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