1. ## (Basic) Integral

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

2. I expect you would have to use Contour Integration and the Residue Theorem to evaluate this...

3. 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.

4. 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.

5. 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.

6. Originally Posted by TheCoffeeMachine
You might be right, but I don't have enough complex analysis knowledge to pull that through.
I can't remember much from Complex Analysis either, I just remember "when all else fails, try contour integration" hahaha.

7. 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!

8. 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

$\displaystyle = \frac{1}{a} \frac{\pi}{\sin \frac{\pi}{a}} = \frac{\pi}{a} \csc (\frac{\pi}{a})$

9. 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.

10. Thanks for the proof, RV. Looks like this integral was anything but basic! O_o

11. 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.