# Improper integral using residues

player1_1_1

## Homework Statement

$$\int\limits^{ +\infty }_{0}\frac{ \sqrt{x} \mbox{d} x }{ x^2+1 }$$

## The Attempt at a Solution

I make substitution $$\sqrt{x}=t$$ and then
$$\int^{ +\infty }_{0}\frac{ t^2 \mbox{d} t }{ t^4+1 }$$
and now this function is odd, so I make a half circle and count residues yeah?

dingo_d
Since you have $$\sqrt{x}$$ I believe you have a branch cut along the x-axes. Your integral should be:

$$\oint\frac{z^{1/2}}{z^2+1}dz$$

I'd try to solve it and I'll get back to you :D

EDIT: Yep I'm right, solved it :D

You have a branch cut along the x-axes.

The result is $$\frac{\pi}{\sqrt{2}}$$

If you want I can scan you the procedure ^^

Plus how did you get to the conclusion that the function under the integral is odd? :\

Last edited:
dingo_d
Also the integral $$\int^{ +\infty }_{0}\frac{ t^2 \mbox{d} t }{ t^4+1 }=\frac{\pi }{2 \sqrt{2}}$$

While $$\int^{ +\infty }_{0}\frac{ t^2 \mbox{d} t }{ t^4+1 }=\frac{\pi }{\sqrt{2}}$$

That's why you need the branch cut...

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Also the integral $$\int^{ +\infty }_{0}\frac{ t^2 \mbox{d} t }{ t^4+1 }=\frac{\pi }{2 \sqrt{2}}$$

While $$\int^{ +\infty }_{0}\frac{ t^2 \mbox{d} t }{ t^4+1 }=\frac{\pi }{\sqrt{2}}$$

That's why you need the branch cut...

That was just a small mistake by the OP during the substitution where he left out a 2 ($dx = 2 t \, dt$). The correct integral (after substitution) should have been.

$$\int^{ +\infty }_{0}\frac{ 2 t^2 \mbox{d} t }{ t^4+1 }=\frac{\pi }{\sqrt{2}}$$

So actually no branch cut was required.

BTW. Normally in the homework section it's better to give hints and advice on how to proceed rather than answers or worked solutions.

dingo_d
Yeah, I had some time to spend...