Page 471 - Handbook Of Integral Equations

8.6-2. Kernel With Logarithmic Singularity

Consider the equation
x
ln(x – t)y(t) dt = f(x), f(0) = 0. (4)
0
Let us apply the Laplace transform to solve this equation. Note that
∞

ν
–px ν Γ(ν +1)
L x = e x dx = ν+1 , ν > –1. (5)
0 p
Let us differentiate relation (5) with respect to ν. We obtain

Γ(ν +1) Γ (ν +1) 1
ν
ν
L x ln x = +ln . (6)
p ν+1 Γ(ν +1) p
For ν = 0, it follows from (6) that
Γ (1)

ν
= –C,
Γ(1)
where C = 0.5772... is the Euler constant. With regard to the last relation, formula (6) becomes
ln p + C

L ln x = – . (7)
p
Applying the Laplace transform to Eq. (4) and taking into account (7), we obtain

ln p + C
˜
– ˜ y(p)= f(p),
p
and hence
˜
pf(p)
˜ y(p)= – . (8)
ln p + C
Now let us express ˜y(p) in the form

2 ˜
p f(p) – f (0) f (0)

x
x
˜ y(p)= – – . (9)
p(ln p + C) p(ln p + C)
Since f(0) = 0, it follows that

2 ˜
L f (x) = p f(p) – f (0). (10)

x
xx
Let us rewrite formula (5) as
ν
x 1
L = (11)
Γ(ν +1) p ν+1
and integrate (11) with respect to ν from 0 to ∞. We obtain
∞ ν ∞
x dν 1

L dν = ν+1 = .
0 Γ(ν +1) 0 p p ln p
Applying the scaling formula for the Laplace transform (see Table 1 in Subsection 7.2-5) we see
that
∞ ν ∞
(x/a) dν 1 1

L dν = ν+1 = = .
0 Γ(ν +1) 0 p p ln ap p (ln p +ln a)
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 453