Show that $\displaystyle \displaystyle \int_0^{\infty} \frac{1}{1+x^n}\;{dx} = \frac{\pi}{n}\csc\left(\frac{\pi}{n}\right)$.

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- Apr 9th 2011, 01:12 PMTheCoffeeMachine(Basic) Integral
Show that $\displaystyle \displaystyle \int_0^{\infty} \frac{1}{1+x^n}\;{dx} = \frac{\pi}{n}\csc\left(\frac{\pi}{n}\right)$.

- Apr 9th 2011, 05:18 PMProve It
I expect you would have to use Contour Integration and the Residue Theorem to evaluate this...

- Apr 9th 2011, 07:01 PMKrizalid
For $\displaystyle n\ge2,$ of course.

Just put $\displaystyle t=x^n$ 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 PMTheCoffeeMachine
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 PMTheCoffeeMachine
- Apr 9th 2011, 07:52 PMProve It
- Apr 9th 2011, 07:58 PMTheCoffeeMachine
- Apr 9th 2011, 08:27 PMRandom Variable
I have this written down in my notebook. It can be used to evaluate some other integrals.

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

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

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

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

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

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

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

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

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

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

and by the definition of the Beta function

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

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

now use the reflection formula for the gamma function

$\displaystyle \displaystyle = \frac{1}{a} \frac{\pi}{\sin \frac{\pi}{a}} = \frac{\pi}{a} \csc (\frac{\pi}{a}) $ - Apr 9th 2011, 08:45 PMRandom 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 $\displaystyle a $. Perhaps there's a more appropriate contour that I'm unware of.

- Apr 9th 2011, 09:39 PMTheCoffeeMachine
Thanks for the proof, RV. Looks like this integral was anything but basic! O_o

- May 30th 2011, 08:44 PMTheCoffeeMachine
See this.