# Contour integral with absolute value

Homework Helper
Suppose I want to compute tthe integral:
$$\int_{-\infty}^{\infty}\frac{\textrm{sech}\hspace{0.1cm} x}{x^{2}-2|x|+1}dx$$
Can I compute this integral via contour integration? The only way that I have thought of is to split up the domain:
$$\int_{-\infty}^{\infty}\frac{\textrm{sech}\hspace{0.1cm} x}{x^{2}-2|x|+1}dx=\int_{-\infty}^{0}\frac{\textrm{sech}\hspace{0.1cm} x}{x^{2}+2x+1}dx+\int_{0}^{\infty}\frac{\textrm{sech}\hspace{0.1cm} x}{x^{2}-2x+1}dx$$
Is this the best way I can go about things for is there a better way?

## Answers and Replies

mathman
Science Advisor
So far OK, but you have singularities at |x|=1.

Homework Helper
I am aware of the singularities, this was just an example. The other integral doesn't have singularities, I just wanted to get my ideas accross.