How to compute $\int_{-\infty}^{\infty} 1/(x^2+1)^s dx$?

Question

How does one evaluate

$$\int_{-\infty}^{\infty} (x^2+1)^{-s} dx$$ in the domain $\operatorname{Re}(s) > \frac{1}{2}$? Specifically, I would like to show that

$$\int_{-\infty}^{\infty} (x^2+1)^{-(s+1)/2} dx = \frac{\pi^{\frac{1}{2}}\Gamma(\frac{s}{2})}{\Gamma(\frac{s+1}{2})}$$

where $\Gamma(s) = \int_0^{\infty} e^{-t} t^{s-1} \space dt$ is the gamma function.

This integral shows up in the intertwining operators on $\operatorname{GL}_2(\mathbb R)$, and I expect a quotient of L-functions to show up.

Answer 1

Writing $d^{\ast}y = \frac{dy}{y}$, we have

$$\Gamma(\frac{s+1}{2})\int_{-\infty}^{\infty} (x+1)^{-(s+1)/2}dx = \int_0^{\infty} e^{-y}y^{(s+1)/2}d^{\ast}y \int_{-\infty}^{\infty} (x+1)^{-(s+1)/2}dx$$

$$= \int_{-\infty}^{\infty} \int_0^{\infty} \Big(\frac{y}{x^2+1} \Big)^{-(s+1)/2} e^{-y} d^{\ast}ydx = \int_0^{\infty} \int_{-\infty}^{\infty} y^{(s+1)/2}e^{-y(x^2+1)} d^{\ast}y dx$$

$$ = \int_0^{\infty} y^{(s+1)/2}\Big( \int_{-\infty}^{\infty} e^{-y(x^2+1)}dx \Big) d^{\ast}y$$

$$ = \int_0^{\infty} y^{(s+1)/2} \Big(y^{-1/2} \sqrt{\pi}\Big) = \pi^{1/2}\Gamma(\frac{s}{2}).$$

Answer 2

$$\begin{align} I=\int_{-\infty}^{\infty} (x^2+1)^{-a} dx=2\int_0^{\infty} \frac1{(x^2+1)^a} dx \end{align}$$

Let $t=x^2$

$$\begin{align} I=\int_0^{\infty} \frac{t^{-\frac12}}{(t+1)^a} dt=B\left(\frac12, a-\frac12\right)=\frac{\Gamma(\frac12)\Gamma(a-\frac12)}{\Gamma(a)}=\frac{\sqrt\pi\cdot\Gamma(a-\frac12)}{\Gamma(a)} \end{align}$$

Next, you plug in $a=\frac{s+1}2$ and get

$$\int_{-\infty}^{\infty} (x^2+1)^{-\frac{s+1}2} dx = \frac{\sqrt\pi\cdot \Gamma(\frac{s}{2})}{\Gamma(\frac{s+1}{2})}$$