1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/ Applications 715

On the other hand, if we let x = -x' (x' > 0) in the first integral

of (9.11-43), we obtain

J: f(-x')(ln I - x'I + i7rt(-dx') = 1R f(x')(lnx' + i7l't dx' (9.11-46)

so (9.11-43) can be written as

1R f(x)(lnx + i7l't dx + j f(z)(logzt dz+ 1R f(x)(Inxt dx

"'(-

+ j J(z)(logz)n dz= 27l'if !,~~ f(z)(logzt

r+ k=1

Hence, by taking (9.11-44) and (9.11-45) into account, we obtain, by letting

r ---t 0, then R ---t oo,

1= f(x)(lnx + i7l't dx + 1= f(x)(lnxt dx
m
= 27l'i 2.:::: ~~ 1cz)(log zr (9.11-47)
k=l
provided that the improper integrals converge. However, (9.11-47) is a
recursive formula and the convergence for a certain exponent n follows from
the convergence of the integrals for the exponents :::; n - 1. For instance,
for n = 1 (9.11-47) gives

2 r= f(x)lnxdx+i7l' r= f(x)dx=27l'if Resf(z)logz (9.11-48)

lo lo k=l z=bk

and the convergence of ft f( x) ln x dx follows from that of ft f( x) dx,

this last integral being proper at the lower limit and· convergent at the

upper limit in view of IJ(x)I :::; K/x°' for x 2:: M.

Next, for n = 2 we have

21= f(x)(lnx)^2 dx + 2i7l' 1= f(x)lnx dx

0 0.

r= m

-7!'^2 lo f(x)dx = 27l'iLResf(z)(logz)^2

(^0) k=l
(9.11-49)
and the convergence of f 0 = f(x)(lnx)^2 dx follows from the convergence of
f 0 = f(x)lnxdx and f 0 = f(x)dx, and similarly for higher exponents.