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.

Hint: Let $x=\dfrac1{t^2+1}$ and then recognize the expression of the beta function in the new integral. 
But first, using the parity of the integrand, write $\displaystyle\int_{-\infty}^\infty f(t)~dt~=~2\int_0^\infty f(t)~dt$.
