Computing explicitly three integrals involving radial functions How can I compute the following three (similar)
integrals?
$$
\int_{\mathbb R^N} \left(\frac{1}{1+|x|^2} \right)^{\beta} dx
$$
$$
\int_{\mathbb R^N} \left(\frac{1}{1+|x|^2} \right)^{(N+2\alpha)/2} \left(\frac{1}{|x|} -\frac{1}{(1+|x|^2)^{1/2}} \right) dx
$$
$$
\int_{\mathbb R^N} \left(\frac{1}{1+|x|^2} \right)^{(N+2\alpha)/2} |x|^{4\alpha - N} dx
$$
where $\alpha, \beta >0$. Since the involved functions are radial, one can make a change of variables to reduce the problems to 1-d problems, but then I don't know how the integral can be computed explicitly or it if can be done in an easy way at all or not.
 A: All these integral can be expressed through Beta function. Let's take, for instance, the first one: $I(n,\beta)=\int_{\mathbb R^N} \left(\frac{1}{1+|x|^2} \right)^{\beta} dx_1..dx_n=\int_{\mathbb \Omega}d{\Omega}\int^\infty_0\left(\frac{1}{1+r^2} \right)^{\beta} r^{n-1}dr$, where we integrate over all angles ($d{\Omega}$) in a polar system of coordinate.
Because the integrand is a scalar (and does not depend on angles), integration over $d{\Omega}$ can be done straightforward. We can use, for instance, the following trick:
$\int_{\mathbb x_1}..\int_{\mathbb x_n}e^\left(-x_1^2-..-x_n^2\right)d{x_1}..d{x_n}=(\int_{\mathbb x}e^{-x^2}dx)^n=\left(\sqrt{\pi}\right)^n=\int_{\mathbb \Omega}d{\Omega}\int^\infty_0e^{-r^2} r^{n-1}dr=\frac{1}{2}\int_{\mathbb \Omega}d{\Omega}\int^\infty_0e^{-t} t^{\frac{n}{2}-1}dt  (t=r^2)=\frac{1}{2}\Gamma(\frac{n}{2})\int_{\mathbb \Omega}d{\Omega} \Rightarrow $$\int_{\mathbb \Omega}d{\Omega}=\frac{2(\sqrt{\pi})^n}{\Gamma(\frac{n}{2})}$ (where ${\Gamma}(x)$ is Gamma function).
$I(n,\beta)=\frac{2(\sqrt{\pi})^n}{\Gamma(\frac{n}{2})}\int^\infty_0\left(\frac{1}{1+r^2}\right)^{\beta} r^{n-1}dr= \frac{(\sqrt{\pi})^n}{\Gamma(\frac{n}{2})}\int^1_0(\frac{1-t}{t})^{\frac{n}{2}-1}t^{\beta-2} dt (t=\frac{1}{1+r^2})=\frac{(\sqrt{\pi})^n}{\Gamma(\frac{n}{2})}B(\frac{n}{2};\beta-\frac{n}{2})=(\sqrt{\pi})^n\frac{\Gamma(\beta-\frac{n}{2})}{\Gamma(\beta)}$
The formula gives itself the the answer when it is valid: ${\beta}>\frac{n}{2}$.
The second and third integral can be evaluated in the same fashion - just make designated substitutions and use formula for Beta function (and its relation with Gamma function) https://en.wikipedia.org/wiki/Beta_function.
PS  Third integral $=\frac{(\sqrt{\pi})^n}{\Gamma(\frac{n}{2})}B(2{\alpha};\frac{n}{2}-{\alpha})=(\sqrt{\pi})^n\frac{\Gamma(2\alpha)\Gamma(\frac{n}{2}-\alpha)}{\Gamma(\frac{n}{2})\Gamma(\frac{n}{2}+\alpha)}$ - valid for $0<\alpha<\frac{n}{2}$
Second integral $=\frac{(\sqrt{\pi})^n}{\Gamma(\frac{n}{2})}\left(B({\alpha}+\frac{1}{2};\frac{n}{2}-\frac{1}{2})-B({\alpha}+\frac{1}{2};\frac{n}{2})\right)=(\sqrt{\pi})^n\frac{\Gamma(\frac{1}{2}+\alpha)}{\Gamma(\frac{n}{2})}\left(\frac{\Gamma(\frac{n}{2}-\frac{1}{2})}{\Gamma(\frac{n}{2}+\alpha)}-\frac{\Gamma(\frac{n}{2})}{\Gamma(\frac{n}{2}+\alpha+\frac{1}{2})}\right)$. In this case, we have uncertainty at $\alpha =-\frac{1}{2}$, so it is easier to get from the original integral at $x\rightarrow \infty$ that integral is valid for $\alpha>-\frac{3}{2}$ and $n>1$.
