Jtdylktuy

I was playing around with glasser's master theorem and integrals of the form $$int_{-infty}^infty frac{1}{x^n+1}dx$$ I observed that for positive, even values of n, the solution to the integral followed the form $$frac{pi}{frac{n}{2}sin(frac{pi}{n})}$$, the obvious example being at n = 2, the integral equals $pi$. I've yet to prove this pattern, but I recognized it after some fiddling around. Is this integral part of some larger concept that is already documented? For example, the way that $int_0^infty frac{x^{p-1}}{e^x-1}dx = zeta(p)Gamma(p), p>1$, I can't help but think about why the general solution for even values has a sine function in it.

For any complex number $alpha$ such that $0<text{Re}(alpha)<1$, let
$$I(alpha):=int_0^infty,frac{t^{alpha-1}}{t+1},text{d}t,,$$
where the branch cut of the map $zmapsto z^{alpha-1}$ is taken to be the positive-real axis.
For a real number $epsilonin(0,1)$, consider the positively oriented keyhole contour $Gamma_epsilon$ given by
$$begin{align}left[epsilon,exp(text{i}epsilon),frac{1}{epsilon},exp(text{i}epsilon)right]&cupleft{frac{1}{epsilon},exp(text{i}t),Big|,tin[epsilon,2pi-epsilon]right}
\&cupleft[frac{1}{epsilon},expbig(text{i}(2pi-epsilon)big),epsilon,expbig(text{i}(2pi-epsilon)big)right]cupBig{epsilon,exp(text{i}t),Big|,tin[2pi-epsilon,epsilon]Big},.end{align}$$
Then,
$$lim_{epsilonto0^+},oint_{Gamma_epsilon},frac{z^{alpha-1}}{z+1},text{d}z=I(alpha)-expbig(2pitext{i}(alpha-1)big),I(alpha)=-2text{i},exp(pitext{i}alpha),sin(pialpha),I(alpha),.$$
Via the Residue Theorem,
$$lim_{epsilonto0^+},oint_{Gamma_epsilon},frac{z^{alpha-1}}{z+1},text{d}z=2pitext{i},text{Res}_{z=-1}left(frac{z^{alpha-1}}{z+1}right)=2pitext{i},(-1)^{alpha-1}=-2pitext{i},exp(pitext{i}alpha),.$$
Hence, $$int_0^infty,frac{t^{alpha-1}}{t+1},text{d}t=I(alpha)=frac{pi}{sin(pialpha)},.$$

Now, take $alpha:=dfrac1n$ for some integer $ngeq 2$. Then,
$$frac{pi}{sinleft(frac{pi}{n}right)}=int_0^infty,frac{t^{frac1n-1}}{t+1},text{d}t=int_0^infty,frac{n}{x^n+1},text{d}x,,$$
by setting $x:=t^{frac1n}$. This proves the equality
$$int_0^infty,frac{1}{x^n+1},text{d}x=frac{pi}{n,sinleft(frac{pi}{n}right)},.$$ If $n$ is even, then
$$int_{-infty}^{+infty},frac{1}{x^n+1},text{d}x=2,int_0^infty,frac{1}{x^n+1},text{d}x=frac{2pi}{n,sinleft(frac{pi}{n}right)}=frac{pi}{frac{n}{2},sinleft(frac{pi}{n}right)},.$$

In fact, it can be seen that $I(alpha)=text{B}(alpha,1-alpha)=Gamma(alpha),Gamma(1-alpha)$, where $text{B}$ and $Gamma$ are the usual beta and gamma functions, respectively. Therefore, this gives a proof of the Reflection Formula for complex numbers $alpha$ such that $0<text{Re}(alpha)<1$. Then, using analytic continuation, we prove the Reflection Formula for all $alphainmathbb{C}$.

Using Mason's tip in the comments you can say

$$ int_{-infty}^{infty} frac{1}{x^n+1}dx = 2int_0^infty frac{1}{x^n+1}dx. $$

And now substitute $x^n + 1 mapsto 1/u$, and you should get something in the form of the beta function, which has a known representation in terms of the gamma function.