Question

Show that ∫_{0,∞}(dx/(1+x^n))=π/(nsin(π/n)) for ∀n∈ℝ.

Repost, but I have no idea what to do, and the last one drowned.

Answer 1

This?\[\int\limits \limits_{0}^{\infty} \frac{1}{1+x^{n}}dx = \frac{\pi}{n \sin\frac{\pi}{n}}\]

Answer 2

Yep. This question was asked in context of the residue theorem http://en.wikipedia.org/wiki/Residue_theorem. Analysis and all that.

Answer 3

http://en.wikipedia.org/wiki/Residue_theorem

The link, since the previous one apparently doesn't work. Not like it's too hard to Google it, or anything.

Answer 4

OK - I'll read up on the residue theorem and come back once I've understood it. Looks interesting.

Answer 5

Thanks, asna!

Answer 6

Is \(x\) a real variable?

Answer 7

I don't believe I'm told if it is, but in context it's a safe assumption.

Answer 8

is it?
the wiki link suggest to me otherwise

Answer 9

Point taken.

Answer 10

I believe it's a real variable. Anyways, the relation above doesn't hold for \(n=1\) since
\[\int_0^{\infty} \frac{1}{1+x}=\ln(1+x)|_0^{\infty} \to \infty.\]

Answer 11

I'll be honest--I can't make heads or tails of the residue theorem.

Answer 12

and I'm running out of tricks....
I think this is beyond me

Answer 13

I'm not mastering the residue theorem, so I will first try to use it on the special case with \(n=2\), since we already know that the integral would be \(\frac{\pi}{2}\).
\[\int_0^{\infty} \frac{1}{1+x^2}=\frac{1}{2}\int_{-\infty}^{\infty}\frac{dx}{1+x^2}=\frac{1}{2}\text{Re}\int_{-\infty}^{\infty}\frac{1}{1+z^2}dz.\]
Now I will apply the theorem to evaluate \(\int_{-\infty}^{\infty}\frac{1}{1+z^2}dz\).

Answer 14

I found a you tube video where the person explains almost exactly this integral using the residue theorem: http://www.youtube.com/watch?v=XRFn0ioBPXI

Answer 15

Whoa, nice find, asna. And clever algebra, math.

Answer 16

@badreferences: I tried to follow the reasoning in the video - it sort of makes sense but I will need to review some of his earlier videos in order to understand this completely. That will take me some time - so don't hold your breath! :)

In the meantime, I'll leave you with the expert Mr.Math.

Answer 17

I still don't see what you're trying to get at, though, Math. What can we determine by examining this special case? Looks like a fairly straightforward arctan.

Answer 18

Yeah I know. I just wanted to do it using the residue theorem to see how it works with this simple case (that we already know its value). That's why I changed it to complex variable integral.

Answer 19

This is a related video to the one that naseer posted above, where the man evaluated the integral of 1/(1+x^4) using the residue theorem. http://www.youtube.com/watch?v=MRLa5bk3_R4&feature=related

Answer 20

I think it would help us understand the general case.

Answer 21

Let's see if I can make myself understand this.