I'm redoing some calculations in the book (E. Elizalde, 10 physical applications of spectral zeta functions, 2nd ed., equations 5.36-5.37) for my thesis and can't solve this integral.
$\int_0^{\infty} x^{d-2} (x^2+1)^{-s/2} dx$
Where $d$ is a positive integer and $s$ can be whatever complex number we want that makes this integral convergent. By reverse engineering from the book the solution should be (unless I dropped some factors)
$\int_0^{\infty} x^{d-2} (x^2+1)^{-s/2} dx=\frac{\Gamma(\frac{d-1}{2})\Gamma(\frac{s-d+1}{2})}{2\Gamma(\frac{s}{2})}$
I tried to merge the three gamma functions into something simpler but can't manage to do so. It doesn't seem to be a binomial coefficient either. Also tried changing variables from $x^2+1$ to $z$ so that I could series expand $(1-z)^{(d-3)/2}$ and while that solves the integral, it leaves us with a very difficult series that seems unrelated to the gamma functions that should appear.
So... I'm pretty lost. Any ideas?
Hint. By the change of variable $$ t=\frac{1}{1+x^2}, \quad x=\sqrt{\frac{1}{t}-1}, \quad dx=-\frac{1}{t^2}\left(\frac{1}{t}-1\right)^{-1/2}dt, $$ one gets $$ \int_0^{\infty} x^{d-2} (x^2+1)^{-s/2} dx=\int_0^1 t^{s/2-d/2-1/2} (1-t)^{d/2-3/2} \,dt $$ then one may use the standard Euler beta function.