be obtained by setting x = a and x = b in the original equation (a similar procedure is used in 3.2.3).

5. $∫ a b [ ∑ k = 1 n A k e x p ( λ k | x − t | ) ] y ( t ) d t = f ( x ) , − ∞ < a < b < ∞$

.

1 °. Let us remove the modulus in the kth summand of the integrand:

1 $I k ( x ) = ∫ a b exp ( λ k | x − t | ) y ( t ) d t = ∫ a x exp [ λ k ( x − t ) ] y ( t ) d t + ∫ x b exp [ λ k ( t − x ) ] y ( t ) d t .$

Differentiating (1) with respect to x twice yields

2 $I ′ k = λ k ∫ a x exp [ λ k ( x − t ) ] y ( t ) d t − λ k ∫ x b exp [ λ k ( t − x ) ] y ( t ) d t , I ″ k = 2 λ k y ( x ) + λ k 2 ∫ a x exp [ λ k ( x − t ) ] y ( t ) d t + λ k 2 ∫ x b exp [ λ k ( t − x ) ] y ( t ) d t ,$

where the primes denote the derivatives with respect to x.By comparing formulas (1) and (2), we find the relation between $I k ″$

and I k :

3 $I ″ k = 2 λ k y ( x ) + λ k 2 I k , I k = I k ( x ) .$

2°.With the aid of (1), the integral equation can be rewritten in the form

4 $∑ k = 1 n A k I k = f ( x ) .$

Differentiating (4) with respect to x twice and taking into account (3), we obtain

5 $σ 1 y ( x ) + ∑ k = 1 n A k λ k 2 I k = f ″ x x ( x ) , σ 1 = 2 ∑ k = 1 n A k λ k .$

Eliminating the integral In from (4) and (5) yields

6 $σ 1 y ( x ) + ∑ k = 1 n − 1 A k ( λ k 2 − λ n 2 ) I k = f ″ x x ( x ) − λ n 2 f ( x ) .$

Differentiating (6) with respect to x twice and eliminating I n–1 from the resulting equation with theaid of (6), we obtain a similar equation whose right-hand side is a second-order linear differential operator (acting on y) with constant coefficients plus the sum $∑ k = 1 n − 2 B k I k$

If we successively eliminate I n−2, I n−3, …, I 1 with theaid of double differentiation, then we finally arrive at a linear nonhomogeneous ordinary differential equation of order 2(n −1) with constant coefficients.

3°.The right-hand side f (x) must satisfy certain conditions. To find these conditions, one must set x = a in the integral equation and its derivatives. (Alternatively, these conditions can be found by setting x = a and x = b in the integral equation and all its derivatives obtained by means of double differentiation.)

#### 3.2-2  Kernels Containing Exponential Functions of the Forms e λx and e μt .

6. $∫ a b | e λ x − e λ t | y ( t ) d t = f ( x ) , λ > 0$

.

This is a special case of equation 3.8.3 with g(x) = e λx .

Solution:

$y ( x ) = 1 2 λ d d x [ e − λ x f ′ x ( x ) ] .$

Theright-handside f (x) of the integral equation must satisfy certain relations (seeitem2° of equation 3.8.3).

7. $∫ 0 a | e β x − e μ t | y ( t ) d t = f ( x ) , β > 0 , μ > 0$

.

This is a special case of equation 3.8.4 with g ( x) = e βx and λ = μ/β.

8. $∫ a b y ( t ) d t | e λ x − e λ t | k = f ( x ) , 0 < k < 1$

.

The transformation z = e λx , T = e λ t , ω(T) = e λt y(t) leads to an equation of the form 3.1.31:

$∫ A B w ( τ ) | z − τ | k d τ = F ( z ) ,$

where $A = e λ a , B = e λ b , F ( z ) = λ f ( 1 λ ln ⁡ z )$

.

9. $∫ 0 ∞ y ( t ) d t ( e λ x + e λ t ) k = f ( x ) , λ > 0 , k > 0$

.

This equation can be rewritten as an equation with difference kernel in the form 3.8.16:

$∫ 0 ∞ w ( t ) d t cosh k [ 1 2 λ ( x − t ) ] = g ( x ) ,$

where $w ( t ) = 2 − k exp ⁡ ( − 1 2 λ k t ) y ( t ) and g ( x ) = exp ⁡ ( 1 2 λ k x ) f ( x )$

.

#### 3.2-3  Kernels Containing Exponential Functions of the Form e λxt .

10. $∫ − ∞ ∞ e − x t y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = 1 2 π i ∫ c − i ∞ c + i ∞ e s t f ( s ) d s = 1 2 π 3 ∫ 0 ∞ e − ξ 2 / 2 d ξ ∫ − ∞ ∞ e − x 2 / 2 cos ( ξ ( x + t ) ) f ( x ) d x .$

The integral equation and its solution form a two-side Laplace transform pair.

⊙ References: B. Van der Pol and H. Bremmer (1955), A. P.Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 433).

11. $∫ − ∞ ∞ e λ x t y ( t ) d t = f ( x ) , λ ≠ 0$

.

1 ° . The transformation

$x = − 1 λ z , f ( x ) = F ( z )$

leads to anequation of the form 3.2.10:

$∫ − ∞ ∞ e − z t y ( t ) d t = F ( z ) .$

2°.The transformation

$y ( t ) = exp ( − t 2 ) Y ( t ) , x = 2 λ ζ , f ( x ) = exp ( ζ 2 ) Φ ( ζ )$

leads to anequation of the form 3.2.17:

$∫ − ∞ ∞ e − ( ζ − t ) 2 Y ( t ) d t = Φ ( ζ )$

12. $∫ − ∞ ∞ e − i x t y ( t ) d t = f ( x ) , i 2 = − 1$

.

Solution:

$y ( t ) = 1 2 π ∫ − ∞ ∞ e i x t f ( x ) d x .$

Up to constant factors, the function f (x)and the solution y(t) are the Fourier transform pair.

13. $∫ 0 ∞ e − z t y ( t ) d t = f ( z )$

.

The left-hand side of the equation is the Laplace transform of y(t)(z is treated as a complex variable).

1°.Solution:

$y ( t ) = 1 2 π i ∫ c − i ∞ c + i ∞ e z t f ( z ) d z , i 2 = − 1.$

For specific functions f (z), one may use tables of inverse Laplace transforms to calculate theintegral (e.g., see Supplement 6).

2°.Forreal z = x, under some assumptions the solution of the original equation can be represented in the form

$y ( x ) = lim n → ∞ ( − 1 ) n n ! ( n x ) n + 1 f x ( n ) ( n x ) ,$

which is the real inversion of the Laplace transform. To calculate the solution approximately, one should restrict oneself to a specific value of n in this formula instead of taking the limit.

⊙ References: G. Doetsch (1950, 1956, 1958, 1974), H. Bateman and A. Erdelyi (vol. 1, 1954), I. I. Hirschman and D. V. Widder (1955), V. A. Ditkin and A. P. Prudnikov (1965), J. W. Miles (1971), F. Oberhettinger (1973), B. Davis (1978), W. R. LePage (1980), R. Bellman and R. Roth (1984), Yu. A. Brychkov and A. P. Prudnikov (1989), W. H. Beyer (1991), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, Vols 4 and 5), R. J. Beerends, H. G. ter Morschem, and J. C. van den Berg (2003).

#### 3.2-4  Kernels Containing Power-Law and Exponential Functions.

14. $∫ 0 a | k e λ x − k − t | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.5 with g(x) = ke λx k.

15. $∫ 0 a | x − k e λ t − k | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.6 with g(t) = ke λt + k.

16. $∫ − ∞ ∞ t − i x − 1 / 2 e x p ( 2 x − i 4 π ) y ( t ) d t = f ( x ) , i 2 = − 1$

.

Solution:

$y ( x ) = 1 4 π ∫ − ∞ ∞ x i t − 1 / 2 exp ( 2 t + i 4 π ) f ( t ) cosh ( π t ) d t .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 463).

#### 3.2-5  Kernels Containing Exponential Functions of the Form eλ(x±t)2.

17. $∫ − ∞ ∞ e − ( x − t ) 2 y ( t ) d t = f ( x )$

.

1 °. The transformation

$Y ( t ) = exp ( − t 2 ) y ( t ) , z = − 2 x , F ( z ) = exp ( x 2 ) f ( x )$

leads to anequation of the form 3.2.10:

$∫ − ∞ ∞ e − z t Y ( t ) d t = F ( z ) .$

2°.Solution:

$y ( t ) = 1 π 3 / 2 ∫ 0 ∞ e s 2 / 4 d s ∫ − ∞ ∞ cos ( s ( t − x ) ) f ( x ) d x = exp [ − 1 4 π d 2 d t 2 f ( t ) ] ≡ ∑ k = 0 ∞ 1 k ! ( − 1 4 π ) k d 2 k f ( t ) d t 2 k .$

(See equation 3.2.18 for λ =1.)

3°.Solution:

$y ( x ) = 1 π ∑ n = 0 ∞ f x ( n ) ( 0 ) 2 n n ! H n ( x ) ,$

where H n (x) are the Hermite polynomials (see Supplement11.17-3)

$H m ( x ) = ( − 1 ) m exp ( x 2 ) d m d x m exp ( − x 2 ) .$

⊙ References: P. M. Morse and H. Feshbach (1953), 1.1. Hirschman and D. V. Widder (1955), P. G. Rooney (1963), M. L. Krasnov (1975).

18. $1 πλ ∫ − ∞ ∞ e x p [ − ( x − t ) 2 λ ] y ( t ) d t = f ( x )$

.

It is the Gauss transform (the Weierstrass transform for λ = 4).

Solution:

$y ( t ) = 1 π ∫ 0 ∞ e λ s 2 / 4 d s ∫ − ∞ ∞ cos ⁡ ( s ( t − x ) ) f ( x ) d x = exp ⁡ [ − λ 4 d 2 d t f ( t ) ] ≡ ∑ k = 0 ∞ 1 k ! ( − λ 4 ) k d 2 k f ( t ) d t 2 k$

⊙ References: I.I. Hirschman and D. V. Widder (1955),P.G.Rooney (1963), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 435).

19. $∫ − ∞ ∞ e i ( x + t ) 2 y ( t ) d t = f ( x ) , i 2 = − 1$

.

Solution:

$y ( x ) = 1 π ∫ − ∞ ∞ e − i ( x + t ) 2 f ( t ) d t .$

⊙ References: E. A. C. Paley and N. Wiener (1934), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 435).

#### 3.2-6  Other Kernels.

20. $∫ a b | e x p ( λ x 2 ) − e x p ( λ t 2 ) | y ( t ) d t = f ( x ) , λ > 0$

.

This is a special case of equation 3.8.3 with g(x)=exp(λx 2).

Solution:

$y ( x ) = 1 4 λ d d x [ 1 x exp ⁡ ( − λ x 2 ) f ′ x ( x ) ] .$

Theright-handside f(x) of the integral equation must satisfy certain relations (seeitem2° of equation 3.8.3).

21. $1 π x ∫ 0 ∞ e x p ( − t 2 4 x ) y ( t ) d t = f ( x )$

.

Applying the Laplace transformation to the equation, we obtain

$y ˜ ( p ) p = f ˜ ( p ) , f ˜ ( p ) = ∫ 0 ∞ e − p t f ( t ) d t .$

Substituting p by p 2 and solving for the transform $y ˜$

we find that $y ˜ ( p ) = p f ˜ ( p 2 )$ The inverse Laplace transform provides the solution of the original integral equation:

$y ( t ) = L − 1 { p f ˜ ( p 2 ) } , L − 1 { g ( p ) } ≡ 1 2 π i ∫ c − i ∞ c + i ∞ e p t g ( p ) d p .$

#### 3.3-1  Kernels Containing Hyperbolic Cosine.

1. $∫ a b | c o s h ( λ x ) − c o s h ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x)=cosh(λx).

Solution:

$y ( x ) = 1 2 λ d d x [ f ′ x ( x ) sinh ⁡ ( λ x ) ]$

Theright-handside f(x) of the integral equation must satisfy certain relations (seeitem2° of equation 3.8.3).

2. $∫ 0 a | c o s h ( β x ) − c o s h ( μ t ) | y ( t ) d t = f ( x ) , β > 0 , μ > 0$

.

This is a special case of equation 3.8.4 with g(x)=cosh(βx)and λ = μ/β.

3. $∫ a b | c o s h k x − c o s h k t | y ( t ) d t = f ( x ) , 0 < k < 1$

.

This is a special case of equation 3.8.3 with g(x)=cosh k x.

Solution:

$y ( x ) = 1 2 λ d d x [ f ′ x ( x ) sinh ⁡ x cosh ⁡ k − 1 x ] .$

The right-hand side f(x) of the integral equation must satisfy certain relations (see item 2° of equation 3.8.3).

4. $∫ a b y ( t ) | c o s h ( λ x ) − c o s h ( λ t ) | k d t = f ( x ) , 0 < k < 1$

.

This is a special case of equation 3.8.7 with g(x)=cosh(λx),where β is an arbitrary number.

#### 3.3-2  Kernels Containing Hyperbolic Sine.

5. $∫ a b s i n h ( λ | x − t | ) y ( t ) d t = f ( x ) , − ∞ < a < b < ∞$

.

1 °. Let us remove the modulus in the integrand:

1 $∫ a x sinh ⁡ [ λ ( x − t ) ] y ( t ) d t + ∫ x b sinh ⁡ [ λ ( t − x ) ] y ( t ) d t = f ( x ) .$

Differentiating (1) with respect to x twice yields

2 $2 λ y ( x ) + λ 2 ∫ a x sinh ⁡ [ λ ( x − t ) ] y ( t ) d t + λ 2 ∫ x b sinh ⁡ [ λ ( t − x ) ] y ( t ) d t = f ″ x x ( x ) .$

Eliminating the integral terms from (1) and (2), we obtain the solution

3 $y ( x ) = 1 2 λ [ f ″ x x ( x ) − λ 2 f ( x ) ] .$

2°.Theright-handsidef(x) of the integral equation must satisfy certain relations. By setting x = a and x = b in (1), we obtain two corollaries

4 $∫ a b sinh ⁡ [ λ ( t − a ) ] y ( t ) d t = f ( a ) , ∫ a b sinh ⁡ [ λ ( b − t ) ] y ( t ) d t = f ( b ) .$

Substituting solution(3) into(4)and integrating by parts yields the desired conditions for f(x):

5 $sinh ⁡ [ λ ( b − a ) ] f ′ x ( b ) − λ cosh ⁡ [ λ ( b − a ) ] f ( b ) = λ f ( a ) , sinh ⁡ [ λ ( b − a ) ] f ′ x ( a ) + λ cosh ⁡ [ λ ( b − a ) ] f ( a ) = − λ f ( b ) .$

The general form of the right-hand side is given by

6 $f ( x ) = F ( x ) + A x + B ,$

where F(x) is an arbitrary bounded twice differentiable function, and the coefficients A and B are expressed in terms of F (a), F (b), $F x ′$

(a), and $F x ′$ (b) and can be determined by substituting formula (6) into conditions (5).

6. $∫ a b { A s i n h ( λ | x − t | ) + B s i n h ( μ | x − t | ) } y ( t ) d t = f ( x ) , − ∞ < a < b < ∞$

.

Let us remove the modulus in the integrand and differentiate the equation with respect to x twice to obtain

1 $2 ( A λ + B μ ) y ( x ) + ∫ a b { A λ 2 sinh ⁡ ( λ | x − t | ) + B μ 2 sinh ⁡ ( μ | x − t | ) } y ( t ) d t = f ″ x x ( x ) .$

Eliminating the integral term with sinh (μ|xt|)from (1) yields

2 $2 ( A λ + B μ ) y ( x ) + A ( λ 2 − μ 2 ) ∫ a b sinh ⁡ ( λ | x − t | ) y ( t ) d t = f ″ x x ( x ) − μ 2 f ( x ) .$

For + = 0, this is anequation of the form 3.3.5, and for + ≠ 0, this is an equation of the form 4.3.26.

The right-hand side f(x) must satisfy certain relations, which can be obtained by setting x = a and x = b in the original equation (a similar procedure is used in 3.3.5).

7. $∫ a b | s i n h ( λ x ) − s i n h ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x)=sinh(λx).

Solution:

$y ( x ) = 1 2 λ d d x [ f ′ x ( x ) cosh ⁡ ( λ x ) ] .$

Theright-hand side f(x) of the integral equation must satisfy certain relations (seeitem2° of equation 3.8.3).

8. $∫ 0 a | s i n h ( β x ) − s i n h ( μ t ) | y ( t ) d t = f ( x ) , β > 0 , μ > 0$

.

This is a special case of equation 3.8.4 with g(x) = sinh (βx) and λ = μ/β.

9. $∫ a b s i n h 3 ( λ | x − t | ) y ( t ) d t = f ( x )$

.

Using the formula sinh3 $β = 1 4$

sinh $β − 3 4$ sinh β, we arrive at an equation of the form 3.3.6:

$∫ a b [ 1 4 A sinh ⁡ ( 3 λ | x − t | ) − 3 4 A sinh ⁡ ( λ | x − t | ) ] y ( t ) d t = f ( x ) .$

10. $∫ a b [ ∑ k = 1 n A k s i n h ( λ k | x − t | ) ] y ( t ) d t = f ( x ) , − ∞ < a < b < ∞$

.

1 °. Let us remove the modulus in the kth summand of the integrand:

1 $I k ( x ) = ∫ a b sinh ⁡ ( λ k | x − t | ) y ( t ) d t = ∫ a x sinh ⁡ [ λ k ( x − t ) ] y ( t ) d t + ∫ x b sinh ⁡ [ λ k ( t − x ) y ( t ) d t ] .$

Differentiating (1) with respect to x twice yields

2 $I ′ k = λ k ∫ a x cosh ⁡ [ λ k ( x − t ) ] y ( t ) d t − λ k ∫ x d cosh ⁡ [ λ k ( t − x ) ] y ( t ) d t , I ″ k = 2 λ k y ( x ) + λ k 2 ∫ a x sinh ⁡ [ λ k ( x − t ) ] y ( t ) d t + λ k 2 ∫ x b sinh ⁡ [ λ k ( t − x ) ] y ( t ) d t ,$

where the primes denote the derivatives with respect to x.By comparing formulas (1) and (2), we find the relation between $I k ″$

and I k :

3 $I ″ k = 2 λ k y ( x ) + λ k 2 I k , I k = I k ( x ) .$

2°.With the aid of (1), the integral equation can be rewritten in the form

4 $∑ k = 1 n A k I k = f ( x ) .$

Differentiating (4) with respect to x twice and taking into account (3), we find that

5 $σ 1 y ( x ) + ∑ k = 1 n A k λ k 2 I k = f ″ x x ( x ) , σ 1 = 2 ∑ k = 1 n A k λ k .$

Eliminating the integral I n from (4) and (5) yields

6 $σ 1 y ( x ) + ∑ k = 1 n − 1 A k ( λ k 2 − λ n 2 ) I k = f ″ x x ( x ) − λ n 2 f ( x ) .$

Differentiating (6) with respect to x twice and eliminating I n−1 from the resulting equation with the aid of (6), we obtain a similar equation whose right-hand side is a second-order linear differential operator (acting on y) with constant coefficients plus the sum $∑ k = 1 n − 2 B k I k$

. If we successively eliminate I n–2, I n–3,…, with the aid of double differentiation, then we finally arrive at a linear nonhomogeneous ordinary differential equation of order 2(n –1) with constant coefficients.

