I’ve been looking at

∞∫0xn1+xmdx

It seems that it always evaluates in terms of sinX and π, where X is to be determined. For example:

∞∫0x11+x3dx=π31sinπ3=2π3√3

∞∫0x11+x4dx=π4

∞∫0x21+x5dx=π51sin2π5

So I guess there must be a closed form – the use of Γ(x)Γ(1−x) first comess to my mind because of the πxsinπx appearing. Note that the arguments are always the ratio of the exponents, like 14, 13 and 25. Is there any way of finding it? I’ll work on it and update with any ideas.

UPDATE:

The integral reduces to finding

∞∫−∞eatet+1dt

With a=n+1m which converges only if

0<a<1

Using series I find the solution is

∞∑k=−∞(−1)ka+k

Can this be put it terms of the Digamma Function or something of the sort?

**Answer**

I would like to make a supplementary calculation on BR's answer.

Let us first assume that 0<μ<ν so that the integral

∫∞0xμ−11+xνdx

converges absolutely. By the substitution x=tan2/νθ, we have

dx1+xν=2νtan(2/ν)−1θdθ.

Thus

∫∞0xμ−11+xνdx=∫π202νtan2μν−1θdθ=1νβ(μν,1−μν)=1νΓ(μν)Γ(1−μν)=πνcsc(πμν),

where the last equality follows from Euler reflexion formula.

**Attribution***Source : Link , Question Author : Pedro left MSE , Answer Author : Sangchul Lee*