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.
 A: 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 $\displaystyle\int_{-\infty}^\infty f(t)~dt~=~2\int_0^\infty f(t)~dt$.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{-\infty}^{\infty}\pars{t^{2} + 1}^{-s}\,\dd t
     =\pi^{1/2}\,{\Gamma\pars{s - 1/2} \over \Gamma\pars{s}}}$

\begin{align}&\color{#00f}{\large%
\int_{-\infty}^{\infty}\pars{t^{2} + 1}^{-s}\,\dd t}
=\ \overbrace{\int_{0}^{\infty}\pars{1 + t}^{-s}t^{-1/2}\,\dd t}
^{\ds{x \equiv {1 \over t + 1}\ \imp\ t = {1 \over x} - 1}}
=
\int_{1}^{0}x^{s}\pars{{1 \over x} - 1}^{-1/2}\,\pars{-\,{\dd x \over x^{2}}}
\\[3mm]&=\int_{0}^{1}x^{s - 3/2}\pars{1 - x}^{-1/2}\,\dd x
={\Gamma\pars{s - 1/2}\Gamma\pars{1/2} \over \Gamma\pars{s}}
=\color{#66f}{\large%
\root{\pi}\,{\Gamma\pars{s - 1/2}\over \Gamma\pars{s}}}\,,
\\[3mm]&\color{#c00000}{\large\Re\pars{s} > \half}
\end{align}

A: $\int_{-\infty}^{+\infty}(t^2 + 1)^{-s}dt = 2\int_{0}^{+\infty}(t^2 + 1)^{-s}dt = 2\int_{0}^{\pi/2}(\sec^2u)^{1-s}du = 2\int_{0}^{\pi/2}\cos^{2s-2}udu$
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,
$$
\int_{-\infty}^{+\infty}(t^2 + 1)^{-s}dt = 2\int_{0}^{\pi/2}\cos^{2s-2}udu = \dfrac{\Gamma(s - 1/2)\Gamma(1/2)}{\Gamma(s)} =\sqrt{\pi}\dfrac{\Gamma(s - 1/2)}{\Gamma(s)}
$$
