1550251515-Classical_Complex_Analysis__Gonzalez_

706 Chapter^9

z =a+ Rei^0 , %7r ~ () ~ -%7r, for the case t < 0. When t = 0 it

suffices to integrate along the vertiq.1,l segment.)

Type VI. J 000 xa f(x) dx, where a is real but not an integer, xa = ealnx(x >

0), and the integrand satisfies the following conditions:

1. f(x) has an extension f(z) that is meromorphic in C.


2. For z = lzlei^8 (z =/:-0), za is defined by za = ealogz where logz =
   ln lzl + i() with 0 ::; () :=:;; 27r.


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


4. za+l f(z) ~ 0 as lzl ---+ 0.


5. zaH f(z)::::; 0 as lzl ---+ oo.

Consider the integral Ia+ za f(z) dz, where c+ = )..t + r+ + A2 + "(-'

with )..t: z = x, 0 < r::; x::; R, r+: z =Reio, 0::; ()::; 27r, A2: z = xe^2 11"i,

R ~ x ;::: r, 1-: z = rei^0 ,27r ~ (} ~ 0, r small enough and R sufficiently

large so as to enclose all poles of f ( z) within c+. Of course, condition

(5) implies that f(z) has finitely many poles in C. This contour is shown
in Fig. 9.20. Note the slit along the real interval [r,R], which is made to

distinguish the path At with argument 0 from the path A2 with argument

27r. As defined, the integrand za f(z) is a single-valued analytic function in

the domain D enclosed by c+, except at the poles bk (k = 1, 2, ... , m) of

f(z). In addition, za f(z) is continuous on D. Hence, by the strong form
of the residue theorem, we have

or

J za J(z) dz= 1R xa f(x) dx + J za f(z) dz+ 1: xae^2 "ia f(x) dx

c+ r+

+ J za f(z)dz = 27ri f ~~~ za f(z)

'Y- k=l

(1-e^2 "ia) {R xaf(x)dx+f zaf(z)dz+jzaf(z)dz = 27rif ~s zaf(z)

Jr z-bk

~ r ~i

. (9.11-31)

In view of (4), given e > 0 there is r 1 > 0 such that r < r 1 implies that

lza+l f(z)I < e. Hence for z = rei^0 , dz= zid(}, r < ri, we have

j z" f(z) dzl :5 J,'

0

1z•+' f(z)I dO <2~•