Page 77 - Handbook Of Integral Equations

1.5-4. Kernels Containing Cotangent

x

82. cot(λx) – cot(λt) y(t) dt = f(x).
a
This is a special case of equation 1.9.2 with g(x) = cot(λx).
1 d 2
Solution: y(x)= – sin (λx)f (x) .
x
λ dx
x

83. A cot(λx)+ B cot(λt) y(t) dt = f(x).
a
For B = –A, see equation 1.5.82. This is a special case of equation 1.9.4 with g(x) = cot(λx).
1 d A x B
Solution: y(x)= tan(λx) A+B tan(λt) A+B f (t) dt .

t
A + B dx a
x

84. A cot(λx)+ B cot(µt)+ C y(t) dt = f(x).
a
This is a special case of equation 1.9.6 with g(x)= A cot(λx) and h(t)= B cot(µt)+ C.
x
2 2
85. cot (λx) – cot (λt) y(t) dt = f(x).
a
2
This is a special case of equation 1.9.2 with g(x) = cot (λx).
3
d sin (λx)f (x)

x
Solution: y(x)= – .
dx 2λ cos(λx)
x

2 2
86. A cot (λx)+ B cot (λt) y(t) dt = f(x).
a
2
For B = –A, see equation 1.5.85. This is a special case of equation 1.9.4 with g(x) = cot (λx).
1 d 2A x 2B

Solution: y(x)= tan(λx) A+B tan(λt) A+B f (t) dt .
t
A + B dx
a
x

2 2
87. A cot (λx)+ B cot (µt)+ C y(t) dt = f(x).
a
2
2
This is a special case of equation 1.9.6 with g(x)= A cot (λx) and h(t)= B cot (µt)+ C.
x

n
88. cot(λx) – cot(λt) y(t) dt = f(x), n =1, 2, ...
a

The right-hand side of the equation is assumed to satisfy the conditions f(a)= f (a)= ··· =
x
f x (n) (a)=0.
(–1) n 2 d n+1
Solution: y(x)= sin (λx) f(x).
n
2
λ n! sin (λx) dx
x

µ
µ
89. (cot x – cot t)y(t) dt = f(x).
a
µ
This is a special case of equation 1.9.2 with g(x) = cot x.
1 d sin µ+1 xf (x)

x
Solution: y(x)= – .
µ dx cos µ–1 x
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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