1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/ Applications 689

quarter-circles), say"(, and again the existence of

lim j f(z)dz

r-+O

7+

is needed for the method to be successful.

The following examples will make clear the technique described above

in general terms. As to the evaluation of real improper integrals of the

second kind, see type X below.

Type L Jo"'' f(x)dx, where f(x) satisfies the following conditions:

1. f ( x) has an extension f ( z) that is meromorphic in C, or at least on

either Imz 2::: 0 or Imz ::; 0.

2. f(z) has no poles on the real axis.

3. f(-z) = f(z): i.e., f is an even function.

4. f(z) = 0(1/zOI), a > 1, as z ---+ oo.

The last condition implies that there exist constants K > 0 and M > 0

such that

K

lf(z)I::; i;F

for all z such that lzl 2::: M. As a consequence, f(z) must be regular at oo
since lf(z)I::; K/MOI for lzl 2::: M, so it will admit only a finite number of
poles. Also, we have zf(z) ~ 0 as lzl = R---+ oo since for R 2::: M we get

K

lzf(z)I::; ROl-1 ---+ 0 asR-+oo (9.11-1)

From conditions (2) and ( 4) it follows that the given integral converges

absolutely, smce

1

00

lf(x)I dx::; 1= -K dx = --K M^1 -1

M M XOI a -1 OI

Now consider the complex integral fc+ f(z)4z, where c+ = [-R,R] U

r+, with r+: z = Re;^6 , 0::; ()::; ?r, and R large enough so all poles of f(z)

in the upper half-plane be enclosed by c+ (Fig. 9.13). If those poles are

denoted by h (k = 1, 2, ... , m), we have, by the residue theorem,

R m

j f(z) dz= 1R f(x) dx + j f(z) dz= 2?ri L ~'1:9,. f(z)

c+ r+ k=1

(9.11-2)