How to show that $\int \limits_{-\infty}^{\infty} (t^2+1)^{-s} dt = \pi^{1/2} \frac{ \Gamma(s-1/2)}{\Gamma(s)}$

Question

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.

Answer 1

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 $\displaystyle\int_{-\infty}^\infty f(t)~dt~=~2\int_0^\infty f(t)~dt$.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{-\infty}^{\infty}\pars{t^{2} + 1}^{-s}\,\dd t =\pi^{1/2}\,{\Gamma\pars{s - 1/2} \over \Gamma\pars{s}}}$

\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}

Answer 3

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

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