# (Basic) Integral

• Apr 9th 2011, 01:12 PM
TheCoffeeMachine
(Basic) Integral
Show that $\displaystyle \int_0^{\infty} \frac{1}{1+x^n}\;{dx} = \frac{\pi}{n}\csc\left(\frac{\pi}{n}\right)$.
• Apr 9th 2011, 05:18 PM
Prove It
I expect you would have to use Contour Integration and the Residue Theorem to evaluate this...
• Apr 9th 2011, 07:01 PM
Krizalid
For $n\ge2,$ of course.

Just put $t=x^n$ and invoke Beta-Gamma Function.

I wouldn't actually call it as a "basic integral," I could post other stuff which I consider basic for me, but hard for others.
• Apr 9th 2011, 07:43 PM
TheCoffeeMachine
Quote:

Originally Posted by Krizalid
I wouldn't actually call it as a "basic integral," I could post other stuff which I consider basic for me, but hard for others.

I'm sorry, I didn't realise it was that hard! I've seen someone use it to calculate a relatively easy sum, so I assumed its proof should at least be easier than the calculation of the sum. It turns out not. I've tried to prove it, but got nowhere, so I assumed I was missing something obvious somewhere, hence the somewhat reckless/misleading title. I'll see what I can do with the beta function.
• Apr 9th 2011, 07:50 PM
TheCoffeeMachine
Quote:

Originally Posted by Prove It
I expect you would have to use Contour Integration and the Residue Theorem to evaluate this...

You might be right, but I don't have enough complex analysis knowledge to pull that through. (Doh)
• Apr 9th 2011, 07:52 PM
Prove It
Quote:

Originally Posted by TheCoffeeMachine
You might be right, but I don't have enough complex analysis knowledge to pull that through. (Doh)

I can't remember much from Complex Analysis either, I just remember "when all else fails, try contour integration" hahaha.
• Apr 9th 2011, 07:58 PM
TheCoffeeMachine
Quote:

Originally Posted by Prove It
I can't remember much from Complex Analysis either, I just remember "when all else fails, try contour integration" hahaha.

Haha, I heard it works like a charm for most integrals!
• Apr 9th 2011, 08:27 PM
Random Variable
I have this written down in my notebook. It can be used to evaluate some other integrals.

$\displaystyle \int^{\infty}_{0} \frac{dx}{1+ x^{a}} \ , a>1$

Let $\displaystyle u = 1+x^{a}$ .

Then $\displaystyle x = (u-1)^{1/a}$ and $dx = \frac{1}{a} \ (u-1)^{\frac{1}{a} -1} \ dx$

$\displaystyle = \frac{1}{a} \int^{\infty}_{1} \frac{1}{u} \ (u-1)^{\frac{1}{a} -1} \ du$

let $t = \frac{1}{u}$

$\displaystyle = \frac{1}{a} \int^{1}_{0} t^{-1} (t^{-1}-1) ^{\frac{1}{a} -1} \ dt$

$= \displaystyle \frac{1}{a} \int^{1}_{0} t^{-1} \Big(t^{-1}(1-t)\Big) ^{\frac{1}{a} -1} \ dt$

$= \displaystyle \frac{1}{a} \int^{1}_{0} t^{-1} t^{-\frac{1}{a} +1} (1-t) ^{\frac{1}{a} -1} \ dt$

$= \displaystyle \frac{1}{a} \int^{1}_{0} t^{-\frac{1}{a}} (1-t) ^{\frac{1}{a} -1} \ dt$

$\displaystyle = \frac{1}{a} \ B(1- \frac{1}{a}, \frac{1}{a} )$

and by the definition of the Beta function

$= \displaystyle \frac{1}{a} \frac{\Gamma(1 - \frac{1}{a}) \Gamma(\frac{1}{a})}{\Gamma(1 - \frac{1}{a} + \frac{1}{a})}$

$= \displaystyle \frac{1}{a} \Gamma(\frac{1}{a}) \Gamma(1 - \frac{1}{a})$

now use the reflection formula for the gamma function

$\displaystyle = \frac{1}{a} \frac{\pi}{\sin \frac{\pi}{a}} = \frac{\pi}{a} \csc (\frac{\pi}{a})$
• Apr 9th 2011, 08:45 PM
Random Variable
I'm confused how you would use contour integration when the number of residues in the upper half plane is dependent upon the value of $a$. Perhaps there's a more appropriate contour that I'm unware of.
• Apr 9th 2011, 09:39 PM
TheCoffeeMachine
Thanks for the proof, RV. Looks like this integral was anything but basic! O_o
• May 30th 2011, 08:44 PM
TheCoffeeMachine
Quote:

Originally Posted by Random Variable
I'm confused how you would use contour integration when the number of residues in the upper half plane is dependent upon the value of $a$. Perhaps there's a more appropriate contour that I'm unware of.

See this.