How to solve the integral using the residue theorem int_(|z|=2)z*cos(z/(z+1))dz ?

1 Answer
Dec 28, 2017

See below.

Explanation:

Making z = 2 e^(i phi) we have

z*cos(z/(z+1))dz = 2e^(i phi) (sum_(k=0)^oo (-1)^k/(2k!)((2e^(i phi))/(2e^(i phi)+1))^(2k))2i e^(i phi) d phi

and now

int_(|z|=2)z*cos(z/(z+1))dz = int_0^(2pi) 2e^(i phi) (sum_(k=0)^oo (-1)^k/(2k!)((2e^(i phi))/(2e^(i phi)+1))^(2k))2i e^(i phi) d phi = sum_(k=0)^oo a_k

where

a_k = 4i int_0^(2pi)(-1)^k/(2k!)((2e^(i phi))/(2e^(i phi)+1))^(2k) e^(i phi) d phi

and then

a_0 = 0
a_1 = -3ipi
a_2 = (5i pi)/6
a_3=-(7ipi)/120
a_4 =(i pi)/560
a_5=-(11i pi)/(362880)
a_6 = (13 i pi)/39916800
cdots

which converges quickly .

And finally

int_(|z|=2)z*cos(z/(z+1))dz approx -(9860533 i pi)/4435200 = -2.22324 i pi