708 Chapter^9
Example To evaluate f 0
00
xa dx/(x + b), where -1 < a < 0 and b > 0.
Here f(z) = 1/(z+b) has the simple pole z =-band since 0 < a+l < 1, the
expression za+l f(z) = za+I /(z + b) tends to 0 as lzl ~ 0 or as izl ~ oo.
We have
a
Res z = (-b)a = ea(lnb+i,,.) = bae71"ia
z=-b Z + b
Hence (9.11-32) gives
roo xa dx -7rbalo x+b=sina7r
The same result holds if a is nonreal with -1 < Re a < 0.
In particular, letting b = 1 and writing a = ( - 1, we get
{oo x<:-I dx -'Ir 7rlo x + 1 = sin((-l)7r = sin(7r
(9.11-33)(9.11-34)with 0 < Re ( < 1, a formula that was used in Theorem 8.48-8. Also, if we
let x = et in (9.11-34), we have, alternatively,
1
(^00) e<:t dt 7r
_ 00 ~t+1 = sin(7r (0 <Re(< 1)
(9.11-35)
Type VII. (PV) J 0
00
xa f(x) dx, where a > -1 is not an integer and the
following conditions are satisfied:
i. f(z) has an extension f(z) that is meromorphic at least on Imz 2:: 0.
2. For z = jzjei^8 (z f. 0), za is defined by za = ealogz where logz is
specified by logz = lnlzl + iO with -%7r < () :5 %7r. Note that byusing this branch of the logarithm we have xa = ea In"' for x > 0.
- f(z) has simple poles on the positive real axis.
- za+lf(z) ~ 0 as lzl ~ 0.
5. za+c+i f(z) = 0(1) as jzj ~ oo for some c > 0.
Condition (5) implies that lza+l f(z)I < A/lzlc for lzl large enough (A
a constant), so jza+l f(z)I ~ 0 as jzj ~ oo. Again, this last condition
means that f has a finite number of poles in C. First we assume that
there is just a simple pole p on the positive real axis (and none on the
negative real axis), and consider the integral fa+ za J(z) dz, where c+ =
[-R, -r]+r:l +[r,p-8]+r2 +[p+8, R]+r+, with 'Y:l: z = rei^0 , 7r 2:: 8 2:: 0,
'Y2: z-p = 8e;o, 7r 2:: 8 2:: 0, r+: z = Re;o, 0 ::::; 8 ::::; 7r, as shown in Fig. 9.21.
It is assumed that r, 8 and R are chosen so that 0 < r < p - 8 < p + 8 < R,
with r and 8 small enough and R sufficiently large so that all poles, if any,