- #1

ShayanJ

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## Homework Statement

Find the value of the integral ## \int_0^\infty dx \frac{\sqrt{x}}{1+x^2} ## using calculus of residues!

## Homework Equations

## The Attempt at a Solution

This is how I did it:

##\int_0^\infty dx \frac{\sqrt{x}}{1+x^2}=\frac 1 2 \int_{-\infty}^\infty dx \frac{\sqrt{|x|}}{1+x^2} ##

##\displaystyle \lim_{R\to \infty} \int_{-R}^R dz \frac{\sqrt{|z|}}{1+z^2}=\lim_{R\to \infty}\oint_{semicircle^+_R} dz \frac{\sqrt{|z|}}{1+z^2}-\underbrace{\lim_{R\to \infty} \int_{C_R}dz \frac{\sqrt{|z|}}{1+z^2}}_{\to 0} \Rightarrow \int_{-\infty}^\infty dz \frac{\sqrt{|z|}}{1+z^2}=2\pi i \frac{\sqrt{|i|}}{2i}=-2\pi i \frac i 2=\pi##

So:

##\int_0^\infty dx \frac{\sqrt{x}}{1+x^2}=\frac \pi 2 ##

I also tried using the keyhole contour because of the branch point at z=0. But this time I got ## \frac{\pi}{1+i} ##.

The problem is, wolframalpha.com gives ## \frac \pi {\sqrt 2} ##.

What did I do wrong?

Thanks