1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/Applications

applying (9.8-2), we have

R z + 1 _ bk + 1 __ ~ (b 2 b )

z=1~ z^4 + 1 - 4bi - 4 k + k

since bt = -1. Hence

1

+^00 x + 1

--dx=

-oo x4+1

7r

= v'2

693

Type III. f~: f(x)cosaxdx and J~: f(x)sinaxdx, where a> 0 and

the function 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) = 0(1/zOI), a 2: 1, as z --> oo.

Consider the complex integral

J eiazf(z)dz

c+

where c+ = [-R, R] + r+, with r+: z = Rei^6 , 0 :$ 0 :$ 7r, and R large
enough so that all poles of f(z) in the upper half-plane, say, bk ( k =

1, 2, ... , m ), be enclosed by c+. Then the residue theorem gives

J eiazf(z)dz = /_: eiaxf(x)dx + J eiazf(z)dz

c+ r+

m

= 27ri L Res eiaz f(z)

k=l z=b1;

(9.11-10)

Since IJ(z)i :$ K/lzlOI for lzl = R 2: M, we have lf(z)I ~ 0 as R --> oo.
Hence by Lemma 9.2;

lim J eiaz f(z) dz= 0

R-+oo

r+

so by taking limits in (9.11-10) as R --> oo, we get

1

+00 m ·

(PV) eiax J(x) dx = 27ri L ~1s eiaz J(z)

-oo k=l k

(9.11-11)