3°.The right-hand side f (x) must satisfy certain conditions. To find these conditions, one should set x = a in the integral equation and its derivatives. (Alternatively, these conditions can be found by setting x = a and x = b in the integral equation and all its derivatives obtained by means of double differentiation.)

11. $∫ 0 b | s i n h k x − s i n h k t | y ( t ) d t = f ( x ) , 0 < k < 1$

.

This is a special case of equation 3.8.3 with g(x)=sinhk x.

Solution:

$y ( x ) = 1 2 k [ f ′ x ( x ) cosh ⁡ x sinh ⁡ k − 1 x ] .$

The right-hand side f(x) must satisfy certain conditions. As follows from item 3° of equation 3.8.3, the admissible general form of the right-hand side is given by

$f ( x ) = F ( x ) + A x + B , A = − F ′ x ( b ) , B = 1 2 [ b F ′ x ( b ) − F ( 0 ) − F ( b ) ] ,$

where F(x) is an arbitrary bounded twice differentiable function (with bounded first derivative).

12. $∫ a b y ( t ) | s i n h ( λ x ) − s i n h ( λ t ) | k d t = f ( x ) , 0 < k < 1$

.

This is a special case of equation 3.8.7 with g(x) = sinh (λx) + β,where β is an arbitrary number.

13. $∫ 0 a | k s i n h ( λ x ) − t | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.5 with g(x)=k sinh(λx).

14. $∫ 0 a | x − k s i n h ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.6 with g(x)=k sinh(λt).

#### 3.3-3  Kernels Containing Hyperbolic Tangent

15. $∫ 0 b | t a n h ( λ x ) − t a n h ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x)=tanh(λx).

Solution:

$y ( x ) = 1 2 λ d d x [ cosh ⁡ 2 ( λ x ) f ′ x ( x ) ] .$

The right-hand side f(x) of the integral equation must satisfy certain relations (see item 2° of equation 3.8.3).

16. $∫ 0 a | t a n h ( β x ) − t a n h ( μ t ) | y ( t ) d t = f ( x ) , β > 0 , μ > 0$

.

This is a special case of equation 3.8.4 with g(x)=tanh(βx)andλ = μ/β.

17. $∫ 0 b | t a n h k x − t a n h k t | y ( t ) d t = f ( x ) , 0 < k < 1$

.

This is a special case of equation 3.8.3 with g(x)=tanh k x.

Solution:

$y ( x ) = 1 2 k d d x [ cosh ⁡ 2 ( x ) coth ⁡ k − 1 x f ′ x ( x ) ] .$

The right-hand side f(x) must satisfy certain conditions. As follows from item 3° of equation 3.8.3, the admissible general form of the right-hand side is given by

$f ( x ) = F ( x ) + A x + B , A = − F ′ x ( b ) , B = 1 2 [ b F ′ x ( b ) − F ( 0 ) − F ( b ) ] ,$

where F(x) is an arbitrary bounded twice differentiable function (with bounded first derivative).

18. $∫ a b y ( t ) | t a n h ( λ x ) − t a n h ( λ t ) | k d t = f ( x ) , 0 < k < 1$

.

This is a special case of equation 3.8.7 with g(x) = tanh(λx) + β,where β is an arbitrary number.

19. $∫ 0 a | k t a n h ( λ x ) − t | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.5 with g(x) = k tanh(λx).

20. $∫ 0 a | x − k t a n h ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.6 with g(x)=k tanh (λt).

#### 3.3-4  Kernels Containing Hyperbolic Cotangent

21. $∫ a b | c o t h ( λ x ) − c o t h ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x)=coth(λx).

22. $∫ 0 b | c o t h k x − c o t h k t | y ( t ) d t = f ( x ) , 0 < k < 1$

.

This is a special case of equation 3.8.3 with g(x)=coth k x.

#### 3.4-1  Kernels Containing Logarithmic Functions.

1. $∫ a b | l n ( x / t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x)=In x.

Solution:

$y ( x ) = 1 2 d d x [ x f ′ x ( x ) ] .$

The right-hand side f(x) of the integral equation must satisfy certain relations (see item 2° of equation 3.8.3).

2. $∫ a b l n | x − t | y ( t ) d t = f ( x )$

.

Carleman’s equation.

1°.Solution with ba ≠ 4:

$y ( x ) = 1 π 2 ( x − a ) ( b − x ) [ ∫ a b ( t − a ) ( b − t ) f ′ t ( t ) d t t − x + 1 In [ 1 4 ( b − a ) ] ∫ a b f ( t ) d t ( t − a ) ( b − t ) ] .$

2°.If b –a =4, then for the equation to be solvable, the condition

$∫ a b f ( t ) ( t − a ) − 1 / 2 ( b − t ) − 1 / 2 d t = 0$

must be satisfied. In this case, the solution has the form

$y ( x ) = 1 π 2 ( x − a ) ( b − a ) [ ∫ a b ( t − a ) ( b − t ) f ′ t ( t ) d t t − x + C ] ,$

where C is an arbitrary constant.

⊙ Reference: F. D. Gakhov (1977).

3. $∫ a b ( l n | x − t | + β ) y ( t ) d t = f ( x )$

.

By setting

$x = e − β z , t = e − β τ , y ( t ) = Y ( τ ) , f ( x ) = e − β g ( z ) ,$

we arrive at an equationoftheform 3.4.2:

$∫ A B In| z − τ | Y ( τ ) d τ = g ( z ) , A = a e β , B = b e β .$

4. $∫ − a a ( l n A | x − t | ) y ( t ) d t = f ( x ) , − a ≤ x ≤ a$

.

This is a special case of equation 3.4.3 with b =–a.Solution with 0 < a <2A:

$y ( x ) = 1 2 M ′ ( a ) [ d d a ∫ − a a w ( t , a ) f ( t ) d t ] w ( x , a ) − 1 2 ∫ | x | a w ( x , ξ ) d d ξ [ 1 M ′ ( ξ ) d d ξ ∫ − ξ ξ w ( t , ξ ) f ( t ) d t ] d ξ − 1 2 d d x ∫ | x | a w ( x , ξ ) M ′ ( ξ ) [ ∫ − ξ ξ w ( t , ξ ) d f ( t ) ] d ξ ,$

where

$M ( ξ ) = ( In 2 A ξ ) − 1 , w ( x , ξ ) = M ( ξ ) π ξ 2 − x 2 ,$

and the prime stands for the derivative.

⊙ Reference: I. C. Gohberg and M. G. Krein (1967).

5. $∫ 0 a l n | x + t x − t | y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = − 2 π 2 d d x ∫ x a F ( t ) d t t 2 − x 2 , F ( t ) = d d t ∫ 0 t s f ( s ) d s t 2 − s 2 .$

⊙ Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975).

6. $∫ a b | l n 1 + λ x 1 − λ t | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x)=In(1 +λx).

Solution:

$y ( x ) = 1 2 λ d d x [ ( 1 + λ x ) f ′ x ( x ) ] .$

Theright-handside f(x) of the integral equation must satisfy certain relations (seeitem2° of equation 3.8.3).

7. $∫ a b | l n β x − l n β t | y ( t ) d t = f ( x ) , 0 < β < 1$

.

This is a special case of equation 3.8.3 with g(x) = In β x.

8. $∫ a b y ( t ) | l n ( x / t ) | β d t = f ( x ) , 0 < β < 1$

.

This is a special case of equation 3.8.7 with g(x)=Inx + A,where A is an arbitrary number.

#### 3.4-2  Kernels Containing Power-Law and Logarithmic Functions

9. $∫ 0 1 ( l n | x − t | + β t k ) y ( t ) d t = f ( x )$

.

See Example 3 in Subsection 12.6-2 with ψ(t) = βt k .

10. $∫ 0 a | k l n ( 1 + λ x ) − t | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.5 with g(x) = k In(1 + λx).

11. $∫ 0 a | x − k l n ( 1 + λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.6 with g(x)=k In(1 + λt).

12. $∫ 0 ∞ 1 t l n | x + t x − t | y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = x π 2 d d x ∫ 0 ∞ d f ( t ) d t ln ⁡ | 1 − x 2 t 2 | d t .$

⊙ Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975).

13. $∫ 0 ∞ ln ⁡ x − ln ⁡ t x − t y ( t ) d t = f ( x )$

.

The left-hand side of this equation is the iterated Stieltjes transform.

Under some assumptions, the solution of the integral equation can be represented in the form

$y ( x ) = 1 4 π 2 lim ⁡ n → ∞ ( e n ) 4 n D n x 2 n D 2 n x 2 n D n f ( x ) , D = d d x .$

To calculate the solution approximately, one should restrict oneself to a specific value of n in this formula instead of taking the limit.

⊙ Reference: 1.1. Hirschman and D. V. Widder (1955).

14. $∫ 0 ∞ ln ⁡ | x β − t β | y ( t ) d t = f ( x ) , β > 0$

.

The transformation

$z = x β , τ = t β , w ( τ ) = t 1 − β y ( t )$

$∫ A B ln ⁡ | z − τ | w ( τ ) d τ = F ( z ) , A = a β , B = b β ,$

where F(z) = βf (z 1/β ).

15. $∫ 0 ∞ ln ⁡ | x β − t μ | y ( t ) d t = f ( x ) , β > 0 , μ > 0$

.

The transformation

$z = x β , τ = t μ , w ( τ ) = t 1 − μ y ( t )$

leads to an equation of the form 3.4.2:

$∫ 0 1 ln ⁡ | z − τ | w ( τ ) d τ = F ( z ) , F ( z ) = μ f ( z 1 / β ) .$

16. $∫ 0 ∞ 1 x t ln ⁡ ( x t ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = − 1 π 2 ∫ 0 ∞ 1 x t ln ⁡ ( x t ) f ( t ) d t .$

⊙ References: E. C. Titchmarsh (1986), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 450).

17. $d d x ∫ − ∞ ∞ | 1 − x t | y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = − 1 π 2 d d x ∫ − ∞ ∞ ln ⁡ | 1 − x t | f ( t ) d t .$

⊙ References: E. C. Titchmarsh (1986), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 450).

18. $∫ 0 ∞ ( x t ) − [ 1 + i ln ⁡ ( x t ) ] / 2 y ( t ) d t = f ( x ) , i 2 = − 1$

.

Solution:

$y ( x ) = 1 2 π ∫ 0 ∞ ( x t ) − [ 1 − i ln ⁡ ( x t ) ] / 2 f ( t ) d t .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 452).

#### 3.4-3  Equation Containing the Unknown Function of a Complicated Argument.

19. $∫ 0 − 1 ( A ln ⁡ t + B ) y ( x t ) d t = f ( x )$

.

The substitution ξ = xt leads to an equation of the form 1.9.3 with g(x) = −A In x:

$∫ 0 x ( A ln ⁡ ξ − A ln ⁡ x + B ) y ( ξ ) d ξ = x f ( x ) .$

#### 3.5-1  Kernels Containing Cosine.

1. $∫ 0 ∞ cos ⁡ ( x t ) y ( t ) d t = f ( x )$

.

Solution: $y ( x ) = 2 π ∫ 0 ∞ cos ⁡ ( x t ) f ( t ) d t$

.

Up to constant factors, the function f (x) and the solution y(t) are the Fourier cosine transform pair.

⊙ References: E. A. C. Paley and N. Wiener (1934), S. Bochner and K. C. Chandrasekharan (1949), G. N. Watson (1952), H. Bateman and A. Erdelyi (Vol. 1, 1954), S. Bochner (1959), V. A. Ditkin and A. P. Prudnikov (1965), B. Davis (1978), F. Oberhettinger (1980), E. C. Titchmarsh (1986), Ya. A. Brychkov and A. P. Prudnikov (1989), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 440), I. Sneddon (1995), A. D. Poularikas (2000).

2. $∫ a b cos ⁡ ( x t ) y ( t ) d t = f ( x ) , 0 ≤ x < ∞$

.

Solution:

$y ( t ) = { 2 π ∫ 0 ∞ cos ⁡ ( x t ) f ( x ) d x if a < t < b , 0 if 0 < t < a or t > b ,$

where $0 ≤ a ≤ b ≤ ∞$

3. $∫ a b | cos ⁡ ( λ x ) − cos ⁡ ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x)=cos(λx).

Solution:

$y ( x ) = − 1 2 λ d d x [ f ′ x ( x ) sin ⁡ ( λ x ) ] .$

The right-hand side f(x) of the integral equation must satisfy certain relations (see item 2° of equation 3.8.3).

4. $∫ 0 a | cos ⁡ ( β x ) − cos ⁡ ( μ t ) | y ( t ) d t = f ( x ) , β > 0 , μ > 0$

.

This is a special case of equation 3.8.4 with g(x)=cos(βx) and λ = μ/β.

5. $∫ a b | cos ⁡ k x − cos ⁡ k t | y ( t ) d t = f ( x ) , 0 < k < 1$

.

This is a special case of equation 3.8.3 with g(x)=cos k x.

Solution:

$y ( x ) = − 1 2 λ d d x [ f ′ x ( x ) sin ⁡ x cos ⁡ k − 1 x ] .$

The right-hand side f(x) of the integral equation must satisfy certain relations (see item 2° of equation 3.8.3).

6. $∫ a b y ( t ) | cos ⁡ ( λ x ) − cos ⁡ ( λ t ) | k d t = f ( x ) , 0 < k < 1$

.

This is a special case of equation 3.8.7 with g(x)=cos(λx),whereβ is an arbitrary number.

7. $∫ 0 ∞ t − i x − 1 / 2 cos ⁡ ( 1 + 2 i x 4 π ) y ( t ) d t = f ( x ) , i 2 = − 1$

.

Solution:

$y ( t ) = 1 π ∫ − ∞ ∞ t i x − 1 / 2 cos ⁡ ( 1 − 2 i x 4 π ) f ( x ) cosh ⁡ ( π x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 463).

#### 3.5-2  Kernels Containing Sine.

8. $∫ 0 ∞ sin ⁡ ( x t ) y ( t ) d t = f ( x )$

.

Solution: $y ( x ) = 2 π ∫ 0 ∞ sin ⁡ ( x t ) f ( t ) d t$

sm(xt)f(t) dt.

Up to constant factors, the function f(x) and the solution y(t) are the Fourier sine transform pair.

⊙ References: E. A. C. Paley and N. Wiener (1934), S. Bochner and K. C. Chandrasekharan (1949), G. N. Watson (1952), H. Bateman and A. Erdelyi (Vol. 1, 1954), S. Bochner (1959), V. A. Ditkin and A. P. Prudnikov (1965), B. Davis (1978), F. Oberhettinger (1980), E. C. Titchmarsh (1986), Ya. A. Brychkov and A. P. Prudnikov (1989), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 440), I. Sneddon (1995), A. D. Poularikas (2000).

9. $∫ a b sin ⁡ ( x t ) y ( t ) d t = f ( x ) , 0 ≤ x < ∞$

.

Solution:

$y ( t ) = { 2 π ∫ 0 ∞ cos ⁡ ( x t ) f ( x ) d x if a < t < b , 0 if 0 < t < a or t > b ,$

where $0 ≤ a ≤ b ≤ ∞$

10. $∫ − ∞ − ∞ sin ⁡ ( λ | x − t | ) y ( t ) d t = f ( x ) , f ( ± ∞ ) = 0$

.

Solution:

$y ( x ) = 1 2 λ [ f ″ x x ( x ) + λ 2 f ( x ) ] .$

11. $∫ − ∞ − ∞ sin ⁡ ( λ | x − t | ) y ( t ) d t = f ( x ) , − ∞ < a < b < ∞$

.

1 °. Let us remove the modulus in the integrand:

1 $∫ a x sin ⁡ [ λ ( x − t ) ] y ( t ) d t + ∫ x b sin ⁡ [ λ ( t − x ) ] y ( t ) d t = f ( x ) .$

Differentiating (1) with respect to x twice yields

2 $2 λ y ( x ) − λ 2 ∫ a x sin ⁡ [ λ ( x − t ) ] y ( t ) d t − λ 2 ∫ x b sin ⁡ [ λ ( t − x ) ] y ( t ) d t = f ″ x x ( x ) .$

Eliminating the integral terms from (1) and (2), we obtain the solution

3 $y ( x ) = 1 2 λ [ f ″ x x ( x ) + λ 2 f ( x ) ] .$

2°.Theright-handside f(x) of the integral equation must satisfy certain relations. By setting x = a and x = b in (1), we obtain two corollaries

4 $∫ a b sin ⁡ [ λ ( t − a ) ] y ( t ) d t = f ( a ) , ∫ a b sin ⁡ [ λ ( b − t ) ] y ( t ) d t = f ( b ) .$

Substitutingsolution(3)into(4)followedbyintegratingbypartsyieldsthedesiredconditions for f(x):

5 $sin ⁡ [ λ ( b − a ) ] f ′ x ( b ) − λ cos ⁡ [ λ ( b − a ) ] f ( b ) = λ f ( a ) , sin ⁡ [ λ ( b − a ) ] f ′ x ( a ) + λ cos ⁡ [ λ ( b − a ) ] f ( a ) = − λ f ( b ) .$

The general form of the right-hand side of the integral equation is given by

6 $f ( x ) = F ( x ) + A x + B ,$

where F(x) is an arbitrary bounded twice differentiable function, and the coefficients A and B are expressed in terms of F (a), F (b), $F ′ x$

(a), and $F ′ x$ (b)and can be determined by substituting formula (6) into conditions (5).

12. $∫ a b { A sin ⁡ ( λ | x − t | ) + B sin ⁡ ( μ | x − t | ) } y ( t ) d t = f ( x ) , − ∞ < a < b < ∞$

.

