How to show that $\int \limits_{-\infty}^{\infty} (t^2+1)^{-s} dt = \pi^{1/2} \frac{ \Gamma(s-1/2)}{\Gamma(s)}$ I want to compute the integral 
$$
\int \limits_{-\infty}^{\infty} (t^2+1)^{-s} dt
$$
for $s \in \mathbb{C}$ such that the integral converges ($\mathrm{Re}(s) > 1/2$ I think) in terms of the Gamma function. If I'm not mistaken, the answer I'm looking for is
$$
\int \limits_{-\infty}^{\infty} (t^2+1)^{-s} dt = \pi^{1/2} \frac{\Gamma(s-1/2)}{\Gamma(s)}
$$
so my question is how to prove the above formula?
Motivation
Since people sometimes ask for motivation, I'm reading a paper in which the author gives the evaluation of a certain Fourier coefficient but doesn't show the computations, just states the result. I was able to reduce the problem of checking the author's evaluation of the Fourier coefficient to proving the above formula for the integral.
Thank you for any help.

$\int_{-\infty}^{+\infty}(t^2 + 1)^{-s}dt = 2\int_{0}^{+\infty}(t^2 + 1)^{-s}dt = 2\int_{0}^{\pi/2}(\sec^2u)^{1-s}du = 2\int_{0}^{\pi/2}\cos^{2s-2}udu$
But, 
$$
2\int_{0}^{\pi/2}\cos^{2m+1}\theta\sin^{2n+1}\theta d\theta = \dfrac{\Gamma(m+1)\Gamma(n+1)}{\Gamma(m+n+2)}
$$
Thus,
$$
\int_{-\infty}^{+\infty}(t^2 + 1)^{-s}dt = 2\int_{0}^{\pi/2}\cos^{2s-2}udu = \dfrac{\Gamma(s - 1/2)\Gamma(1/2)}{\Gamma(s)} =\sqrt{\pi}\dfrac{\Gamma(s - 1/2)}{\Gamma(s)}
$$
