I'm trying to evaluate the following integral but I'm stuck. $$I(\alpha,k)=\int_{0}^{\infty}\frac{1}{(1+x^\alpha)^k} \ dx$$ Using complex analysis I know that for $k=1$ it's equal to $$\frac{\pi}{\alpha\sin(\frac{\pi}{\alpha})}$$ (I used a contour shaped like a circle sector around the pole at $z=\exp(\frac{i\pi}{\alpha})$ and let the radius $R$ approach infinity)
I however don't know how to evaluate it for a general $k$ as it generates a residue of a pole of order $k$ which would require a $(k-1)$-th derivative that I don't know how to evaluate.
I tried putting in some numbers in WolframAlpha and it I got the pattern that $$I(\alpha,k) = \frac{\Gamma(\frac{\alpha+1}{\alpha})\Gamma(k-\frac{1}{\alpha})}{\Gamma(k)}, \ \ k>\frac{1}{\alpha}$$ which is correct for the $k=1$ case, but I have no idea why it is the case.
(The formula above is also defined fo fractional $k$ which I originally didn't consider to avoid nasty branch cuts)
Start with the following substitution $$u=\frac{1}{1+x^a}$$ $$x=\left(\frac{1}{u}-1\right)^{\frac{1}{a}}=\left(\frac{1-u}{u}\right)^{\frac{1}{a}}$$ $$dx=\frac{-1}{au^2}\left(\frac{1-u}{u}\right)^{\frac{1}{a}-1}du$$ $$u(0)=1, u(\infty)=0$$ Plugging it in we get $$I(a,k)=\int_0^\infty\frac{1}{(1+x^a)^k}dx=\int_1^0u^k\frac{-1}{au^2}(1-u)^{\frac{1}{a}-1}u^{1-\frac{1}{a}}du=\frac{1}{a}\int_0^1x^{k-\frac{1}{a}-1}(1-x)^{\frac{1}{a}-1}dx$$Using the Beta Function we get $$=\frac{1}{a}B\left(k-\frac{1}{a},\frac{1}{a}\right)=\frac{1}{a}\frac{\Gamma(k-\frac{1}{a})\Gamma(\frac{1}{a})}{\Gamma(k)}=\frac{\Gamma(k-\frac{1}{a})\Gamma(1+\frac{1}{a})}{\Gamma(k)}$$ I'm unsure how to approach this with complex analysis. Maybe I would suggest substituting so we only have one pole, then maybe use a keyhole contour. For the $k-1$th derivative, use the General Leibniz Rule