Let us remove the modulus in the integrand and differentiate the equation with respect to x twice to obtain

1 $2 ( A λ + B μ ) y ( x ) − ∫ a b { A λ 2 sin ⁡ ( λ | x − t | ) + B μ 2 sin ⁡ ( μ | x − t | ) } y ( t ) d t = f ″ x x ( x ) .$

Eliminating the integral term with sin (μ|x – t|) from (1) with the aid of the original equation, we find that

2 $2 ( A λ + B μ ) y ( x ) + A ( μ 2 − λ 2 ) ∫ a b sin ⁡ ( λ | x − t | ) y ( t ) d t = f ″ x x ( x ) + μ 2 f ( x ) .$

For + = 0, this is an equation of the form 3.5.11 and for + ≠ 0, this is an equation of the form 4.5.29.

The right-hand side f(x) must satisfy certain relations, which can be obtained by setting x = a and x = b in theoriginal equation (a similar procedure is used in 3.5.11).

13. $∫ a b | sin ⁡ ( λ x ) − sin ⁡ ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x)=sin(λx).

Solution:

$y ( x ) = 1 2 λ d d x [ f ′ x ( x ) cos ⁡ ( λ x ) ] .$

The right-hand side f(x) of the integral equation must satisfy certain relations (see item 2° of equation 3.8.3).

14. $∫ 0 a | sin ⁡ ( β x ) − sin ⁡ ( μ t ) | y ( t ) d t = f ( x ) , β > 0 , μ > 0$

.

This is a special case of equation 3.8.4 with g(x)=sin(βx)andλ = μ/β.

15. $∫ a b sin ⁡ 3 ( λ | x − t | ) y ( t ) d t = f ( x )$

.

Using the formula sin3 $β = − 1 4$

sin β, we arrive at an equation of the form 3.5.12:

$∫ a b [ − 1 4 A sin ⁡ ( 3 λ | x − t | ) + 3 4 A sin ⁡ ( λ | x − t | ) ] y ( t ) d t = f ( x ) .$

16. $∫ a b [ ∑ k = 1 n A k sin ⁡ ( λ k | x − t | ) ] y ( t ) d t = f ( x ) , − ∞ < a < b < ∞$

.

1 °. Let us remove the modulus in the kth summand of the integrand:

1 $I k ( x ) = ∫ a b sin ⁡ ( λ k | x − t | ) y ( t ) d t = ∫ a x sin ⁡ [ λ k ( x − t ) ] y ( t ) d t + ∫ x b sin ⁡ [ λ k ( t − x ) ] y ( t ) d t .$

Differentiating (1) with respect to x yields

2 $I ′ k = λ k ∫ a x cos ⁡ [ λ k ( x − t ) ] y ( t ) d t − λ k ∫ x b cos ⁡ [ λ k ( t − x ) ] y ( t ) d t , I ″ k = 2 λ k y ( x ) − λ k 2 ∫ a x sin ⁡ [ λ k ( x − t ) ] y ( t ) d t − λ k 2 ∫ x b sin ⁡ [ λ k ( t − x ) ] y ( t ) d t ,$

where the primes denote the derivatives with respect to x.Bycomparing formulas (1) and (2), we find the relation between $I ″ k$

and I k :

3 $I ″ k = 2 λ k y ( x ) − λ k 2 I k , I k = I k ( x ) .$

2°.With the aid of (1), the integral equation can be rewritten in the form

4 $∑ k = 1 n A k I k = f ( x ) .$

Differentiating (4) with respect to x twiceand taking into account (3), we find that

5 $σ 1 y ( x ) − ∑ k = 1 n A k λ k 2 I k = f ″ x x ( x ) , σ 1 = 2 ∑ k = 1 n A k λ k .$

Eliminating the integral In from (4) and (5) yields

6 $σ 1 y ( x ) + ∑ k = 1 n − 1 A k ( λ n 2 − λ k 2 ) I k = f ″ x x ( x ) + λ n 2 f ( x ) .$

Differentiating (6) with respect to x twice and eliminating I n−1 from the resulting equation with theaid of (6), we obtain a similar equation whose left-hand side is a second-order linear differential operator (acting on y) with constant coefficients plus the sum $∑ k = 1 n − 2 B k I k$

If we successively eliminate I n−2, I n−3,…,with the aid of double differentiation, then we finally arrive at a linear nonhomogeneous ordinary differential equation of order 2(n −1) with constant coefficients.

3°.The right-hand side f (x) must satisfy certain conditions. To find these conditions, one should set x = a in the integral equation and its derivatives. (Alternatively, these conditions can be found by setting x = a and x = b in the integral equation and all its derivatives obtained by means of double differentiation.)

17. $∫ a b | sin ⁡ k x − sin ⁡ k t | y ( t ) d t = f ( x ) , 0 < k < 1$

.

This is a special case of equation 3.8.3 with g(x)=sin k x.

Solution:

$y ( x ) = 1 2 k d d x [ f ′ x ( x ) cos ⁡ x sin ⁡ k − 1 x ] .$

The right-hand side f(x) must satisfy certain conditions. As follows from item 3° of equation 3.8.3, the admissible general form of the right-hand side is given by

$f ( x ) = F ( x ) + A x + B , A = − F ′ x ( b ) , B = 1 2 [ b F ′ x ( b ) − F ( 0 ) − F ( b ) ] ,$

where F(x) is an arbitrary bounded twice differentiable function (with bounded first derivative).

18. $∫ a b y ( t ) | sin ⁡ ( λ x ) − sin ⁡ ( λ t ) | k d t = f ( x ) , 0 < k < 1$

.

This is a special case of equation 3.8.7 with g(x) = sin(λx) + β,where β is an arbitrary number.

19. $∫ 0 a | k sin ⁡ ( λ x ) − t | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.5 with g(x)=k sin(λx).

20. $∫ 0 a | x − k sin ⁡ ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.6 with g(t)=k sin(λt).

21. $∫ 0 ∞ sin ⁡ t t 2 [ y ( x − t ) − y ( x − t ) ] d t = f ( x )$

.

Solution:

$y ( x ) = 1 π ∫ 0 ∞ [ cos ⁡ t t + Si ( t ) [ f ( x − t ) − f ( x + t ) ] d t , ]$

where Si(t) is sine integral (see Supplement 11.3-1).

The integral equation and its solution form the Boas transform pair.

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 442).

22. $∫ 0 ∞ t − i x − 1 / 2 sin ⁡ ( 1 + 2 i x 4 ) y ( t ) d t = f ( x ) , i 2 = − 1$

.

Solution:

$y ( t ) = 1 π ∫ − ∞ ∞ t i x − 1 / 2 sin ⁡ ( 1 − 2 i x 4 π ) f ( x ) cosh ⁡ ( π x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 463).

#### 3.5-3  Kernels Containing Tangent.

23. $∫ a b | tan ⁡ ( λ x ) − tan ⁡ ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x)=tan (λx).

Solution:

$y ( x ) = 1 2 λ d d x [ cos ⁡ 2 ( λ x ) f ′ x ( x ) ] .$

The right-hand side f(x) of the integral equation must satisfy certain relations (see item 2° of equation 3.8.3).

24. $∫ 0 a | tan ⁡ ( β x ) − tan ⁡ ( μ t ) | y ( t ) d t = f ( x ) , β > 0 , μ > 0$

.

This is a special case of equation 3.8.4 with g(x)=tan(βx) and λ = μ/β.

25. $∫ 0 a | tan ⁡ k x − tan ⁡ k t | y ( t ) d t = f ( x ) , 0 < k < 1$

.

This is a special case of equation 3.8.3 with g(x)=tank x.

Solution:

$y ( x ) = 1 2 k d d x [ cos ⁡ 2 x cot ⁡ k − 1 x f ′ x ( x ) ] .$

The right-handside f(x) must satisfy certain conditions. As follows from item 3° of equation 3.8.3, the admissible general form of the right-hand side is given by

$f ( x ) = F ( x ) + A x + B , A = − F ′ x ( b ) , B = 1 2 [ b F ′ x ( b ) − F ( 0 ) − F ( b ) ] ,$

where F(x) is an arbitrary bounded twice differentiable function (with bounded first derivative).

26. $∫ a b y ( t ) | tan ⁡ ( λ x ) − tan ⁡ ( λ t ) | k d t = f ( x ) , 0 < k < 1$

.

This is a special caseofequation3.8.7withg(x)=tan(λx),where β is an arbitrary number.

27. $∫ 0 a | k tan ⁡ ( λ x ) − t | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.5 with g(x)=k tan(λx).

28. $∫ 0 a | x − k tan ⁡ ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.6 with g(t)=k tan(λt).

#### 3.5-4  Kernels Containing Cotangent.

29. $∫ a b | cot ⁡ ( λ x ) − cot ⁡ ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x) = cot(λx).

30. $∫ a b | cot ⁡ k x − cot ⁡ k t | y ( t ) d t = f ( x ) , 0. < k < 1$

.

This is a special case of equation 3.8.3 with g(x) = cot k x.

#### 3.5-5  Kernels Containing a Combination of Trigonometric Functions.

31. $∫ − ∞ − ∞ [ cos ⁡ ( x t ) + sin ⁡ ( λ t ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 1 2 π ∫ − ∞ ∞ [ cos ⁡ ( x t ) + sin ⁡ ( x t ) ] f ( t ) d t .$

Up to constant factors, the function f (x) and the solution y(t) are the Hartley transform pair.

⊙ Reference: D. Zwillinger (1989).

32. $∫ 0 ∞ [ sin ⁡ ( x t ) − x t cos ⁡ ( x t ) ] y ( t ) d t = f ( x )$

.

This equation can be reduced to a special case of equation 3.7.17 with $ν = 3 2$

Solution:

$y ( x ) = 2 π ∫ 0 ∞ sin ⁡ ( x t ) − x t cos ⁡ ( x t ) x 2 t 2 f ( t ) d t .$

33. $∫ 0 ∞ [ sin ⁡ ( x t ) + x t cos ⁡ ( x t ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = − 2 π ∫ 0 ∞ si ( x t ) y ( t ) d t ,$

where si(z) is the sine integral (see Supplement 11.3-1).

⊙ References: H. M. Srivastava and R. G. Buschman (1977),A.P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 457).

34. $∫ 0 ∞ [ 1 − cos ⁡ ( x t ) + x t sin ⁡ ( x t ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 2 π ∫ 0 ∞ ci ( x t ) f ( t ) d t ,$

where ci(z) is the cosine integral (see Supplement 11.3-2).

⊙ References: H. M. Srivastava and R. G. Buschman (1977),A.P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 457).

35. $∫ 0 ∞ ( x t ) 1 / 2 [ sin ⁡ ( x t ) x t + 2 cos ⁡ ( x t ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 2 π ∫ 0 ∞ [ 1 2 − S ( x t ) ] f ( t ) d t .$

where S (z ) is the Fresnel sine integral (see Supplement 11.3-3).

⊙ References: H. M. Srivastava and R. G. Buschman (1977),A.P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 459).

36. $∫ 0 ∞ ( x t ) 1 / 2 [ cos ⁡ ( x t ) − 1 x t − 2 sin ⁡ ( x t ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 2 π ∫ 0 ∞ [ 1 2 − C ( x t ) ] f ( t ) d t ,$

where C( z ) is the Fresnel cosine integral (see Supplement 11.3-3).

⊙ References: H. M. Srivastava and R. G. Buschman (1977),A.P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 460).

37. $∫ 0 ∞ ( 1 − ν ) sin ⁡ ( x t ) + x t cos ⁡ ( x t ) ( x t ) ν y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 2 π ∫ 0 ∞ S ( x t , ν ) f ( t ) d t ,$

where S(z, ν) is the generalized Fresnel sine integral (see Supplement 11.3-3).

⊙ References: H. M. Srivastava and R. G. Buschman (1977),A.P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 461).

38. $∫ 0 ∞ ( 1 − ν ) cos ⁡ ( x t ) − x t sin ⁡ ( x t ) ( x t ) ν y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 2 π ∫ 0 ∞ C ( x t , ν ) y ( t ) d t ,$

where C(z, ν) is the generalized Fresnel cosine integral (see Supplement 11.3-3).

⊙ References: H. M. Srivastava and R. G. Buschman (1977),A.P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 461).

39. $∫ 0 π [ a sin ⁡ ( x + t ) 1 + 2 a cos ⁡ ( x + t ) + a 2 + a sin ⁡ ( x − t ) 1 − 2 a cos ⁡ ( x − t ) + a 2 ] y ( t ) d t = f ( x ) , 0 < a < 1$

.

Solution:

$y ( x ) = C + 2 π 2 ∑ n = 1 ∞ f n a n cos ⁡ ( n x ) , f n = ∫ 0 π f ( x ) sin ⁡ ( n x ) d x ,$

where C is an arbitrary constant.

Remark. The kernel of the integral equation can be represented as a series in powers of a:

$K ( x , t ) = a sin ⁡ ( x + t ) 1 − 2 a cos ⁡ ( x + t ) + a 2 + a sin ⁡ ( x − t ) 1 − 2 a cos ⁡ ( x − t ) + a 2 = 2 ∑ n = 1 ∞ a n sin ⁡ ( n x ) cos ⁡ ( n t ) .$

⊙ References: W. Schmeidler (1950, p. 169), S. Fenyo and H. W. Stolle (1984, pp. 18–19).

#### 3.5-6  Equations Containing the Unknown Function of a Complicated Argument.

40. $∫ 0 π / 2 y ( ξ ) d t = f ( x ) , ξ = x sin ⁡ t$

.

Schlomilch equation.

Solution:

$y ( x ) = 2 π [ f ( 0 ) + x ∫ 0 π / 2 f ′ ξ ( ξ ) d t ] , ξ = x sin ⁡ t .$

⊙ References: E. T. Whittaker and G. N. Watson (1958), F. D. Gakhov (1977).

41. $∫ 0 π / 2 y ( ξ ) d t = f ( x ) , ξ = x sin ⁡ k t$

.

Generalized Schlomilch equation.

This is a special case of equation 3.5.43 for λ =0 and m =0.

Solution:

$y ( x ) = 2 k π x k − 1 k d d x [ x 1 k ∫ 0 x sin ⁡ t f ( ξ ) d t ] , ξ = x sin ⁡ k t .$

42. $∫ 0 π / 2 sin ⁡ λ t y ( ξ ) d t = f ( x ) , ξ = x sin ⁡ k t$

.

This is a special case of equation 3.5.43 for m =0.

Solution:

$y ( x ) = 2 k π x k − λ − 1 k d d x [ x λ + 1 k ∫ 0 x sin ⁡ λ + 1 t f ( ξ ) d t ] , ξ = x sin ⁡ k t .$

43. $∫ 0 π / 2 sin ⁡ λ t cos ⁡ m t y ( ξ ) d t = f ( x ) , ξ = x sin ⁡ k t$

.

1°.Let λ > −1, m > −1, and k >0. The transformation

$z = x 2 k , ζ = z sin ⁡ 2 t , w ( ζ ) = ζ λ − 1 2 y ( ζ k 2 )$

leads to an equation of the form 1.1.44:

$∫ 0 z ( z − ζ ) m − 1 2 w ( ζ ) d ζ = F ( z ) , F ( z ) = 2 z λ + m 2 f ( z k 2 ) .$

2°.Solution with −1 < m <1:

$y ( x ) = 2 k π sin ⁡ [ π ( 1 − m ) 2 ] x k − λ − 1 k d d x [ x λ + 1 k ∫ 0 π / 2 sin ⁡ λ + 1 t tan ⁡ m t f ( ξ ) d t ] ,$

where ξ = x sin k t.

#### 3.5-7  Singular Equations.

44. $∫ 0 2 π cot ⁡ ( t − x 2 ) y ( t ) d t = f ( x ) , 0 ≤ x ≤ 2 π$

.

Here the integral is understood in the sense of the Cauchy principal value and the right-hand side is assumed to satisfy the condition $∫ 0 2 π f ( t ) d t = 0$

Solution:

$y ( x ) = − 1 4 π 2 ∫ 0 2 π cot ⁡ ( t − x 2 ) f ( t ) d t + C ,$

where C is an arbitrary constant.

It follows from the solution thaw $∫ 0 2 π y ( t ) d t = 2 π C$

The equation and its solution form a Hilbert transform pair (in the asymmetric form).

⊙ Reference: F. D. Gakhov (1977).

45. $∫ 0 2 π [ 1 + cot ⁡ ( x − t 2 ) ] y ( t ) d t = f ( x ) , − π ≤ x ≤ π$

.

Hilbert–Plessner equation.

Solution:

$y ( x ) = 1 4 π 2 ∫ − π π [ 1 + cot ⁡ ( x − t 2 ) ] f ( t ) d t .$

⊙Reference: S. Fenyö and H. W. Stolle (1984, pp. 36–38).

46. $∫ 0 2 π [ sin ⁡ ( ξ − x 2 ) ] − 2 y ( ξ ) d ξ = f ( x ) , 0 ≤ x ≤ 2 π$

.

The simple hypersingular equation of the first kind with Hilbert-type kernel.

Let the periodic conditions y(0) = y(2π) be satisfied. Then the solution is

$y ( x ) = − 1 4 π 2 ∫ 0 2 π f ( ξ ) ln ⁡ | sin ⁡ ( ξ − x 2 ) | d ξ + C ,$

where C is an arbitrary constant.

This equation is discussed in Subsection 14.6-4 in detail.

⊙ Reference: I. K. Lifanov, L. N. Poltavskii, and G. M. Vainikko (2004, p. 8).

#### 3.6-1  Kernels Containing Hyperbolic and Logarithmic Functions.

1. $∫ a b ln ⁡ | cosh ⁡ ( λ x ) − cosh ⁡ ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.9 with g(x) = cosh(λx).

2. $∫ a b ln ⁡ | sinh ⁡ ( λ x ) − sinh ⁡ ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.9 with g(x)=sinh(λx).

3. $∫ a b ln ⁡ [ sinh ⁡ ( 1 2 A ) 2 sinh ⁡ ( 1 2 | x − t | ) ] y ( t ) d t = f ( x ) , − a ≤ x ≤ a$

.

Solution with 0 < a < A:

$y ( x ) = 1 2 M ′ ( a ) [ d d a ∫ − a a w ( t , a ) f ( t ) d t ] w ( x , a ) − 1 2 ∫ | x | a w ( x , ξ ) d d ξ [ 1 M ′ ( ξ ) d d ξ ∫ − ξ ξ w ( t , ξ ) f ( t ) d t ] d ξ − 1 2 d d x ∫ | x | a w ( x , ξ ) M ′ ( ξ ) [ ∫ − ξ ξ w ( t , ξ ) d f ( t ) ] d ξ ,$

where the prime stands for the derivative with respect to the argument and

$M ( ξ ) = [ ln ⁡ ( sinh ⁡ ( 1 2 A ) sinh ⁡ ( 1 2 ξ ) ) ] − 1 , w ( x , ξ ) = cosh ⁡ ( 1 2 x ) M ( ξ ) π 2 cosh ⁡ ξ − 2 cosh ⁡ x .$

⊙ Reference: I. C. Gohberg and M. G. Krein (1967).

