1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/Applications 691

Hence

3
For integrals of this type the following alternative method is available:
Consider the arc r+: z = Rei^6 , 0 ::;; (} ::;; n:/3, R > 1, and let c+ =
[O, R] + r+ + 1'+, where 1'+: z = xei1r/^3 , R ~ x ~ 0 (Fig. 9.14). Inside this
contour the integrand has just one pole, namely, bo = ei'fr/^5 • Hence we have

J

z

4

dz = {R x

4

dx + J z

4

dz + J z

4

dz ( 9. 11 _ 5 )

z^6 + 1 Jo x^6 + 1 z^6 + 1 z^6 + 1

c+ r+ ~+

=^2 n:i ·(1) 6 e -i7r/6 = n:i(.tn3 6 v J - i ')

But

------= -ei57r/3 __

J

z4 dz =lo x4ei47r/3ei7r/3 dx 1R x4 dx

z^6 + 1 R x^6 + 1 0 x^6 + 1

~+

so that (9.11-5) becomes

(1 - ei57r/3) 1R :64 :xl + J ;4 :zl = i (1 + v13i) (9.11-6)

r+
Letting R -t oo, the second integral in (9.11-6) tends to zero, and we
obtain

{'"° x^4 dx n: 1 + J3i n:

Jo x^6 + 1 = 6 %(1 + J3i) = 3

0 -

Fig. 9.14