Prove $\int_{-\infty}^{\infty} \frac{du}{(1+u^{2})^{\frac{7-p}{2}}} =\frac{\sqrt{\pi}\Gamma[\frac{1}{2}(6-p)]}{\Gamma[\frac{1}{2}(7-p)]}$

Question

I try to evaluate following integral

$\int_{-\infty}^{\infty} \frac{du}{(1+u^{2})^{\frac{7-p}{2}}} =\frac{\sqrt{\pi}\Gamma[\frac{1}{2}(6-p)]}{\Gamma[\frac{1}{2}(7-p)]}$

It seems okay to extend

$\int_{-\infty}^{\infty} \frac{du}{(1+u^{2})^{\frac{p}{2}}} =\frac{\sqrt{\pi}\Gamma[\frac{1}{2}(p-1)]}{\Gamma[\frac{p}{2}]}$

It seems it is also related with Beta function, but i could not find how to do.

Answer 1

First of all, your integrand is even, i.e., symmetrical with regards to the origin, so your integral can

be written as $2\displaystyle\int_0^\infty f(u)du.~$ Secondly, all integrals of the form $\displaystyle\int_0^\infty\frac{u^{n-1}}{\big(a^m+u^m\big)^k}du$ can be

evaluated in terms of the beta function as follows: Let $u=at$, and then $x=\dfrac1{\big(1+t^m\big)^k}$ , yielding

$\dfrac{a^{n-km}}m\cdot B\bigg(\dfrac nm~,~k-\dfrac nm\bigg)$. Then, for integer values of k, Euler's reflection formula can be used

to simplify the latter expression. In this particular case, $n=1,~$ $m=2,~$ $k=\dfrac{7-p}2~$ and $~a=1$.

Answer 2

The integrand is even, so we can change the limits to $0 \le u \le \infty$ and double the answer.

$I = 2\displaystyle\int\limits_0^{\infty} \dfrac{1}{(1 + u^2)^{(7 - p)/2}} du$

Let $u = \tan x$, then $du = \sec^2 x \,dx$.

$\begin{align} \therefore I & = 2\int_0^{\pi/2}\dfrac{1}{(\sec x)^{(7 - p)}}\sec^2 x \,dx\\ & = 2\int_0^{\pi/2} (\sec x)^{p - 5} \,dx\\ & = 2\int_0^{\pi/2} \sin^0 x \,\cos^{5 - p} x \,dx\\ & = \beta\left(\dfrac 1 2, \dfrac{6 - p}{2} \right)\\ & = \dfrac{\Gamma\left( \dfrac 1 2 \right) \Gamma \left( \dfrac{6 - p}{2} \right)}{\Gamma\left(\dfrac{7 - p}{2} \right)}\\ & = \boxed{\sqrt{\pi} \dfrac{\Gamma\left[ \frac 1 2 (6 - p) \right]}{\Gamma\left[ \frac 1 2 (7 - p) \right]}} \end{align}$

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Note:

1. $\displaystyle\int\limits_0^{\pi/2} \cos^p \theta \sin^q \theta \,d\theta = \beta\left( \dfrac{p + 1}{2}, \dfrac{q + 1}{2} \right)$
2. $\beta(m, n) = \dfrac{\Gamma(m)\Gamma(n)}{\Gamma(m + n)}$