4. $∫ a b ln ⁡ | tanh ⁡ ( λ x ) − tanh ⁡ ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.9 with g(x)=tanh(λx).

5. $∫ − a b ln ⁡ | coth ⁡ ( 1 4 | x − t | ) | y ( t ) d t = f ( x ) , − a ≤ x ≤ a$

.

Solution:

$y ( x ) = 1 2 M ′ ( a ) [ d d a ∫ − a a w ( t , a ) f ( t ) d t ] w ( x , a ) − 1 2 ∫ | x | a w ( x , ξ ) d d ξ [ 1 M ′ ( ξ ) d d ξ ∫ − ξ ξ w ( t , ξ ) f ( t ) d t ] d ξ − 1 2 d d x ∫ | x | a w ( x , ξ ) M ′ ( ξ ) [ ∫ − ξ ξ w ( t , ξ ) d f ( t ) ] d ξ ,$

where the prime stands for the derivative with respect to the argument and

$M ( ξ ) = P − 1 / 2 ( cosh ⁡ ξ ) Q − 1 / 2 ( cosh ⁡ ξ ) , w ( x , ξ ) = 1 π Q − 1 / 2 ( cosh ⁡ ξ ) 2 cosh ⁡ ξ − 2 cosh ⁡ x ,$

and P −1/2(cosh ξ) and Q−1/2(cosh ξ) are the Legendre functions of the first and second kind, respectively.

⊙ Reference: I. C. Gohberg and M. G. Krein (1967).

#### 3.6-2  Kernels Containing Logarithmic and Trigonometric Functions.

6. $∫ a b ln ⁡ | cos ⁡ ( λ x ) − cos ⁡ ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.9 with g(x)=cos(λx).

7. $∫ a b ln ⁡ | sin ⁡ ( λ x ) − sin ⁡ ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.9 with g(x) = sin(λx).

8. $∫ 0 π ln ⁡ 1 − cos ⁡ ( x + t ) 1 − cos ⁡ ( x − t ) y ( t ) d t = f ( x ) , 0 ≤ x ≤ π$

.

Solution:

$y ( x ) = 2 π 2 ∑ n = 1 ∞ n f n sin ⁡ ( n x ) , f n = ∫ 0 π f ( x ) sin ⁡ ( n x ) d x .$

⊙ Reference: S. Fenyö and H. W. Stolle (1984, p. 44).

9. $∫ 0 π ln ⁡ [ sin ⁡ ( 1 2 A ) 2 sin ⁡ ( 1 2 | x − t | ) ] y ( t ) d t = f ( x ) , − a ≤ x ≤ a$

.

Solution with 0 < a < A:

$y ( x ) = 1 2 M ′ ( a ) [ d d a ∫ − a a w ( t , a ) f ( t ) d t ] w ( x , a ) − 1 2 ∫ | x | a w ( x , ξ ) d d ξ [ 1 M ′ ( ξ ) d d ξ ∫ − ξ ξ w ( t , ξ ) f ( x ) d t ] d ξ − 1 2 d d x ∫ | x | a w ( x , ξ ) M ′ ( ξ ) [ ∫ − ξ ξ w ( t , ξ ) d f ( t ) ] d ξ ,$

where the prime stands for the derivative with respect to the argument and

$M ( ξ ) = [ ln ⁡ ( sin ⁡ ( 1 2 A ) sin ⁡ ( 1 2 ξ ) ) ] − 1 , w ( x , ξ ) = cos ⁡ ( 1 2 ξ ) M ( ξ ) π 2 cos ⁡ x − 2 cos ⁡ ξ .$

⊙ Reference: I. C. Gohberg and M. G. Krein (1967).

10. $d d x ∫ − π π ln ⁡ ( 2 | sin ⁡ x − t 2 | ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = − 1 π 2 d d x ∫ − π π ln ⁡ ( 2 | sin ⁡ x − t 2 | ) f ( t ) d t , ∫ − π π y ( x ) d t = 0.$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 452).

#### 3.6-3  Kernels Containing Combinations of Exponential and Other Elementary Functions.

11. $∫ a b ( ln ⁡ | x − t | + A e − α x − β t ) y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.28 with ϑ(x) = Ae ax and ψ(t) = eβt .

12. $∫ 0 ∞ [ sin ⁡ ( x t ) + A e − α x − β t ] y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.29 with ψ(x) = Ae ax and ψ(t) = e −βt .

13. $∫ 0 ∞ [ cos ⁡ ( x t ) + A e − α x − β t ] y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.30 with ψ(x) = Ae αx and ψ(t) = e −βt .

#### 3.7-1  Kernels Containing Error Function, Exponential Integral or Logarithmic Integral.

1. $∫ 0 ∞ [ exp ( i ( x + t ) 2 ) erf ( e π i / 4 ( x + t ) ) + exp ⁡ ( i ( x − t ) 2 ) erf ( e π i / 4 ( x − t ) ) ] y ( t ) d t = f ( x )$

.

Here erf z is theerror function (see Supplement 11.2-1) and i 2 = −1.

Solution:

$y ( x ) = − 1 π ∫ 0 ∞ [ exp ⁡ ( − i ( t + x ) 2 ) erf ( e 3 π i / 4 ( t + x ) ) + exp ⁡ ( − i ( t − x ) 2 ) erf ( e 3 π i / 4 ( t − x ) ) ] f ( t ) d t .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 459).

2. $∫ 0 ∞ e − i x t E i ( i x t ) y ( t ) d t = f ( x ) , i 2 = − 1$

.

Here Ei(z) is the exponential integral (see Supplement 11.2-2).

Solution:

$y ( t ) = 1 2 π 2 ∫ − ∞ ∞ [ e i x t erf ( e π i / 4 x t ) − 1 + i 2 π x t ] f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 456).

3. $∫ 0 ∞ l i ( x t ) y ( t ) d t = f ( x ) , f ( 1 ) = f ′ ( 1 ) = 0$

.

Here li(z) is the logarithmic integral (see Supplement 11.2-3).

Solution:

$y ( t ) = − ∫ 1 x t − 2 v ( ln ⁡ t x ) [ ( t d d t ) 2 − t d d t ] f ( x ) d t ,$

where $ν ( z ) = ∫ 0 ∞ z ξ d ξ Γ ( ξ + 1 )$

⊙ References: H. M. Srivastava and R. G. Buschman (1977),A.P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 457).

#### 3.7-2  Kernels Containing Sine Integrals, Cosine Integrals, or Fresnel Integrals.

4. $∫ 0 ∞ s i ( x t ) y ( t ) d t = f ( x )$

.

Here si(z) is the sine integral (see Supplement 11.3-1).

Solution:

$y ( x ) = − 2 π ∫ 0 ∞ [ sin ⁡ ( x t ) + x t cos ⁡ ( x t ) ] f ( t ) d t .$

⊙ References: H. M. Srivastava and R. G. Buschman (1977),A.P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 457).

5. $∫ 0 ∞ c i ( x t ) y ( t ) d t = f ( x )$

.

Here ci(z) is the cosine integral (see Supplement 11.3-2).

Solution:

$y ( x ) = 2 π ∫ 0 ∞ [ 1 − cos ⁡ ( x t ) + x t sin ⁡ ( x t ) ] f ( t ) d t .$

⊙ References: H. M. Srivastava and R. G. Buschman (1977),A.P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 457).

6. $∫ 0 ∞ [ 1 2 − S ( x t ) ] y ( t ) d t = f ( x )$

.

Here S(z) is the Fresnel sine integral (see Supplement 11.3-3).

Solution:

$y ( x ) = 2 π ∫ 0 ∞ ( x t ) 1 / 2 [ sin ⁡ ( x t ) x t + 2 cos ⁡ ( x t ) ] f ( t ) d t .$

⊙ References: H. M. Srivastava and R. G. Buschman (1977),A.P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 459).

7. $∫ 0 ∞ [ 1 2 − C ( x t ) ] y ( t ) d t = f ( x )$

.

Here C(z) is the Fresnel cosine integral (see Supplement 11.3-3).

Solution:

$y ( x ) = 2 π ∫ 0 ∞ ( x t ) 1 / 2 [ cos ⁡ ( x t ) − 1 x t − 2 sin ⁡ ( x t ) ] f ( t ) d t .$

⊙ References: H. M. Srivastava and R. G. Buschman (1977),A.P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 460).

8. $∫ 0 ∞ S ( x t , ν ) y ( t ) d t = f ( x )$

.

Here S(z,v) is the generalized Fresnel sine integral (see Supplement 11.3-3).

Solution:

$y ( x ) = 2 π ∫ 0 ∞ ( 1 − v ) sin ⁡ ( x t ) + x t cos ⁡ ( x t ) ( x t ) v f ( t ) d t .$

⊙ References: H. M. Srivastava and R. G. Buschman (1977),A.P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 461).

9. $∫ 0 ∞ C ( x t , ν ) y ( t ) d t = f ( x )$

.

Here C(z, v) is the generalized Fresnel cosine integral (see Supplement 11.3-3).

Solution:

$y ( x ) = 2 π ∫ 0 ∞ ( 1 − v ) cos ⁡ ( x t ) − x t sin ⁡ ( x t ) ( x t ) v f ( t ) d t .$

⊙ References: H. M. Srivastava and R. G. Buschman (1977),A.P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 461).

#### 3.7-3  Kernels Containing Gamma Functions.

10. $∫ 0 ∞ ( x t ) − ( π + 1 ) / 2 Γ ( ± i ln ⁡ ( x t ) ) y ( t ) d t = f ( x ) , i 2 = − 1$

.

Here Γ( z ) is the incomplete gamma function (see Supplement 11.4-1).

Solution:

$y ( x ) = 1 4 π 2 ∫ 0 ∞ ( x t ) − ( π + 1 ) / 2 Γ ( ∓ i ln ⁡ ( x t ) ) f ( t ) d t .$

The integral equation and its solution form a Paley–Wiener transform pair (in the asymmetric form).

⊙ References: E. C. Titchmarsh (1986), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 453).

11. $∫ − ∞ ∞ e − π ( x + t ) / 2 Γ ( ± i ( x + t ) ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 1 4 π 2 ∫ − ∞ ∞ e − π ( x + 1 ) / 2 Γ ( ∓ i ( x + t ) ) f ( t ) d t ,$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 453).

12. $∫ − ∞ ∞ Γ ( α + i ( x + t ) ) Γ ( α − i ( x + t ) ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = − α sin ⁡ ( 2 π α ) 2 π 3 ∫ − ∞ ∞ Γ ( − α + i ( x + t ) ) Γ ( − α − i ( x + t ) ) f ( t ) d t ,$

where Re α <0 (2α ≠ −1, −2, …).

⊙ References: J. Wimp (1971), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 453).

#### 3.7-4  Kernels Containing Incomplete Gamma Functions.

13. $∫ − ∞ ∞ ( t − x ) α − 1 γ ( 1 − α , 2 i ( t − x ) ) y ( t ) d t = f ( x ) , i 2 = − 1$

.

Here γ(ν, z) is the incomplete gamma function (see Supplement 11.5-1).

Solution:

$y ( x ) = − 1 4 π 2 ∫ − ∞ ∞ ( t − a ) − α − 1 γ ( 1 + α , 2 i ( t − x ) ) f ( t ) d t .$

where −1/2< Re α ≤ 0.

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 462).

14. $∫ − ∞ ∞ [ exp ⁡ ( 2 x − i 4 π ) t − i x − 1 / 2 + ( b − a ) a i x − 1 / 2 e i a t Γ ( 1 2 i x , i a t ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 1 4 π ∫ − ∞ ∞ [ exp ⁡ ( 2 t + i 4 π ) x i t − 1 / 2 + ( a − b ) b − i t − 1 / 2 e − i b x Γ ( 1 2 + i t , − i b x ) ] f ( t ) cosh ⁡ ( π t ) d t .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 463).

15. $∫ 0 ∞ { t − i x − 1 / 2 sin ⁡ ( 1 + 2 i x 4 π ) + i 2 ( b − a ) a i x − 1 / 2 [ e − i a t Γ ( 1 2 − i x , − i a t ) − e i a t Γ ( 1 2 − i x , i a t ) ] } y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = 1 π ∫ − ∞ ∞ { t i x − 1 / 2 sin ⁡ ( 1 − 2 i x 4 π ) + i 2 ( a − b ) b − i x − 1 / 2 [ e − i b t Γ ( 1 2 + i x , − i b t ) − e i b t Γ ( 1 2 + i x , i b t ) ] } f ( x ) cosh ⁡ ( π , x ) d x ,$

where a, b ∉ (−∞,0) are complex numbers.

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 463).

16. $∫ 0 ∞ { t − i x − 1 / 2 cos ⁡ ( 1 + 2 i x 4 π ) + i 2 ( b − a ) a i x − 1 / 2 [ e − i a t Γ ( 1 2 − i x , − i a t ) + e i a t Γ ( 1 2 − i x , i a t ) ] } y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = 1 π ∫ − ∞ ∞ { t i x − 1 / 2 cos ⁡ ( 1 − 2 i x 4 π ) + i 2 ( a − b ) b − i x − 1 / 2 [ e − i b t Γ ( 1 2 + i x , − i b t ) + e i b t Γ ( 1 2 + i x , i b t ) ] } f ( x ) cosh ⁡ ( π , x ) d x ,$

where a, b ∉ (−∞,0) are complex numbers.

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 463).

#### 3.7-5  Kernels Containing Bessel Functions of the First Kind.

17. $∫ 0 ∞ t J ν ( x t ) y ( t ) d t = f ( x )$

.

Here J ν (z) is the Bessel function of the first kind (see Supplement 11.6-1).

Solution:

$y ( x ) = { ∫ 0 ∞ t J v ( x t ) f ( t ) d t if Re ⁡ v ≥ − 1 or v = − 2 , − 3 , … , ∫ 0 ∞ t [ J v ( x t ) − ∑ k = 0 n − 1 ( − 1 ) ( x t / 2 ) 2 k + v k ! Γ ( v + k + 1 ) ] f ( t ) d t if Re ⁡ v < − 1 or v ≠ − 2 , − 3 , … , w h e r e − n − 1 < Re ⁡ v < − n , n = 1 , 2 , …$

The functions f (x) and y(x) are the Hankel transform pair.

⊙ References: E. C. Titchmarsh (1923), J. L. Griffith (1958), V. A. Ditkin and A. P. Prudnikov (1965), F. Oberhet-tinger (1972), I. Sneddon (1972), H. M. Srivastava and R. G. Buschman (1977), B. Davis (1978), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 468), S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993), I. Sneddon (1995).

18. $∫ a b t J ν ( x t ) y ( t ) d t = f ( x ) , 0 ≤ x < ∞$

.

Solution:

$y ( t ) = { ∫ 0 ∞ x J v ( x t ) f ( x ) d x if a < t < b , 0 if 0 < t < a or t > b ,$

where 0 ≤ ab ≤ ∞ and Re ν < –1.

⊙ References: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 468), I. N. Sneddon (1995).

19. $∫ a b t J 0 ( x t ) y ( t ) d t = 0 , a ≤ x < ∞$

.

Homogeneous integral equation of the first kind.

Solution:

$y ( t ) = ∫ 0 a c cos ⁡ ( x t ) φ ( x ) d x ,$

where ϑ(x) is an arbitrary continuously differentiable function.

⊙ Reference: Ya. S. Uflyand (1977).

20. $∫ a b t J ν ( x t ) y ( t ) d t = 0 , a ≤ x < ∞$

.

Homogeneous integral equation of the first kind, Re ν >−1/2.

Solution:

$y ( t ) = π t 2 ∫ 0 a x J v − 1 / 2 ( x t ) φ ( x ) d x ,$

where ϑ(x) is an arbitrary continuously differentiable function.

⊙ Reference: Ya. S. Uflyand (1977).

21. $∫ a b | J ν ( λ t ) − J ν ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x) = J v (λx), where J v (z) is the Bessel function of the first kind.

22. $∫ 0 ∞ J ν ( λ ( x − t ) ) y ( t ) d t = f ( x )$

.

1½. If | Re ν| < 1 and f(0) = f′(0) = 0 then

$y ( x ) = ∫ 0 a J − v ( λ ( x − t ) ) ( d 2 d t 2 + λ 2 ) f ( t ) d t .$

2°.If ν = n is a positive integer number and f (0) = f’(0) = …= f (n+1)(0) = 0 then

$y ( x ) = 1 λ n ∑ k = 0 [ ( n − 1 ) / 2 ] C n 2 k + 1 ( d d x ) n − 2 k − 1 ( d 2 d x 2 + λ 2 ) k + 1 f ( x ) + 1 λ n ∫ 0 x J 0 ( λ ( x − t ) ) ∑ k = 0 [ n / 2 ] C n 2 k ( d d t ) n − 2 k ( d 2 d x 2 + λ 2 ) k + 1 f ( t ) d t ,$

where [A] stands for the integer part of the number A and $C n k = n ! k ! ( n − k ) !$

are binomial coefficients (0! = 1).

3°.If ν is not an integer, m −1 < Re ν < m (m = 0, 1,2, …), and f (0) = f’(0) = …= f (m+1)(0) = 0 then

$y ( x ) = m − v λ m ∫ 0 x J m − v ( λ ( x − t ) ) x − t ∑ k = 0 [ ( m − 1 ) / 2 ] C n 2 k + 1 ( d d x ) m − 2 k − 1 ( d 2 d x 2 + λ 2 ) k + 1 f ( t ) d t + 1 λ m ∫ 0 x J m − v ( λ ( x − t ) ) ∑ k = 0 [ m / 2 ] C n 2 k ( d d t ) m − 2 k ( d 2 d x 2 + λ 2 ) k + 1 f ( t ) d t .$

⊙ References: H. M. Srivastava and R. G. Buschman (1977),A.P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 470), S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

23. $∫ − ∞ ∞ | x − t | ν J ν ( λ | x − t | ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = − λ cos ⁡ ( v π ) 4 sin ⁡ 2 ( v π ) ∫ − ∞ ∞ sin ⁡ ( t − x ) | t − x | 2 v + 1 d d t [ | t − x | v + 1 J − v − 1 ( λ | t − x | ) f ( t ) ] d t ,$

where 0 < Re ν <1/2.

