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3.4-2. Kernels Containing Power-Law and Logarithmic Functions
a
9. k ln(1 + λx) – t y(t) dt = f(x).
0
This is a special case of equation 3.8.5 with g(x)= k ln(1 + λx).
a
10. x – k ln(1 + λt) y(t) dt = f(x).
0
This is a special case of equation 3.8.6 with g(x)= k ln(1 + λt).
∞ 1 x + t
11. ln y(t) dt = f(x).
t x – t
0
Solution:
2
x d ∞ df(t) x
y(x)= ln 1 – dt.
2
π dx 0 dt t 2
•
Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975).
∞ ln x – ln t
12. y(t) dt = f(x).
0 x – t
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
1 4n n 2n 2n 2n n d
e
y(x)= lim D x D x D f(x), D = .
4π 2 n→∞ n dx
To calculate the solution approximately, one should restrict oneself to a specific value of n in
this formula instead of taking the limit.
•
Reference: I. I. Hirschman and D. V. Widder (1955).
b
β
β
13. ln |x – t | y(t) dt = f(x), β >0.
a
The transformation
β
β
z = x , τ = t , w(τ)= t 1–β y(t)
leads to Carleman’s equation 3.4.2:
B
β
β
ln |z – τ|w(τ) dτ = F(z), A = a , B = b ,
A
1/β
where F(z)= βf z .
1
β
µ
14. ln |x – t | y(t) dt = f(x), β >0, µ >0.
0
The transformation
µ
β
z = x , τ = t , w(τ)= t 1–µ y(t)
leads to an equation of the form 3.4.2:
1
1/β
ln |z – τ|w(τ) dτ = F(z), F(z)= µf z .
0
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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