# Complex integration problem using residues .

1. Dec 9, 2011

### KNUCK7ES

1. The problem statement, all variables and given/known data

I =$\int \frac{cosx}{x^{2}-2x+2}dx$ the integral runs from -inf to inf

evaluate the integral using the calculus of residues.

2. Relevant equations

shown in my attempt

3. The attempt at a solution

Re $\oint\frac{e^{iz}}{z^{2}-2z+2}$

with singularities at z=1+i and z-i they are both simple poles since we can turn

$z^{2}-2z+2$

to

$(z-2)z+2$

now all we need to do is find the residues

residue=(z-pole)f(z)

the problem is i get an answer with imaginary numbers.

my contour uses a semi circle above the real axis, so i only need to use the 1+i pole.

2. Dec 9, 2011

### jackmell

$$\int_{-\infty}^{\infty}\frac{\cos(z)}{z^2-2z+2}=\text{Re}\left\{\int_{-\infty}^{\infty} \frac{e^{iz}}{z^2-2z-2}\right\}$$

so that when you compute the e^(iz) integral and obtain an answer in the form of a+bi, a is the value of the cosine integral. Also, need to write the integrand in factored form as

$$\frac{e^{iz}}{(z-(1+i))(z-(1+i))}$$

then compute the residue.