⊙ References: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 469), S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

24. $∫ 0 ∞ J n / 2 − 1 ( 2 π x t ) G ( x , t ) y ( t ) d t = f ( x ) , G ( x , t ) = 2 π x ( t / x ) n / 2 , n = 1 , 2 …$

.

Solution:

$y ( x ) = ∫ 0 ∞ J n / 2 − 1 ( 2 π x t ) G ( x , t ) f ( t ) d t .$

The functions f (x)and y(t) are the Bochner transform pair.

⊙ Reference: Yu. A. Brychkov and A. P. Prudnikov (1979).

25. $∫ 0 ∞ d d x [ x J ν 2 ( x t ) ] t y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = − 2 π ∫ 0 ∞ t v ( x , t ) Y v ( x t ) f ( x ) d t = π ∫ 0 ∞ t { sin ⁡ ( 2 v π ) [ J − v 2 ( x t ) − Y v 2 ( x t ) ] − 2 cos ⁡ ( 2 v π ) J − v ( x t ) Y − v ( x t ) } f ( x ) d t .$

⊙ References: I. I. Hirschman and D. V. Widder (1955), A.P.Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 474).

26. $∫ 0 ∞ t [ J − μ ( x t ) J − ν ( x t ) ± J μ ( x t ) J ν ( x t ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = π 2 cos ⁡ [ π 2 ( v ± μ ) ] sin ⁡ [ π 2 ( v ± μ ) ] ∫ 0 ∞ t d d t [ t ( J μ ( x t ) J − v ( x t ) ∓ J μ ( x t ) J v ( x t ) ) ] f ( t ) d t ,$

where Re(Ϝ + Î½) < 3/2.

⊙ References: 1.1. Hirschman and D. V. Widder (1955), E. C.Titchmarsh (1986), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 475).

27. $∫ 0 ∞ [ J i x ( t ) + J − i x ( t ) ] y ( t ) d t = f ( x ) , i 2 = − 1$

.

Solution:

$y ( x ) = 1 2 x ∫ 0 ∞ t [ J i t ( x ) + J − i t ( x ) ] sinh ⁡ ( π t ) f ( t ) d t .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 469).

28. $∫ 0 ∞ [ J i t ( t ) + J − i t ( t ) ] y ( t ) d t = f ( x ) , i 2 = − 1$

.

Solution:

$y ( x ) = x 2 sinh ⁡ ( π x ) ∫ 0 ∞ J i t ( t ) + J − i t ( t ) t f ( t ) d t .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 469).

#### 3.7-6  Kernels Containing Bessel Functions of the Second Kind.

29. $∫ 0 ∞ t Y ν ( x t ) y ( t ) d t = f ( x )$

.

Here Y ν (z) is the Bessel function of the second kind (see Supplement 11.6-1).

1 °. If Re ν∣<1 then

$y ( x ) = ∫ 0 ∞ t H v ( x t ) f ( t ) d x ,$

where H ν (x) is the Struve function, which is defined as

$Η v ( x ) = ∑ j = 0 ∞ ( − 1 ) j ( x / 2 ) v + 2 j + 1 Γ ( j + 3 2 ) Γ ( v + j + 3 2 ) .$

The function f (x) and the solution y(x) are the Y ν -transform pair.

2°.If 1 < ∣Reν∣ <3 then

$y ( x ) = ∫ 0 ∞ t [ H v ( x t ) − ( x t ) v − 1 2 v − 1 π Γ ( v + 1 / 2 ) ] f ( t ) d t .$

⊙ References: E. C. Titchmarsh (1948), G. N. Watson (1952), J. L. Griffith (1958), F. Oberhettinger (1972), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 475).

30. $∫ a b | Y ν ( λ x ) − Y ν ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x) = Y ν (λx), where Y ν (z) is the Bessel function of the second kind.

#### 3.7-7  Kernels Containing Combinations of the Bessel Functions.

31. $∫ 0 ∞ [ cos ⁡ ( p π ) J ν ( x t ) + sin ⁡ ( p π ) Y ν ( x t ) ] t y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = ∫ 0 ∞ Φ ( x t ) t f ( x ) d t , Φ ( z ) = ∑ n = 0 ∞ ( − 1 ) n ( z / 2 ) v + 2 p + 2 n Γ ( p + n + 1 ) Γ ( v + p + n + 1 ) .$

The functions f(x) and y(x) are the Hardy transform pair.

⊙ Reference: Yu. A. Brychkov and A. P. Prudnikov (1989).

32. $∫ 0 ∞ t J ν ( x t ) Y ν ( x t ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 2 π ∫ 0 ∞ t d d t [ t J v 2 ( x t ) ] f ( x ) d t ,$

where Re ν >−1/4.

⊙ References: E. C. Titchmarsh (1948), I.I. Hirschman and D. V. Widder (1955), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 476).

33. $∫ a ∞ t [ J ν ( a x ) Y ν ( x t ) − Y ν ( a x ) J ν ( x t ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = ∫ 0 ∞ t [ J v ( a t ) Y v ( x t ) − Y v ( a t ) J v ( x t ) ] J v 2 ( a t ) + Y v 2 ( a t ) f ( x ) d t .$

The function f(x) and the solution y(x) are the Weber transform pair.

⊙ References: G. N. Watson (1952), Yu. A. Brychkov and A. P. Prudnikov (1979, 1989), A. P. Prudnikov, Yu. A.Brychkov, and O. I. Marichev (1992, p. 477).

34. $∫ a ∞ t [ J ν ( a t ) Y ν ( x t ) − Y ν ( a t ) J ν ( x t ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = x J v 2 ( a x ) + Y v 2 ( a x ) ∫ 0 ∞ t [ J v ( a x ) Y v ( x t ) − Y v ( a x ) J v ( x t ) ] f ( x ) d t .$

⊙ References: G. N. Watson (1952), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 477).

35. $∫ − ∞ ∞ e ± π ( x − t ) / 2 H i ( t − x ) ( 1 ) ( a ) y ( t ) d t = f ( x ) , i 2 = − 1$

.

Here $H ν ( 1 ) ( z ) = J ν ( z ) + i Y ν ( z )$

is the Hankel function of the first kind (see Supplement 11.6-5).

Solution:

$y ( x ) = 1 4 ∫ − ∞ ∞ e ± π ( t − x ) / 2 H i ( t − x ) ( 1 ) ( a ) f ( t ) d t ,$

where a >0.

⊙ References: Vu Kim Tuan (1988), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 479).

36. $∫ − ∞ ∞ e ± π ( x − t ) / 2 H i ( t − x ) ( 2 ) ( a ) y ( t ) d t = f ( x )$

.

Here $H ν ( 2 ) ( z ) = J ν ( z ) − i Y ν ( z )$

is the Hankel function of the second kind (see Supplement 11.6-5).

Solution:

$y ( x ) = 1 4 ∫ − ∞ ∞ e ± π ( t − x ) / 2 H i ( t − x ) ( 2 ) ( a ) f ( t ) d t ,$

where a >0.

⊙ References: Vu Kim Tuan (1988), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 479).

#### 3.7-8  Kernels Containing Modified Bessel Functions of the First Kind.

37. $∫ a b | I ν ( λ x ) − I ν ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x) = I v (λx), where I ν (z) is the modified Bessel function of the first kind (see Supplement 11.7-1).

38. $∫ 0 ∞ d d x I i t 2 ( x ) y ( t ) d t = f ( x ) , i 2 = − 1$

.

Solution:

$y ( x ) = 2 i π x ∫ 0 ∞ K i x 2 ( t ) f ( t ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 485).

39. $∫ − ∞ ∞ A i ( t + x ) y ( t ) d t = f ( x )$

.

Here Ai(x) $= 1 3 x [ I − 1 / 3 ( z ) − I 1 / 3 ( z ) ]$

is the Airy function (see Supplement 11.8-1).

Solution:

$y ( x ) = ∫ ∞ ∞ Ai ( x + t ) f ( t ) d t .$

⊙ References: Vu Kim Tuan (1988), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 485).

#### 3.7-9  Kernels Containing Modified Bessel Functions of the Second Kind.

40. $∫ − ∞ ∞ K 0 | ( x − t ) | y ( t ) d t = f ( x )$

.

Here K 0(z) is the modified Bessel function of the second kind (the MacDonald function), see Supplement 11.7-1.

Solution:

$y ( x ) = − 1 π 2 ( d 2 d x 2 − 1 ) ∫ − 0 ∞ K 0 ( | x − t | ) f ( t ) d t .$

⊙ Reference: D. Naylor (1986).

41. $∫ a b K 0 | K ν ( λ x ) − K ν ( λ t ) | y ( t ) d t = f ( x )$

.

This is a special case of equation 3.8.3 with g(x) = K v (λx).

42. $∫ 0 ∞ z t K ν ( z t ) y ( t ) d t = f ( z )$

.

Here K ν (z) is the modified Bessel function of the second kind.

Up to a constant factor, the left-hand side of this equation is the Meijer transform of y(t) (z is treated as a complex variable).

Solution:

$y ( t ) = 1 π i ∫ c − i ∞ c + i ∞ z t I v ( z t ) f ( z ) d z .$

For specific f (z), one may use tables of Meijer integral transforms to calculate the integral.

⊙ Reference: V. A. Ditkin and A. P. Prudnikov (1965).

43. $∫ 0 ∞ K i x ( t ) y ( t ) d t = f ( x ) , i 2 = − 1$

.

Solution:

$y ( x ) = 2 π 2 x ∫ 0 ∞ t sinh ⁡ ( π t ) K i t ( x ) f ( t ) d t .$

The function f (x) and the solution y(x) are the Kontorovich-Lebedev transform pair.

⊙ References: V. A. Ditkin and A. P. Prudnikov (1965), F. Oberhettinger (1972), Yu. A. Brychkov and A. P. Prud-nikov (1989), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 487).

44. $∫ 0 ∞ K i t ( x ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 2 x sinh ⁡ ( π x ) π 2 ∫ 0 ∞ K i x ( t ) t f ( t ) d t .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 487).

45. $∫ 0 ∞ K i t 2 ( x ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 4 x sinh ⁡ ( π x ) π 2 ∫ 0 ∞ d d t { [ I i x ( t ) + I − i x ( t ) ] K i x ( t ) } f ( t ) d t .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 492).

46. $∫ 0 ∞ Re ⁡ K i x + 1 / 2 ( t ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 4 π 2 ∫ 0 ∞ cosh ⁡ ( π t ) Re ⁡ K i t + 1 / 2 ( x ) f ( t ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 488).

47. $∫ 0 ∞ Im ⁡ K i x + 1 / 2 ( t ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 4 π 2 ∫ 0 ∞ cosh ⁡ ( π t ) Im ⁡ K i t + 1 / 2 ( x ) f ( t ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 488).

48. $∫ 0 ∞ Re ⁡ K i x + 1 / 2 ( x ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 4 π 2 cosh ⁡ ( π x ) ∫ 0 ∞ Re ⁡ K i x + 1 / 2 ( t ) f ( t ) d t .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 488).

49. $∫ 0 ∞ Im ⁡ K i t + 1 / 2 ( x ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 4 π 2 cosh ⁡ ( π x ) ∫ 0 ∞ Im ⁡ K i x + 1 / 2 ( t ) f ( t ) d t .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 488).

50. $∫ − ∞ ∞ e π ( x + t ) / 2 K i ( x + t ) ( a ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 1 π 2 ∫ − ∞ ∞ e π ( x + t ) / 2 K i ( x + t ) ( a ) f ( t ) d t ,$

where a >0. The function f ( x ) and the solution y ( x ) are a Crum transform pair (in the asymmetric form).

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 488).

51. $∫ − ∞ ∞ K i ( x + t ) ( ± i a ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 1 π 2 ∫ − ∞ ∞ K i ( x + t ) ( ∓ i a ) f ( t ) d t .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, pp. 488–489).

52. $∫ − ∞ ∞ t − 1 4 ( 2 i x + 1 ) K 1 2 + i x ( 2 i λ t ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = λ π 2 ∫ − ∞ ∞ x 1 4 ( 2 i t − 1 ) K 1 2 − i t ( 2 i λ x ) f ( t ) d t ,$

where λ > 0 and $x = i | x | for x < 0$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 489).

53. $∫ 0 ∞ [ ( a + t ) − 1 4 ( 2 i x + 1 ) K 1 2 + i x ( 2 i λ a + t ) + ( a − t ) − 1 4 ( 2 i x + 1 ) K 1 2 + i x ( 2 i λ a − t ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = λ π 2 ∫ − ∞ ∞ [ ( a + t ) 1 4 ( 2 i x − 1 ) K 1 2 − i x ( − 2 i λ a + t ) + ( a − t ) 1 4 ( 2 i x − 1 ) K 1 2 − i x ( − 2 i λ a − t ) ] f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 489).

54. $∫ 0 ∞ x 1 4 ( 2 i x − 1 ) K 1 2 − i t ( 2 i λ x ) y ( t ) d t = f ( x ) , λ > 0$

.

Solution:

$y ( x ) = λ π 2 ∫ − ∞ ∞ t − 1 4 ( 2 i x + 1 ) K 1 2 + i x ( 2 i λ t ) f ( t ) d t .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 489).

55. $∫ 0 ∞ [ ( a + t ) 1 4 ( 2 i t − 1 ) K 1 2 − i t ( − 2 i λ a + x ) + ( a − t ) 1 4 ( 2 i t − 1 ) K 1 2 − i t ( − 2 i λ a − x ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = λ π 2 ∫ 0 ∞ [ ( a + x ) − 1 4 ( 2 i t + 1 ) K 1 2 + i t ( 2 i λ a + x ) + ( a − x ) − 1 4 ( 2 i x + 1 ) K 1 2 + i t ( 2 i λ a − x ) ] f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 490).

56. $∫ − ∞ ∞ exp ⁡ ( π x 2 s i g n t ) K i x ( | t | ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = 1 π 2 x ∫ − ∞ ∞ t exp ⁡ ( π t 2 sign x ) K i t ( | x | ) f ( t ) d t .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 490).

#### 3.7-10  Kernels Containing a Combination of Bessel and Modified Bessel Functions.

57. $∫ 0 ∞ [ I i x ( t ) + I − i x ( t ) ] K i x ( t ) y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = − 4 π 2 d d x ∫ 0 ∞ t sinh ⁡ ( π t ) K i t 2 ( x ) f ( t ) d t .$

The integral equation and its solution form the Lebedev transform pair.

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 493).

58. $∫ 0 ∞ [ K i t ( a ) I i x ( x ) − I i x ( a ) K i x ( x ) ] y ( t ) d t = f ( x ) , 0 < x < a$

.

Solution:

$y ( t ) = 2 t sinh ⁡ ( π t ) π 2 | I i a ( a ) | 2 ∫ 0 a x − 1 [ K i t ( a ) I i t ( x ) − I i t ( a ) K i t ( x ) ] f ( x ) d x , t > 0.$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 494).

59. $∫ 0 ∞ [ Y 0 ( x t ) − 2 π K 0 ( x ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = ∫ 0 ∞ t [ Y 0 ( x t ) − 2 π K 0 ( x t ) ] f ( t ) d t .$

The integral equation and its solution form the divisor transform pair.

⊙ References: F. Oberhettinger (1973), E. C. Titchmarsh (1986), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 492).

60. $∫ 0 ∞ t [ Y 2 n + 1 ( x t ) ± 2 π K 2 n + 1 ( x t ) ] y ( t ) d t = f ( x ) , n = 1 , 2 , …$

.

Solution:

$y ( x ) = ∫ 0 ∞ t [ Y 2 n + 1 ( x t ) ∓ 2 π K 2 n + 1 ( x t ) ] f ( t ) d t .$

⊙ References: E. C. Titchmarsh (1986), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 493).

61. $∫ 0 ∞ t [ Y 2 n ( x t ) + 2 π K 2 n ( x t ) ] y ( t ) d t = f ( x ) , n = 1 , 2 , …$

.

Solution:

$y ( x ) = ∫ 0 ∞ t [ Y 2 n ( x t ) + 2 π K 2 n ( x t ) ] f ( t ) d t .$

⊙ References: E. C. Titchmarsh (1986), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 493).

#### 3.7-11  Kernels Containing Legendre Functions.

62. $∫ 1 ∞ P − 1 2 + i x ( t ) y ( t ) d t = f ( x ) , 0 ≤ x < ∞$

.

Here P v (x) is the Legendre function of the first kind (see Supplement 11.11-3) and i 2 = −1.

Solution:

$y ( t ) = ∫ 0 ∞ x tanh ⁡ ( π x ) P i x − 1 / 2 ( t ) f ( x ) d x .$

The functions f (x) and y(t) are the Mehler–Fock transform pair.

Remark. The Legendre function of the first kind can be represented in the form

$P − 1 2 + i x ( t ) = 2 π cosh ⁡ ( π x ) ∫ 0 ∞ cos ⁡ ( x s ) d s 2 ( t + cosh ⁡ s ) , 1 ≤ t < ∞ .$

⊙ References: N. N. Lebedev (1965), V. A. Ditkin and A.P.Prudnikov (1965), Yu. A. Brychkov and A. P. Prudnikov (1989), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 512).

63. $∫ 0 ∞ P − 1 2 + i t ( x ) y ( t ) d t = f ( x ) , 1 ≤ x < ∞$

.

Solution:

$y ( t ) = t tanh ⁡ ( π t ) ∫ 1 ∞ P − 1 2 + i t ( x ) f ( x ) d x .$

64. $∫ 0 ∞ [ P − 1 2 + i x ( i t ) ± P − 1 2 + i x ( − i t ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = 1 2 ∫ 0 ∞ sinh ⁡ ( π x ) cosh ⁡ 2 ( π x ) [ P − 1 2 + i x ( − i t ) ± P − 1 2 + i x ( i t ) ] f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 513).

65. $∫ 0 ∞ [ P − 1 2 + i t ( i x ) ± P − 1 2 + i t ( − i x ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = t sinh ⁡ ( π t ) 2 cosh ⁡ 2 ( π t ) ∫ 0 ∞ [ P − 1 2 + i t ( − i x ) ± P − 1 2 + i t ( i x ) ] f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 514).

66. $∫ − ∞ ∞ [ i e − i π x P − 1 2 + x ( cos ⁡ t ) + P − 1 2 + x ( − cos ⁡ t ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = 1 2 sin ⁡ t ∫ − ∞ ∞ x sinh ⁡ ( 2 π x ) [ i e i π x P − 1 2 + x ( cos ⁡ t ) + P − 1 2 + x ( − cos ⁡ t ) ] f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 513).

