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