1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

8.5 • INDENTED CONTOUR INTEGRALS 319

and the Cauchy principal limit at t = 2 as r --> 0 is

lim [g(2 +r)-g(2-r)) = O.

r-o+

Therefore, the Cauchy principal value of the improper integral is

00

P.V. j

• oo

[

2-r oo l

_!__!!!__ - lim _!__!!!__ + _!__!!!__

t^3 - 8 - r -o+ f t^3 - 8 f t^3 - 8

-oo 2+ r

= limg(t)- lim [g(2+r)-g(2-r)]- lim g(t)

t-oo r - o+ t- - oo

= 1Tv'3 - 0 + 1Tv'3 = 1Tv'3.

12 12 6

• EXAMPLE (^8). 22 E va (^1) ua t e. Pv. J-oo oo (x-l)(x'+4). oinx dx
Solution The integrand f (z) = (z~~f{;i~ 4 ) has simple poles at the points

ti = 1 on the x-axis and z 1 = 2i in the upper half-plane. By Theorem 8.6,

The proofs of Theorems 8.5 and 8.6 depend on the following result.

t Lem ma 8.2 Suppose that f has a simple pole at the point to on the x-axis. If

Cr is the contour Cr : z = to + rei^9 , for 0 ::; 8 ::; 1T, then

lim f f (z) dz= i11"Res (!,to].

r-oJcr

Proof The Laurent series for f at z = to has the form

I ( )

Res [/, to] ( )

z = + g z,

z -to

(8- 23 )