67. $∫ 0 ∞ [ P − 1 2 + i t ( x ) ] 2 y ( t ) d t = f ( x ) , 1 ≤ x < ∞$

.

Solution:

$y ( t ) = t tanh ⁡ ( π t ) ∫ 1 ∞ P − 1 2 + i t ( x ) [ Q − 1 2 + i t ( x ) + Q − 1 2 − i t ( x ) ] ( x 2 − 1 ) 1 / 2 d d x [ ( x 2 − 1 ) 1 / 2 f ( x ) ] d x ,$

where Q ν (x) is the Legendre function of the second kind.

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 514).

68. $∫ 1 ∞ P − 1 2 + i x ( t ) [ Q − 1 2 + i x ( t ) + Q − 1 2 − i x ( t ) ] y ( t ) d t = f ( x ) , 0 ≤ x < ∞$

.

Here Q ν (x) is the Legendre function of the second kind.

Solution:

$y ( t ) = ( t 2 − 1 ) 1 / 2 d d t [ ( t 2 − 1 ) 1 / 2 ∫ 0 ∞ x tanh ⁡ ( π x ) [ P − 1 2 + i x ( t ) ] 2 f ( x ) d x ] .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 519).

#### 3.7-12  Kernels Containing Associated Legendre Functions.

69. $∫ 1 ∞ P − 1 2 + i x μ ( t ) y ( t ) d t = f ( x ) , 0 ≤ x < ∞$

.

Here $P v μ ( x )$

is the associated Legendre function of the first kind (see Supplement 11.11-3) and i 2 = −1.

Solution:

$y ( t ) = 1 π ∫ 0 ∞ x sinh ⁡ ( π x ) Γ ( 1 2 − μ + i x ) Γ ( 1 2 − μ − i x ) P i x − 1 / 2 μ ( t ) f ( x ) d x .$

The functions f (x) and y(t) are the generalized MehlerFock transform pair.

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 518).

70. $∫ 1 ∞ P − 1 2 + i x μ ( x ) y ( t ) d t = f ( x ) , 1 ≤ x < ∞$

.

Solution:

$y ( t ) = 1 π t sinh ⁡ ( π t ) Γ ( 1 2 − μ + i t ) Γ ( 1 2 − μ − i t ) ∫ 1 ∞ P i t − 1 / 2 μ ( x ) f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 519).

71. $∫ − 1 ∞ P − 1 2 + i a i x ( ± t ) y ( t ) d t = f ( x ) , − ∞ < x < ∞$

.

Solution:

$y ( t ) = 1 2 π i ( 1 − t ) ∫ − ∞ ∞ x Γ ( 1 2 + i a − i x ) Γ ( 1 2 − i a − i x ) P − 1 2 + i a i x ( ∓ t ) f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 518).

72. $∫ 0 ∞ [ ( x + t − 1 ) 2 − 4 x t ] − 1 / 2 Q v − 1 2 1 ( x + t − 1 2 x t ) y ( t ) d t = f ( x ) , Re v > − 1$

.

Here $Q v μ ( x )$

is the associated Legendre function of the second kind (see Supplement 11.11-3).

Solution:

$y ( t ) = 1 4 π 2 ∫ 0 ∞ ( x t ) − 1 / 2 [ ( x + t − 1 ) 2 − 4 x t ] − 1 / 2 Q ν − 1 2 1 ( x + t − 1 2 x t ) f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 520).

#### 3.7-13  Kernels Containing Kummer Confluent Hypergeometric Functions.

73. $∫ 0 ∞ F ( a , b ; i x t ) y ( t ) d t = f ( x )$

.

Here F (a, b; x) is the Kummer confluent hypergeometric function (see Supplement 11.9-1) and i 2 = −1.

Let Re(b – a)< n <Re b −1/2. Then the solution is

$y ( t ) = Γ ( a ) 2 π Γ ( b ) t b − 1 ( d d t ) n [ t n − b + 1 ∫ − ∞ ∞ e − i x t Ψ ( n + a − b , n − b + 2 ; i x t ) f ( x ) d x ] .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 530).

74. $∫ 0 ∞ F ( 1 2 b ± i x , b ; − i t ) y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = t b − 1 2 π Γ 2 ( b ) ∫ − ∞ ∞ e ∓ π x Γ ( 1 2 b + i x ) Γ ( 1 2 b − i x ) F ( 1 2 b ∓ i x , b ; i t ) f ( x ) d x ,$

where Re b >0.

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 531).

75. $∫ 0 ∞ t i x F ( 1 2 + i x , b + i x ; i α t ) y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = t b − 1 2 π ( − d d t ) n ∫ − ∞ ∞ t n − b − i x e − i α t Γ ( 1 2 + i x ) Γ ( b + i x ) Ψ ( n − b + 1 2 , n − b + 1 − i x ; i α t ) f ( x ) d x ,$

where Im α = 0 and 0<Re b −1/2< n <Re b.

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 531).

76. $∫ 0 ∞ F ( a , b ; i β ( x − t ) ) y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = β 2 ( a − 1 ) ( a − b + 1 ) sin ⁡ ( π b ) 4 π ( b − 1 ) ( b − 2 ) ( b − 3 ) sin ⁡ ( π a ) sin ⁡ [ π ( b − a ) ] ∫ − ∞ ∞ F ( ( 2 − a , b − a − 1 ; i β ( x − t ) ) f ( x ) d x ,$

where 1 < Re a < 3/2 and −1 < Re(b − a) < −1/2.

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 531).

77. $∫ 0 ∞ F ( 1 2 ± i a , 1 2 ; ± i ( x − t ) 2 ) y ( t ) d t = f ( x ) , a > 0$

.

Solution:

$y ( t ) = e π a π cosh ⁡ ( π a ) ∫ − ∞ ∞ F ( 1 2 ∓ i a , 1 2 ; ∓ i ( x − t ) 2 ) f ( x ) d x .$

⊙ References: Vu Kim Tuan (1988), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 532).

78. $∫ − ∞ ∞ F ( 1 2 b ± i t , b ; i x ) y ( t ) d t = f ( x ) , Re b > 0$

.

Solution:

$y ( t ) = e ± π t 2 π Γ 2 ( b ) Γ ( 1 2 b + i t ) Γ ( 1 2 b − i t ) ∫ 0 ∞ x b − 1 F ( 1 2 b ∓ i t , b ; − i x ) f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 532).

79. $∫ − ∞ ∞ F ( 1 2 b ± i t , b ; − i x ) y ( t ) d t = f ( x ) , Re b > 0$

.

Solution:

$y ( t ) = e ∓ π t 2 π Γ 2 ( b ) Γ ( 1 2 b + i t ) Γ ( 1 2 b − i t ) ∫ 0 ∞ x b − 1 F ( 1 2 b ∓ i t , b ; i x ) f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 532).

80. $∫ − ∞ ∞ x − i t F ( 1 2 − i t , b − i t ; i β x ) y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = Γ ( 1 2 ( 1 − i t ) ) 2 π Γ ( b − 1 2 i t ) ∫ 0 ∞ x n − b + i t e − i β x Ψ ( n + 1 2 − b , n + 1 − b + i t ; i β x ) ( d d x ) n [ x b − 1 f ( x ) ] d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 533).

#### 3.7-14  Kernels Containing Tricomi Confluent Hypergeometric Functions.

81. $∫ 0 ∞ t i x Ψ ( a + i x , 2 i x + 1 ; t ) y ( t ) d t = f ( x )$

.

Here Ψ(a, b; x) is the Tricomi confluent hypergeometric function (see Supplement 11.9-1) and i 2 = −1.

Solution:

$y ( t ) = e − t π 2 t ∫ 0 ∞ x sinh ⁡ ( 2 π x ) Γ ( a − i x ) Γ ( a + i x ) t i x Ψ ( a + i x , 2 i x + 1 ; t ) f ( x ) d x .$

⊙ References: J. Wimp (1971), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 534).

82 . $∫ 0 ∞ x i x Ψ ( a + i x , 2 i x + 1 ; t ) y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = t π 2 sinh ⁡ ( 2 π t ) Γ ( a − i t ) Γ ( a + i t ) ∫ 0 ∞ x − 1 + i t e − x Ψ ( a + i t , 2 i t + 1 ; x ) f ( x ) d x .$

⊙ References: J. Wimp (1971), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 535).

83. $∫ 0 ∞ Ψ ( 1 2 + i x , 3 2 − i β + i x ; ± i t ) y ( t ) d t = f ( x ) Im ⁡ β = 0$

.

Solution:

$y ( t ) = 1 4 π ∫ − ∞ ∞ 1 cosh ⁡ ( π x ) Ψ ( 1 2 − i x , 3 2 + i β − i x ; ∓ i t ) f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 536).

#### 3.7-15  Kernels Containing Whittaker Confluent Hypergeometric Functions.

84. $∫ 0 ∞ M ± i x , v ( i t ) y ( t ) d t = f ( x ) , Re ⁡ v > − 1 2$

.

Here M μ,v (z) is the Whittaker confluent hypergeometric function (see Supplement 11.9-3) and i 2 = −1.

Solution:

$y ( t ) = 1 2 π Γ 2 ( 2 ν + 1 ) t ∫ − ∞ ∞ e ∓ π x Γ ( 1 2 + ν + i x ) Γ ( 1 2 + ν − i x ) M ± i x , ν ( − i t ) f ( x ) d x .$

The integral equation and its solution form the Buchholz transform pair.

⊙ References: H. Buchholz (1969), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 523).

85. $∫ 0 ∞ M ± i x , v ( − i t ) y ( t ) d t = f ( x ) , Re ⁡ v > − 1 2$

.

Solution:

$y ( t ) = 1 2 π Γ 2 ( 2 ν + 1 ) t ∫ − ∞ ∞ e ± π x Γ ( 1 2 + ν + i x ) Γ ( 1 2 + ν − i x ) M ∓ i x , ν ( i t ) f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, pp. 523–524).

86. $∫ − ∞ ∞ M ± i x , v ( i x ) y ( t ) d t = f ( x ) , Re ⁡ v > − 1 2$

.

Solution:

$y ( t ) = e ∓ π t 2 π Γ 2 ( 2 ν + 1 ) Γ ( 1 2 + ν + i t ) Γ ( 1 2 + ν − i t ) ∫ 0 ∞ x − 1 M ∓ i t , ν ( − i x ) f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 524).

87. $∫ − ∞ ∞ M ± i x , v ( − i x ) y ( t ) d t = f ( x ) , Re ⁡ v > − 1 2$

.

Solution:

$y ( t ) = e ± π t 2 π Γ 2 ( 2 ν + 1 ) Γ ( 1 2 + ν + i t ) Γ ( 1 2 + ν − i t ) ∫ 0 ∞ x − 1 M ∓ i t , ν ( i x ) f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, pp. 524–525).

88. $∫ − ∞ ∞ Γ ( 1 2 + v + i x − i t ) Γ ( 1 2 + v − i x + i t ) M i t − i x , v ( a ) y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = ( 2 ν + 1 ) sin ⁡ ( 2 π ν ) 4 π 3 ∫ − ∞ ∞ Γ ( − 1 2 − ν + i x − i t ) Γ ( − 1 2 − ν − i x + i t ) M i t − i x , − ν − 1 ( a ) f ( x ) d x .$

⊙ References: J. Wimp (1971), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 526).

89. $∫ 0 ∞ W μ , i x ( t ) y ( t ) d t = f ( x )$

.

Here W μ,ν (z) is the Whittaker confluent hypergeometric function (see Supplement 11.9-3).

Solution:

$y ( t ) = 1 π 2 t 2 ∫ 0 ∞ x sinh ⁡ ( 2 π x ) Γ ( 1 2 − μ − i x ) Γ ( 1 2 − μ + i x ) W μ , i x ( t ) f ( x ) d x .$

⊙ References: J. Wimp (1971), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 527).

90. $∫ 0 ∞ W μ , i t ( x ) y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = t π 2 sinh ⁡ ( 2 π t ) Γ ( 1 2 − μ − i t ) Γ ( 1 2 − μ + i t ) ∫ 0 ∞ x − 2 W μ , i t ( x ) f ( x ) d x .$

⊙ References: J. Wimp (1971), A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 527).

91. $∫ − ∞ ∞ e − i x t / 2 W μ , v ( i x t ) y ( t ) d t = f ( x )$

.

Solution:

$y ( t ) = Γ ( 3 2 − μ − ν ) 2 π Γ ( 1 + n − 2 ν ) ( i t ) − n / 2 − 1 × ∫ 0 ∞ x ( n − 1 ) / 2 − ν e i x t / 2 W μ + n / 2 − 1 , n / 2 − ν ( i x t ) ( d d x ) n [ x ν − 1 / 2 f ( x ) ] d x ,$

where Re μ <Re ν +1/2< 3/4.

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 528).

#### 3.7-16  Kernels Containing Gauss Hypergeometric Functions.

92. $∫ 0 a F ( β 2 , β + 1 2 , μ ; 4 x 2 t 2 ( x 2 + t 2 ) ) y ( t ) d t ( x 2 + t 2 ) β = f ( x )$

.

Here 0< a ≤ ∞,0 < β < μ < β +1, and F (a, b, c; z) is the Gauss hypergeometric function (see Supplement11.10-1).

1.°Solution:

$y ( x ) = x 2 μ − 2 Γ ( 1 + β − μ ) d d x ∫ x a t g ( t ) d t ( t 2 − x 2 ) μ − β , g ( t ) = 2 Γ ( β ) sin ⁡ [ ( β − μ ) π ] π Γ ( μ ) t 1 − 2 β d d t ∫ 0 t s 2 μ − 1 f ( s ) d s ( t 2 − s 2 ) μ − β .$

2°. If a = ∞ and f (x) is a differentiable function, then the solution can be represented in the form

$y ( x ) = A d d t ∫ 0 ∞ ( x t ) 2 μ f ′ t ( t ) ( x 2 + t 2 ) 2 μ − β F ( μ − β 2 , μ + 1 − β 2 , μ + 1 ; 4 x 2 t 2 ( x 2 + t 2 ) 2 ) d t ,$

where $A = Γ ( β ) Γ ( 2 μ − β ) sin ⁡ [ ( β − μ ) π ] π Γ ( μ ) Γ ( 1 + μ )$

.

⊙ Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975).

93. $∫ 0 ∞ F ( a + i x , a − i x , c ; − t ) y ( t ) d t = f ( x ) , a , c > 0$

.

Solution:

$y ( t ) = t c − 1 ( 1 + t ) 2 a − c π 2 Γ 2 ( c ) ∫ 0 ∞ x sinh ⁡ ( 2 π x ) | Γ ( a + i x ) Γ ( c − a + i x ) | 2 F ( a + i x , a − i x , c ; − t ) f ( x ) d x .$

The integral equation and its solution form the Olevskii transform pair.

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 538).

#### 3.7-17  Kernels Containing Parabolic Cylinder Functions.

94. $∫ − ∞ ∞ D − i x − 1 / 2 ( ± e − π i / 4 t ) y ( t ) d t = f ( x ) , i 2 = − 1$

.

Here D v (z) is the parabolic cylinder function (see Supplement 11.12-1).

Solution:

$y ( x ) = 1 4 π ∫ − ∞ ∞ e − π t / 2 cosh ⁡ ( π t ) D i t − 1 / 2 ( ± e π i / 4 x ) f ( t ) d t .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 467).

95. $∫ − ∞ ∞ exp ⁡ [ ± i ( x − t ) 2 4 ] [ D ± i α ( e ∓ π i / 4 ( t − x ) ) − D ± i α ( e ∓ π i / 4 ( x − t ) ) ] y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = e π α / 2 8 π cosh ⁡ 2 ( π α / 2 ) ∫ − ∞ ∞ exp ⁡ [ ∓ i ( x − t ) 2 4 ] [ D ∓ i α ( e ± π i / 4 ( t − x ) ) + D ∓ i α ( e ± π i / 4 ( x − t ) ) ] f ( t ) d t ,$

where α >0.

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 466).

96. $∫ 0 ∞ { exp ⁡ [ i ( x + t ) 2 4 ] [ D 2 i α ( e 3 π i / 4 ( x + t ) ) − D 2 i α ( e − π i / 4 ( x + t ) ) ] + exp ⁡ [ i ( x − t ) 2 4 ] [ D 2 i α ( e 3 π i / 4 ( x + t ) ) − D 2 i α ( e − π i / 4 ( x − t ) ) ] } y ( t ) d t = f ( x )$

.

Solution:

$y ( x ) = e π α 8 π sinh ⁡ 2 ( π α ) ∫ 0 ∞ { exp ⁡ [ − i ( x + t ) 2 4 ] [ D − 2 i α ( − e π i / 4 ( x + t ) ) − D − 2 i α ( e π i / 4 ( x + t ) ) ] + exp ⁡ [ − i ( t − x ) 2 4 ] [ D − 2 i α ( − e π i / 4 ( t − x ) ) − D − 2 i α ( e π i / 4 ( t − x ) ) ] } f ( t ) d t .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, pp. 465–466).

#### 3.7-18  Kernels Containing Other Special Functions.

97. $∫ 0 a K ( 2 x t x + t ) y ( t ) d t x + t = f ( x )$

.

Here $K ( z ) = ∫ 0 1 d t ( 1 − t 2 ) ( 1 − z 2 t 2 )$

is the complete elliptic integral of the first kind (see Supplement 11.13-1).

Solution:

$y ( x ) = − 4 π 2 x ∫ x a t F ( t ) d t t 2 − x 2 , F ( t ) = t ∫ 0 t s f ( s ) d s t 2 − s 2 .$

⊙ Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975).

98. $∫ 0 ∞ [ ζ ( 1 2 + i x , i t ) − ζ ( 1 2 + i x , 1 2 + i t ) ] y ( t ) d t = f ( x )$

.

Here $ζ ( z , υ ) = ∑ k = 0 ∞ 1 ( υ + k ) z$

is the generalized Riemann zeta function (Re z > 1; ν

Solution:

$y ( t ) = e π i / 4 4 π t ∫ − ∞ ∞ e π x / 2 cosh ⁡ ( π x ) [ 1 + ( 1 + i 2 t ) i x − 1 / 2 ] t i x f ( x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 454).

99. $∫ 0 ∞ { t − i x − 1 / 2 sin ⁡ ( 1 + 2 i x ) π 4 + 2 − i x − 3 / 2 [ ζ ( 1 2 + i x , 1 − i t 2 ) − ζ ( 1 2 + i x , 1 + i t 2 ) − ζ ( 1 2 + i x , − i t 2 ) + ζ ( 1 2 + i x , i t 2 ) ] } y ( t ) d t = f ( x ) .$

.

