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? :\

 

[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...

 

[uart]

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...