# Contour integrals

1. Mar 26, 2013

In trying to solve $\int^{\infty}_{-\infty} x + \frac{1}{x} dx$ could it be split up and solved using the Cauchy Principle Value theorem and a contour integral along a semi-circle. Thus;
$PV\int^{\infty}_{-\infty}x dx =0$ $+\int \frac{1}{x} dx = \int^{\pi}_{0} i d\theta$

Is this valid reasoning?

2. Mar 27, 2013

### HallsofIvy

Staff Emeritus
No, it isn't. For one thing, you have a pole at x= 0 which is on your contour. You would need another semicircle around x= 0 to avoid that.

3. Mar 28, 2013

The initial intergal was $f(x)=\int^{\infty}_{-\infty} \sqrt{x^{2}+y^{2}}dx$ so I taylor expanded it to get $f(x) \approx \int^{\infty}_{-\infty} x + \frac{y^{2}}{2x} dx$
I thought one could then justify that the cauchy principle value of $\int^{\infty}_{-\infty} x dx =0$ and then what I have done with the $\frac{1}{x}$ integral. I am doubting my approach because the Taylor series was about x=0 which seems odd, is there a better way? I read that one take x to be complex then contour interagtes it, I am just not sure how?