Here ζ(z, v) is the generalized Riemann zeta function (see Eq. 3.7.98).

Solution:

$y ( t ) = 1 π ∫ − ∞ ∞ { t i x − 1 / 2 sin ⁡ ( 1 − 2 i x ) π 4 + sin ⁡ [ ( 1 2 − i x ) arctan ⁡ t ] ( t 2 + 1 ) i x / 2 − 1 / 4 } f ( x ) cosh ⁡ ( π x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 454).

100. $∫ 0 ∞ { t − i x − 1 / 2 cos ⁡ ( 1 + 2 i x ) π 4 − 2 − i x − 3 / 2 e π x [ ζ ( 1 2 + i x , 1 − i t 2 ) + ζ ( 1 2 + i x , 1 + i t 2 ) − ζ ( 1 2 + i x , − i t 2 ) − ζ ( 1 2 + i x , i t 2 ) ] } y ( t ) d t = f ( x ) .$

.

Here ζ(z, v) is the generalized Riemann zeta function (see Eq. 3.7.98).

Solution:

$y ( t ) = 1 π ∫ − ∞ ∞ { t i x − 1 / 2 cos ⁡ ( 1 − 2 i x ) π 4 + cos ⁡ [ ( 1 2 − i x ) arctan ⁡ t ] ( t 2 + 1 ) i x / 2 − 1 / 4 } f ( x ) cosh ⁡ ( π x ) d x .$

⊙ Reference: A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992, p. 455).

#### 3.8-1  Equations with Degenerate Kernel.

1. $∫ a b [ g 1 ( x ) h 1 ( t ) + g 2 ( x ) h 2 ( t ) ] y ( t ) d t = f ( x )$

.

This integral equation has solutions only if its right-hand side is representable in the form

1 $f ( x ) = A 1 g 1 ( x ) + A 2 g 2 ( x ) , A 1 = c o n s t , A 2 = c o n s t .$

In this case, any function y = y(x) satisfying the normalization type conditions

2 $∫ a b h 1 ( t ) y ( t ) d t = A 1 , ∫ a b h 2 ( t ) y ( t ) d t = A 2$

is a solution of the integral equation. Otherwise, the equation has no solutions.

2. $∫ a b [ ∑ a n g k ( x ) h k ( t ) ] y ( t ) d t = f ( x )$

.

This integral equation has solutions only if its right-hand side is representable in the form

1 $f ( x ) = ∑ k = 0 n A k g k ( x ) ,$

where the A k are some constants. In this case, any function y = y(x) satisfying the normalization type conditions

2 $∫ a b h k ( t ) y ( t ) d t = A k ( k = 1 , … , n )$

is a solution of the integral equation. Otherwise, the equation has no solutions.

#### 3.8-2  Equations Containing Modulus

3. $∫ a b | g ( x ) − g ( t ) | y ( t ) d t = f ( x )$

Let ax < b and atb; it is assumed in items 1° and 2° that 0 < g’ x (x)< ∞

1 °. Let us remove the modulus in the integrand:

1 $∫ a x [ g ( x ) − g ( t ) ] y ( t ) d t + ∫ x b [ g ( t ) − g ( x ) ] y ( t ) d t = f ( x ) .$

Differentiating (1) with respect to x yields

2 $g ′ x ( x ) ∫ a x y ( t ) d t − g ′ x ( x ) ∫ x b y ( t ) d t = f ′ x ( x ) .$

Divide both sides of (2) by g x ’ (x) and differentiate the resulting equation to obtain the solution

3 $y ( x ) = 1 2 d d x [ f ′ x ( x ) g ′ x ( x ) ] .$

2°.Letusdemonstrate that the right-hand side f(x) of the integral equation must satisfy certain relations. By setting x = a and x = b,in(1), we obtain two corollaries

4 $∫ a b [ g ( t ) − g ( a ) ] y ( t ) d t = f ( a ) , ∫ a b [ g ( b ) − g ( t ) ] y ( t ) d t = f ( b ) .$

Substitute y(x) of (3) into(4). Integrating by parts yields the desired constraints for f(x):

5 $[ g ( b ) − g ( a ) ] f ′ x ( b ) g ′ x ( b ) = f ( a ) + f ( b ) , [ g ( a ) − g ( b ) ] f ′ x ( a ) g ′ x ( a ) = f ( a ) + f ( b ) .$

Let us point out a useful property of these constraints: $f ′ x ( b ) g ′ x ( a ) + f ′ x ( a ) g ′ x ( b ) = 0$

.

Conditions (5) make it possible to find the admissible general form of the right-hand side of the integral equation:

6 $f ( x ) = F ( x ) + A x + B ,$

where F(x) is an arbitrary bounded twice differentiable function (with bounded first derivative), and the coefficients A and B are given by

$A = g ′ x ( a ) F ′ x ( a ) g ′ x ( a ) + g ′ x ( b ) , B = − 1 2 A ( a + b ) − 1 2 [ F ( a ) + F ( b ) ] − g ( b ) − g ( a ) 2 g ′ x ( a ) [ A + F ′ x ( a ) ] .$

3°.If g(x) is representable in the form g(x) = O(xa) k with 0 < k <1 in the vicinity of the point x = a (in particular, the derivative g’ x is unbounded as x → a), then the solution of the integral equation is given by formula (3) as well. In this case, the right-hand side of the integral equation must satisfy the conditions

7 $f ( a ) + ( b ) = 0 , f ′ x ( b ) = 0.$

As before, the right-hand side of the integral equation is given by (6), with

$A = - F ′ x ( b ) , B = 1 2 [ ( a + b ) F ′ x ( b ) − F ( b ) − F ( b ) ] .$

4°.For g x (a) = 0, the right-hand side of the integral equation must satisfy the conditions

$f ′ x ( a ) = 0 , [ g ( b ) - g ( a ) ] f ′ x ( b ) = [ f ( a ) + f ( b ) ] g ′ x ( b ) .$

As before, the right-hand side of the integral equation is given by (6), with

$A = − F ′ x ( a ) , B = 1 2 [ ( a + b ) F ′ x ( a ) − F ( a ) − F ( b ) ] + g ( b ) − g ( a ) 2 g ′ x ( b ) [ F ′ x ( b ) − F ′ x ( a ) ] .$

4. $∫ 0 a | g ( x ) − g ( λ t ) | y ( t ) d t = f ( x ) , λ > 0$

.

Assume that 0 ≤ xa,0t ≤ a,and 0 < g xX (x)<∞.

1°.Let us remove the modulus in the integrand:

1 $∫ 0 x / λ [ g ( x ) − g ( λ t ) ] y ( t ) d t + ∫ x / λ a [ g ( λ t ) − g ( x ) ] y ( t ) d t = f ( x ) .$

Differentiating (1) with respect to x yields

2 $g ′ x ( x ) ∫ 0 x / λ y ( t ) d t − g ′ x ( x ) ∫ x / λ a y ( t ) d t = f ′ x ( x ) .$

Let us divide both sides of (2) by g’ x (x) and differentiate the resulting equation to obtain $y ( x / λ ) = 1 2 λ [ f ′ x ( x ) / g ′ x ( x ) ] x$

Substituting x by λx yields the solution

3 $y ( x ) = λ 2 d d z [ f ′ z ( z ) g ′ z ( z ) ] , z = λ x .$

2°. Let us demonstrate that the right-hand side f(x) of the integral equation must satisfy certain relations. By setting x =0 in(1)and(2), we obtain two corollaries

4 $∫ 0 a [ g ( λ t ) − g ( 0 ) ] y ( t ) d t = f ( 0 ) , g ′ x ( 0 ) ∫ 0 a y ( t ) d t = − f ′ x ( 0 ) .$

Substitute y(x) of (3) into (4). Integrating by parts yields the desired constraints for f(x):

5 $f ′ x ( 0 ) g ′ x ( λ a ) + f ′ x ( λ a ) g ′ x ( 0 ) = 0 , [ g ( λ a ) − g ( 0 ) ] f ′ x ( λ a ) g ′ x ( λ a ) = f ( 0 ) + f ( λ a ) .$

Conditions (5) make it possible to find the admissible general form of the right-hand side of the integral equation:

6 $f ( x ) = F ( x ) + A x + B ,$

where F(x) is an arbitrary bounded twice differentiable function (with bounded first derivative), and the coefficients A and B are given by

$A = − g ′ x ( 0 ) F ′ x ( λ a ) + g ′ x ( λ a ) F ′ x ( 0 ) g ′ x ( 0 ) + g ′ x ( λ a ) , B = − 1 2 A a λ − 1 2 [ F ( 0 ) + F ( λ a ) ] − g ( λ a ) − g ( 0 ) 2 g ′ x ( 0 ) [ A + F ′ x ( 0 ) ] .$

3°.If g(x) is representable in the form g(x) = O(x) k with 0 < k <1 in the vicinity of the point x =0 (in particular, the derivative g x’ is unbounded as x → 0), then the solution of the integral equation is given by formula (3) as well. In this case, the right-hand side of the integral equation must satisfy the conditions

7 $f ( 0 ) + f ( λ a ) = 0 , f ′ x ( λ a ) = 0.$

As before, the right-hand side of the integral equation is given by (6), with

$A = − F ′ x ( λ a ) , B = 1 2 [ a λ F ′ x ( λ a ) − F ( 0 ) − F ( λ a ) ] .$

5. $∫ 0 a | g ( x ) − t | y ( t ) d t = f ( x )$

.

Assume that 0 ≤ xa,0≤ ta; g(0) = 0, and 0 < g x ′(x)<∞.

1 °. Let us remove the modulus in the integrand:

1 $∫ 0 g ( x ) [ g ( x ) − t ] y ( t ) d t + ∫ g ( x ) a [ t − g ( x ) ] y ( t ) d t = f ( x ) .$

Differentiating (1) with respect to x yields

2 $g ′ x ( x ) ∫ 0 g ( x ) y ( t ) d t − g ′ x ( x ) ∫ g ( x ) a y ( t ) d t = f ′ x ( x ) .$

Let us divide both sides of (2] by g x ′ (x) and differentiate the resulting equation to obtain $2 g ′ x ( x ) y ( g ( x ) ) = [ f ′ x ( x ) / g ′ x ( x ) ] ′$

.Hence, we find the solution:

3 $y ( x ) = 1 2 g ′ z ( z ) d d z [ f ′ z ( z ) g ′ z ( z ) ] , z = g − 1 ( x ) ,$

where g −1 is the inverse of g.

2°. Let us demonstrate that the right-hand side f(x) of the integral equation must satisfy certain relations. By setting x =0 in (1) and (2), we obtain two corollaries

4 $∫ 0 a t y ( t ) d t = f ( 0 ) , g ′ x ( 0 ) ∫ 0 a y ( t ) d t = − f ′ x ( 0 ) .$

Substitute y(x) of (3) into (4). Integrating by parts yields the desired constraints for f(x):

5 $f ′ x ( 0 ) g ′ x ( x a ) + f ′ x ( x a ) g ′ x ( 0 ) = 0 , x a = g − 1 ( a ) ; g ( x a ) f ′ x ( x a ) g ′ x ( x a ) = f ( 0 ) + f ( x a ) .$

Conditions (5) make it possible to find the admissible general form of the right-hand side of the integral equation in question:

6 $f ( x ) = F ( x ) + A x + B ,$

where F(x) is an arbitrary bounded twice differentiable function (with bounded first derivative), and the coefficients A and B are given by

$A = − g ′ x ( 0 ) F ′ x ( x a ) + g ′ x ( x a ) F ′ x ( 0 ) g ′ x ( 0 ) + g ′ x ( x a ) , x a = g − 1 ( a ) , B = − 1 2 A x a − 1 2 [ ( 0 ) + F ( x a ) ] − g ( x a ) 2 g ′ x ( 0 ) [ A + F ′ x ( 0 ) ] .$

3°.If g(x) is representable in the vicinity of the point x = 0 in the form g(x) = O(x)k with 0< k <1 (i.e., the derivative g x ′ is unbounded as x → 0), then the solution of the integral equation is given by formula (3) as well. In this case, the right-hand side of the integral equation must satisfy the conditions

7 $f ( 0 ) + f ( x a ) = 0 , f ′ x ( x a ) = 0.$

As before, the right-hand side of the integral equation is given by (6), with

$A = − F ′ x ( x a ) , B = 1 2 [ x a F ′ x ( x a ) − F ( 0 ) − F ( x a ) ] .$

6. $∫ 0 a | x − g ( t ) | y ( t ) d t = f ( x )$

.

Assume that 0 ≤ x < a,0 ≤ ta; g(0) = 0, and 0 < g x ′(x)< ∞.

1 °. Let us remove the modulus in the integrand:

1 $∫ 0 g − 1 ( x ) [ x − g ( t ) ] y ( t ) d t + ∫ g − 1 ( x ) a [ g ( t ) − x ] y ( t ) d t = f ( x ) ,$

where g −1 is the inverse of g. Differentiating (1) with respect to x yields

2 $∫ 0 g − 1 ( x ) y ( t ) d t − ∫ g − 1 ( x ) a y ( t ) d t = f ′ x ( x ) .$

Differentiating the resulting equation yields 2y g1(x)) = g x ′ (x)f’ xx ′′(x). Hence, we obtain the solution

3 $y ( x ) = 1 2 g ′ z ( z ) f ′ z ( z ) , z = g ( x ) .$

2°.Letusdemonstrate that the right-hand side f(x) of the integral equation must satisfy certain relations. By setting x =0 in (1) and (2), we obtain two corollaries

4 $∫ 0 a g ( t ) y ( t ) d t = f ( 0 ) , ∫ 0 a y ( t ) d t = − f ′ x ( 0 ) .$

Substitute y(x) of (3) into (4). Integrating by parts yields the desired constraints for f(x):

5 $x a f ′ x ( x a ) = f ( 0 ) + f ( x a ) , f ′ x ( 0 ) + f ′ x ( x a ) = 0 , x a = g ( a ) .$

Conditions (5) make it possible to find the admissible general form of the right-hand side of the integral equation:

$f ( x ) = F ( x ) + A x + B , A = − 1 2 [ F ′ x ( 0 ) + F ′ x ( x a ) ] , B = 1 2 [ x a F ′ x ( 0 ) − F ( x a ) − F ( 0 ) ] , x a = g ( a ) ,$

where F(x) is an arbitrary bounded twice differentiable function (with bounded first derivative).

7. $∫ a b y ( t ) | g ( x ) − g ( t ) | k d t = f ( x ) , 0 < k < 1$

.

Let g x ′≠ 0. The transformation

$z = g ( x ) , τ = g ( t ) , w ( τ ) = 1 g ′ t ( t ) y ( t )$

leads to an equation of the form 3.1.31:

$∫ A B w ( τ ) | z − τ | k d τ = F ( z ) , A=g ( a ) , B = g ( b ) ,$

whereF=F(z) is the function which is obtained from z=g(x) and F=f(x) by eliminating x.

8. $∫ 0 b y ( t ) | g ( x ) − h ( t ) | k d t = f ( x ) , 0 < k < 1$

.

Let g(0) = 0, g(1) =1, g x ′ >0; h(0) = 0,h(1) = 1, and h t ′>0.

The transformation

$z = g ( x ) , τ = h ( t ) , w ( τ ) = 1 h ′ t ( t ) y ( t )$

leads to an equation of the form 3.1.30:

$∫ 0 1 w ( τ ) | z − τ | k d τ = F ( z ) ,$

whereF=F(z) is the function which is obtained fromz=g(x) and F=f(x) by eliminating x.

9. $∫ a b y ( t ) ln ⁡ | g ( x ) − g ( t ) | d t = f ( x )$

.

Let g x ′≠ 0. The transformation

$z = g ( x ) , τ = g ( t ) , w ( τ ) = 1 g ′ t ( t ) y ( t )$

$∫ A B ln ⁡ | z − τ | w ( τ ) d τ = F ( z ) , A = g ( a ) , B = g ( b ) ,$

where F = F(z) is the function which is obtained from z = g(x) and F = f(x) by eliminating x.

10. $∫ 0 1 y ( t ) ln ⁡ | g ( x ) − h ( t ) | d t = f ( x )$

.

Let g(0)=0, g(1) =1,g x ′>0;h(0) =0, h(1) =1,and h t ′ >0.

The transformation

$z = g ( x ) , τ = h ( t ) , w ( τ ) = 1 h ′ t ( t ) y ( t )$

leads to an equation of the form 3.4.2:

$∫ 0 1 ln ⁡ | z − τ | w ( τ ) d τ = F ( z ) ,$

where F = F(z) is the function which is obtained from z = g(x) and F = f(x) by eliminating x.

#### 3.8-3  Equations with Difference Kernel: K(x,t) = K(x – t).

11. $∫ − ∞ ∞ K ( x − t ) y ( t ) d t = A x n , n = 0 , 1 , 2 , …$

.

1°.Solution with n =0:

$y ( x ) = A B , B = ∫ − ∞ ∞ K ( x ) d x .$

2°.Solution with n =1:

$y ( x ) = A B x + A C B 2 , B = ∫ − ∞ ∞ K ( x ) d x , C = ∫ − ∞ ∞ x K ( x ) .$

3°.Solution with n ≥ 2:

$y ( x ) = { d n d λ n [ A e λ x B ( λ ) ] } λ = 0 , B ( λ ) = ∫ − ∞ ∞ K ( x ) e − λ x d x .$

12. $∫ − ∞ ∞ K ( x − t ) y ( t ) d t = A e λ x$

.

Solution:

$y ( x ) = A B e λ x , B = ∫ − ∞ ∞ K ( x ) e − λ x d x .$

13. $∫ − ∞ ∞ K ( x − t ) y ( t ) d t = A e λ x e λ x , n = 1 , 2 , …$

.

1°. Solution with n =1:

$y ( x ) = A B x e λ x + A C B 2 e λ x , B = ∫ − ∞ ∞ K ( x ) e − λ x d x , C = ∫ − ∞ ∞ x K ( x ) e − λ x d x .$

2°.Solution with n ≥ 2:

$y ( x ) = d n d λ n [ A e λ x B ( λ ) ] , B ( λ ) = ∫ − ∞ ∞ K ( x ) e − λ x d x .$

14. $∫ − ∞ ∞ K ( x − t ) y ( t ) d t = A cos ⁡ ( λ x ) + B sin ⁡ ( λ x )$

.

Solution:

$y ( x ) = A I c + B I s I c 2 + I s 2 cos ⁡ ( λ x ) + B I c − A I s I c 2 + I s 2 sin ⁡ ( λ x ) , I c = ∫ − ∞ ∞ K ( z ) cos ⁡ ( λ z ) d z , I s = ∫ − ∞ ∞ K ( z ) sin ⁡ ( λ z ) d z .$

15. $∫ − ∞ ∞ K ( x − t ) y ( t ) d t = f ( x )$

.

