Singularities/Residues/ Applications 6971
(^00) x (^2) dx
10. -oo (x2 + 2x + 2)2 = 71'
11 loo dx = 71'
- _ 00 (x^2 - 4x + 8)^2 16
12.100 dx = 371'
_ 00 (x^2 + 4)(x^2 + 4x + 5) 261
(^00) cosax 71' -ab
13.
0
x 2 + b 2 dx = 2 b e (a > 0, b > 0)
(^14) • 1= xsinxdx 71' ( )
0 (x2+1)(x2+4) = 6e2 e-l
15.1
00
x:in~~ dx= b~e-abf-./2sin(ab/./2) (a>O,b>O)
-oo x +
16 Joo x cos x dx 71' •
·(a) 2
2 5= -
2 2(cosl - 2sml)
_ 00 x - x + e
1
(^00) xsinxdx 71' •
(b) 2
2 5
= -
2 2
(2cosl +sml)
_ 00 x - x + e
Type IV. (PV) J~;: f(z) 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 ~ 0 or Imz ::::; 0).2. , f ( z) has simple poles on the real axis.
- J(z) = 0(1/zOI), a: > 1, as z -t oo.
As before, condition (3) implies that f(z) has a finite number of poles in
C, in particular, on the real axis.
First we examine the case of just a simple pole on the real line. Consider
the integral
J f(z)dz
c+where c+ = [-R,p-r] +7-+ [p+r, R] +r+, with -R < p-r, p+r < R,
7-: z - p = Reio, 71' ~ 8 ~ 0, r+: z = rei^6 , 0 :::; 8 :$ 71', as shown in
Fig. 9.16. Choosing R large enough and r sufficiently small so that all
poles bk (k = 1, ... , m) off in Imz > 0 be enclosed by c+, we have
J f(z)dz= 1::r J(x)dx+ J f(z)dz+ i:rf(x)dx+ J f(z)dz
c+ 'Y- r+
m
= 27l'i""' Res f(z)
L...J z=bk
k=l(9.11-19)