Consider the following function of a real variable $p$ , defined for $p>\frac{1}{2}$:

$$I(p) = \int_{-\infty}^\infty\frac{dx}{\left(1+x^2\right)^p}$$

Playing around in Wolfram Alpha, I have conjectured that we have the following closed form:

$$I(p) = \sqrt{\pi}\frac{\Gamma\left(p-\frac{1}{2}\right)}{\Gamma\left(p\right)}$$

This agrees with a number of known results, such as:

$$I(1) = \pi$$ $$I(2) = \frac{\pi}{2}$$ $$I(3) = \frac{3\pi}{8}$$

It also agrees with $I\left(\frac{1}{2}\right)$ being divergent, which follows from the fact that the antiderivative of $\left(1+x²\right)^{-\frac{1}{2}}$ is $\text{arsinh }(x)$ and the fact that $\lim\limits_{x\rightarrow\pm\infty}{\sinh(x)} = \pm\infty$.

Furthermore, this closed form agrees with any non-integer $p\geq\frac{1}{2}$ that I have tried to evaluate, such as $I\left(\frac{3}{5}\right) = \sqrt{\pi}\frac{\Gamma\left(\frac{1}{10}\right)}{\Gamma\left(\frac{3}{5}\right)}$.

However, I have not been able to prove this result. I'm thinking that perhaps the residue theorem might work, but I have no idea how to treat poles of a non-integer order in the denominator.

Any ideas?


marked as duplicate by Jack D'Aurizio definite-integrals Apr 29 '18 at 17:24

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    $\begingroup$ Surely, with some changes of variable, you can convert it into a beta integral. $\endgroup$ – Lord Shark the Unknown Apr 29 '18 at 12:58
  • $\begingroup$ @Tom See this related question. $\endgroup$ – Toby Mak Apr 29 '18 at 12:59

By substituting into the definition of the $\Gamma$-function, $$ \frac{1}{(1+x^2)^p} = \frac{1}{\Gamma(p)} \int_0^{\infty} \lambda^{p-1} e^{-\lambda(1+x^2)} \, d\lambda. $$ Interchanging the order of integration, $$ I(p) = \frac{1}{\Gamma(p)} \int_0^{\infty} \lambda^{p-1} e^{-\lambda} \left( \int_{-\infty}^{\infty}e^{-\lambda x^2} \, dx \right) d\lambda \\ = \frac{1}{\Gamma(p)} \int_0^{\infty} \lambda^{p-1} e^{-\lambda} \sqrt{\pi}\lambda^{-1/2} d\lambda = \frac{\sqrt{\pi}\Gamma(p-1/2)}{\Gamma(p)}. $$


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