The Fourier transform is used to solve this equation.

1°.Solution:

$y ( x ) = 1 2 π ∫ − ∞ ∞ f ˜ ( u ) K ˜ ( u ) e i u x d u , f ˜ ( u ) = 1 2 π ∫ − ∞ ∞ f ( x ) e − i u x d x , K ˜ ( u ) = 1 2 π ∫ − ∞ ∞ K ( x ) e − i u x d x .$

The following statement is valid. Let f ( x ) ∊ L 2(−∞, ∞)and K ( x ) G L 1(−∞, ∞). Then for a solution y(x) ∊ L 2(−∞, ∞)of the integral equation to exist, it is necessary and sufficient that $f ˜ ( u ) / K ˜ ( u ) ∈ L 2 ( − ∞ , ∞ )$

.

2°. Let the function P(s) defined by the formula

$1 P ( s ) = ∫ − ∞ ∞ e − s t K ( t ) d t$

be a polynomial of degree n with real roots of the form

$P ( s ) = ( 1 − s a 1 ) ( 1 − s a 2 ) … ( 1 − s a n ) .$

Then the solution of the integral equation is given by

$y ( x ) = P ( D ) f ( x ) , D = d d x .$

⊙ References: I.I. Hirschman and D. V. Widder (1955), V. A. Ditkin and A. P. Prudnikov (1965).

16. $∫ 0 ∞ K ( x − t ) y ( t ) d t = f ( x )$

.

The Wiener–Hopf equation of the first kind. This equation is discussed in Subsection 12.8-1 in detail.

#### 3.8-4  Other Equations of the Form K(x,t)y(t) dt = F(x).

17. $∫ − ∞ ∞ K ( a x − t ) y ( t ) d t = A e λ x$

.

Solution:

$y ( x ) = A B exp ⁡ ( λ a x ) , B = ∫ − ∞ ∞ K ( z ) exp ⁡ ( − λ a z ) d z .$

18. $∫ − ∞ ∞ K ( a x − t ) y ( t ) d t = f ( x )$

.

The substitution z = ax leads to an equation of the form 3.8.15:

$∫ − ∞ ∞ K ( z − t ) y ( t ) d t = f ( z / a ) .$

19. $∫ − ∞ ∞ K ( a x + t ) y ( t ) d t = A e λ x$

.

Solution:

$y ( x ) = A B exp ⁡ ( − λ a x ) , B = ∫ − ∞ ∞ K ( z ) exp ⁡ ( − λ a z ) d z .$

20. $∫ − ∞ ∞ K ( a x + t ) y ( t ) d t = f ( x )$

.

The transformation τ =t, z = ax, y(t) = Y(T) leads to an equation of the form 3.8.15:

$∫ − ∞ ∞ K ( z − τ ) Y ( τ ) d t = f ( z / a ) .$

21. $∫ − ∞ ∞ [ e β t K ( a x + t ) + e μ t M ( a x − t ) ] y ( t ) d t = A e λ x$

.

Solution:

$y ( x ) = A I k ( q ) e p x − I m ( p ) e q x I k ( p ) I k ( q ) − I m ( p ) I m ( q ) , p = − λ a − β , q = λ a − μ ,$

where

$I k ( q ) = ∫ − ∞ ∞ K ( z ) e ( β + q ) z d z , I m ( q ) = ∫ − ∞ ∞ M ( z ) e − ( μ + q ) z d z .$

22. $∫ 0 ∞ g ( x t ) y ( t ) d t = f ( x )$

.

By setting

$x = e z , t = e − τ , y ( t ) = e τ w ( τ ) , g ( ξ ) = G ( ln ⁡ ξ ) , f ( ξ ) = F ( ln ⁡ ξ ) ,$

we arrive at an integral equation with differencekernel of the form 3.8.15:

$∫ − ∞ ∞ G ( z − τ ) w ( τ ) d τ = F ( z ) .$

23. $∫ 0 ∞ g ( x t ) y ( t ) d t = f ( x )$

.

By setting

$x = e z , t = e τ , y ( t ) = e − τ w ( τ ) , g ( ξ ) = G ( ln ⁡ ξ ) , f ( ξ ) = F ( ln ⁡ ξ ) ,$

we arrive at an integral equation with difference kernel of the form 3.8.15:

$∫ − ∞ ∞ G ( z − τ ) w ( τ ) d τ = F ( z ) .$

24. $∫ 0 ∞ g ( x β t λ ) y ( t ) d t = f ( x ) , β > 0 , λ > 0$

.

By setting

$x = e z / β , t = e − τ / λ , y ( t ) = e τ / λ w ( τ ) , g ( ξ ) = G ( ln ⁡ ξ ) , f ( ξ ) = 1 λ F ( β ln ⁡ ξ ) ,$

we arrive at an integral equation with difference kernel of the form 3.8.15:

$∫ − ∞ ∞ G ( z − τ ) w ( τ ) d τ = F ( z ) .$

25. $∫ 0 ∞ g ( x β t λ ) y ( t ) d t = f ( x ) , β > 0 , λ > 0$

.

By setting

$x = e z / β , t = e τ / λ , y ( t ) = e − τ / λ w ( τ ) , g ( ξ ) = G ( ln ⁡ ξ ) , f ( ξ ) = 1 λ F ( β ln ⁡ ξ ) ,$

we arrive at an integral equation with difference kernel of the form 3.8.15:

$∫ − ∞ ∞ G ( z − τ ) w ( τ ) d τ = F ( z ) .$

26. $∫ 0 a [ 1 | x − t | k + φ ( x ) ψ ( t ) ] y ( t ) d t = f ( x ) , 0 < k < 1$

.

The solution can be obtained by the methods described in Subsection 12.6-2; it must be taken into account that the truncated equation, with ϑ(x) = 0, coincides with equation 3.1.30.

27. $∫ 0 ∞ exp ⁡ [ − g ( x ) t 2 ] y ( t ) d t = f ( x )$

.

Assume that g (0) = ∞, g(∞) = 0, and g x ′ <0.

The substitution $z = 1 4 g ( x )$

$1 π z ∫ 0 ∞ exp ⁡ ( − t 2 4 z ) y ( t ) d t = F ( z ) ,$

where the function F(z) is determined by the relations $F = 2 π f ( x ) g ( x )$

and $z = 1 4 g ( x )$ by means of eliminating x.

28. $∫ a b [ ln ⁡ | x − t | + φ ( x ) ψ ( t ) ] y ( t ) d t = f ( x )$

The solution can be obtained by the methods described in Subsection 12.6-2; it must be taken into account that the truncated equation, with ϑ(x) = 0, coincides with equation 3.4.2. See also Example 3 in Subsection 12.6-2.

29. $∫ 0 ∞ [ sin ⁡ ( x t ) + φ ( x ) ψ ( t ) ] y ( t ) d t = f ( x )$

.

The solution can be obtained by the methods described in Subsection 12.6-2; it must be taken into account that the truncated equation, with ϑ(x) = 0, coincides with equation 3.5.8.

Solution:

$y ( t ) = y f ( t ) + A y φ ( t ) ,$

where

$y f ( t ) = 2 π ∫ 0 ∞ sin ⁡ ( x t ) f ( x ) d x , y φ ( t ) = 2 π ∫ 0 ∞ sin ⁡ ( x t ) φ ( x ) d x , A = − ∫ 0 ∞ ψ ( t ) y f ( t ) d t 1 + ∫ 0 ∞ ψ ( t ) y φ ( t ) d t .$

30. $∫ 0 ∞ [ cos ⁡ ( x t ) + φ ( x ) ψ ( t ) ] y ( t ) d t = f ( x )$

.

The solution can be obtained by the methods described in Subsection 12.6-2; it must be taken into account that the truncated equation, with ϑ(x) = 0, coincides with equation 3.5.1.

Solution:

$y ( t ) = y f ( t ) + A y φ ( t ) ,$

where

$y f ( t ) = 2 π ∫ 0 ∞ cos ⁡ ( x t ) f ( x ) d x , y φ ( t ) = 2 π ∫ 0 ∞ cos ⁡ ( x t ) φ ( x ) d x , A = − ∫ 0 ∞ ψ ( t ) y f ( t ) d t 1 + ∫ 0 ∞ ψ ( t ) y φ ( t ) d t .$

31. $∫ 0 ∞ t a − 1 cos ⁡ [ φ ( x ) t a ] y ( t ) d t = f ( x ) , a > 0$

.

Transformation

$z = φ ( x ) , τ = t a , Y ( τ ) = y ( t ) , F ( z ) = a f ( x )$

leads to anequation of the form 3.5.1:

$∫ 0 ∞ cos ⁡ ( z τ ) Y ( τ ) d τ = F ( z ) .$

32. $∫ 0 ∞ t a − 1 sin ⁡ [ φ ( x ) t a ] y ( t ) d t = f ( x ) , a > 0$

.

Transformation

$z = φ ( x ) , τ = t a , Y ( τ ) = y ( t ) , F ( z ) = a f ( x )$

leads to an equation of the form 3.5.8:

$∫ 0 ∞ sin ⁡ ( z τ ) Y ( τ ) d τ = F ( z ) .$

33. $∫ 0 ∞ [ t J v ( x t ) + φ ( x ) ψ ( t ) ] y ( t ) d t = f ( x ) , v > − 1$

.

Here J ν (z) is the Bessel function of the first kind. The solution can be obtained by the methods described in Subsection 12.6-2; it must be taken into account that the truncated equation, with ϑ(x) = 0, coincides with equation 3.7.17.

Solution:

$y ( t ) = y f ( t ) + A y φ ( t ) ,$

where

$y f ( t ) = ∫ 0 ∞ x J ν ( x t ) f ( x ) d x , y φ ( t ) = ∫ 0 ∞ x J ν ( x t ) φ ( x ) d x , A = − ∫ 0 ∞ ψ ( t ) y f ( t ) d t 1 + ∫ 0 ∞ ψ ( t ) y φ ( t ) d t .$

#### 3.8-5  Equations of the Form # ∫ a b K ( x , t ) y ( ⋯ ) d t = F ( x )

34. $∫ a b f ( t ) y ( x t ) d t = A x + B$

.

Solution:

$y ( x ) = A I 1 x + B I 0 , I 0 = ∫ a b f ( t ) d t , I 1 = ∫ a b t f ( t ) d t .$

35. $∫ a b f ( t ) y ( x t ) d t = A x β$

.

Solution:

$y ( x ) = A B x β , B = ∫ a b f ( t ) t β d t .$

36. $∫ a b f ( t ) y ( x t ) d t = A ln ⁡ x + B$

.

Solution:

$y ( x ) = p ln ⁡ x + q ,$

where

$p = A I 0 , q = B I 0 − A I l I 0 2 , I 0 = ∫ a b f ( t ) d t , I l = ∫ a b f ( t ) ln ⁡ t d t .$

37. $∫ a b f ( t ) y ( x t ) d t = A x β ln ⁡ x$

.

Solution:

$y ( x ) = p x β ln ⁡ x + q x β ,$

where

$p = A I 1 , q = − A I 2 I 1 2 , I 1 = ∫ a b f ( t ) t β d t , I 2 = ∫ a b f ( t ) t β ln ⁡ t d t .$

38. $∫ a b f ( t ) y ( x t ) d t = A cos ⁡ ( ln ⁡ x )$

.

Solution:

$y ( x ) = A I c I c 2 + I s 2 cos ⁡ ( ln ⁡ x ) + A I s I c 2 + I s 2 sin ⁡ ( ln ⁡ x ) , I c = ∫ a b f ( t ) cos ⁡ ( ln ⁡ t ) d t , I s = ∫ a b f ( t ) sin ⁡ ( ln ⁡ t ) d t .$

39. $∫ a b f ( t ) y ( x t ) d t = A sin ⁡ ( ln ⁡ x )$

.

Solution:

$y ( x ) = − A I s I c 2 + I s 2 cos ⁡ ( ln ⁡ x ) + A I c I c 2 + I s 2 sin ⁡ ( ln ⁡ x ) , I c = ∫ a b f ( t ) cos ⁡ ( ln ⁡ t ) d t , I s = ∫ a b f ( t ) sin ⁡ ( ln ⁡ t ) d t .$

40. $∫ a b f ( t ) y ( x t ) d t = A x β cos ⁡ ( ln ⁡ x ) + B x β sin ⁡ ( ln ⁡ x )$

.

Solution:

$y ( x ) = p x β cos ⁡ ( ln ⁡ x ) + q x β sin ⁡ ( ln ⁡ x ) ,$

where

$p = A I c − B I s I c 2 + I s 2 , q = A I s + B I c I c 2 + I s 2 , I c = ∫ a b f ( t ) t β cos ⁡ ( ln ⁡ t ) d t , I s = ∫ a b f ( t ) t β sin ⁡ ( ln ⁡ t ) d t .$

41. $∫ a b f ( t ) y ( x − t ) d t = A x + B$

.

Solution:

$y ( x ) = p x + q ,$

where

$p = A I 0 , q = A I 1 I 0 2 + B I 0 , I 0 = ∫ a b f ( t ) d t , I 1 = ∫ a b t f ( t ) d t .$

42. $∫ a b f ( t ) y ( x − t ) d t = A e λ x$

.

Solution:

$y ( x ) = A B e λ x , B = ∫ a b f ( t ) exp ⁡ ( − λ t ) d t .$

43. $∫ a b f ( t ) y ( x − t ) d t = A cos ⁡ ( λ x )$

.

Solution:

$y ( x ) = − A I s I c 2 + I s 2 sin ⁡ ( λ x ) + A I c I c 2 + I s 2 cos ⁡ ( λ x ) , I c = ∫ a b f ( t ) cos ⁡ ( λ t ) d t , I s = ∫ a b f ( t ) sin ⁡ ( λ t ) d t .$

44. $∫ a b f ( t ) y ( x − t ) d t = A sin ⁡ ( λ x )$

.

Solution:

$y ( x ) = − A I c I c 2 + I s 2 sin ⁡ ( λ x ) + A I s I c 2 + I s 2 cos ⁡ ( λ x ) , I c = ∫ a b f ( t ) cos ⁡ ( λ t ) d t , I s = ∫ a b f ( t ) sin ⁡ ( λ t ) d t .$

45. $∫ a b f ( t ) y ( x − t ) d t = e μ x ( A sin ⁡ λ x + B cos ⁡ λ x )$

.

Solution:

$y ( x ) = e μ x ( p sin ⁡ λ x + q cos ⁡ λ x ) ,$

where

$p = A I c − B I s I c 2 + I s 2 , q = A I s + B I c I c 2 + I s 2 , I c = ∫ a b f ( t ) e − μ t cos ⁡ ( λ t ) d t , I s = ∫ a b f ( t ) e − μ t sin ⁡ ( λ t ) d t .$

46. $∫ a b f ( t ) y ( x − t ) d t = g ( x )$

.

1°.For g(x)= $∑ k = 1 n A k exp ⁡ ( λ k x )$

the solution of the equation has the form

$y ( x ) = ∑ k = 1 n A k B k exp ⁡ ( λ k x ) , B k = ∫ a b f ( t ) exp ⁡ ( − λ k t ) d t .$

2°.For a polynomial right-hand side, $g ( x ) = ∑ k = 0 n A k x k$

the solution has the form

$y ( x ) = ∑ k = 0 n B k x k ,$

where the constants B k are found by the method of undetermined coefficients.

3°.For $g ( x ) = e λ x ∑ k = 0 n A k x k$

the solution has the form

$y ( x ) = e λ x ∑ k = 0 n B k x k ,$

where the constants B k are found by the method of undetermined coefficients.

4°.Forg(x) $= ∑ k = 1 n A k cos ⁡ ( λ k x )$

the solution has the form

$y ( x ) = ∑ k = 1 n B k cos ⁡ ( λ k x ) + ∑ k = 1 n C k sin ⁡ ( λ k x ) ,$

where the constants B k and C k are found by the method of undetermined coefficients.

5°.For g(x)= $∑ k = 1 n A k sin ⁡ ( λ k x )$

the solution has the form

$y ( x ) = ∑ k = 1 n B k cos ⁡ ( λ k x ) + ∑ k = 1 n C k sin ⁡ ( λ k x ) ,$

where the constants B k and C k are found by the method of undetermined coefficients.

6°.For g ( x) = cos $( λ x ) ∑ k = 0 n A k x k$

the solution has the form

$y ( x ) = cos ⁡ ( λ x ) ∑ k = 0 n B k x k + sin ⁡ ( λ x ) ∑ k = 0 n C k x k ,$

where the constants B k and C k are found by the method of undetermined coefficients.

7°.For g ( x) = sin $( λ x ) ∑ k = 0 n A k x k$

the solution has the form

$y ( x ) = cos ⁡ ( λ x ) ∑ k = 0 n B k x k + sin ⁡ ( λ x ) ∑ k = 0 n C k x k ,$

where the constants B k and C k are found by the method of undetermined coefficients.

8°.For g ( x) = $e μ x ∑ k = 1 n A k$

cos (λ k x), the solution has the form

$y ( x ) = e μ x ∑ k = 1 n B k cos ⁡ ( λ k x ) + e μ x ∑ k = 1 n C k sin ⁡ ( λ k x ) ,$

where the constants B k and C k are found by the method of undetermined coefficients.

9°.For g(x) = $e μ x ∑ k = 1 n A k$

sin(λ k x), the solution has the form

$y ( x ) = e μ x ∑ k = 1 n B k cos ⁡ ( λ k x ) + e μ x ∑ k = 1 n C k sin ⁡ ( λ k x ) ,$

where the constants B k and C k are found by the method of undetermined coefficients.

10°.For g(x) = cos $( λ x ) ∑ k = 1 n A k exp ⁡ ( μ k x )$

the solution has the form

$y ( x ) = cos ⁡ ( λ x ) ∑ k = 1 n B k exp ⁡ ( μ k x ) + sin ⁡ ( λ x ) ∑ k = 1 n B k exp ⁡ ( μ k x ) ,$

where the constants B k and C k are found by the method of undetermined coefficients.

11°.For g(x) = sin $( λ x ) ∑ k = 1 n A k exp ⁡ ( μ k x )$

the solution has the form

$y ( x ) = cos ⁡ ( λ x ) ∑ k = 1 n B k exp ⁡ ( λ x ) + sin ⁡ ( λ x ) ∑ k = 1 n B k exp ⁡ ( μ k x ) ,$

where the constants B k and C