How does exprews $\int_{-\infty}^{\infty} \frac{1}{(1+x^2)^n}dx$ for $1<n \leq 2$ in terms of the gamma function?
E.g. for $n=4/3$, we have $\int_{-\infty}^{\infty}\frac{1}{(1+x^2)^{4/3}}dx=\frac{\sqrt{\pi} \Gamma(5/6)}{\Gamma(4/3)}$.
Why is this true?
Using the following integral representation of the Beta function \begin{align} B(x, y) = \int^\infty_0 \frac{t^{x-1}}{(1+t)^{x+y}}\ dt, \end{align} we see that \begin{align} B\left(\frac{1}{2}, n-\frac{1}{2}\right) = \int^\infty_0 \frac{t^{-1/2}}{(1+t)^{n}}\ dt = \int^\infty_0 \frac{2dx}{(1+x^2)^{n}} = \int^\infty_{-\infty} \frac{dx}{(1+x^2)^{n}}. \end{align} However, since \begin{align} B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \ \ \implies \ \ B\left(\frac{1}{2}, n-\frac{1}{2}\right)=\frac{\Gamma(\frac{1}{2})\Gamma(n-\frac{1}{2})}{\Gamma(n)} = \frac{\sqrt{\pi}\Gamma(n-\frac{1}{2})}{\Gamma(n)}. \end{align}
