Evaluate $\int_{\mathbb{R}^n} \left(\frac{\lambda}{\lambda^2+|x|^2}\right)^N dx$

Question

As part of my thesis I’m reading the following paper: https://arxiv.org/abs/1612.08225 and I have trouble with the following integral calculation in multiple dimensions.

Show that $$\int_{\mathbb{R}^n}\left (\frac{\lambda}{\lambda^2+|x|^2}\right)^N dx=\frac{2^{1-N}\pi^{\frac{N+1}{2}}}{\Gamma\left(\frac{N+1}{2}\right)}$$

I know the primitive of $1/(1+|x|^2)$ is arctan x and I have thought of using Fubini to calculate the integral iteratively but it’s getting complicated. With polar coordinates, the integral is transformed to $2\pi \int_0^{\infty} \frac{\lambda}{\lambda^2+r^2} r^{n-1} dr$ I believe but I cannot find an antiderivative. I don’t see how to get the $\Gamma$ function, there is no exponential in the integral. Is there a clever substitution to use here?

Answer 1

$$I=\int_{R^N}\Big(\frac{\lambda}{\lambda^2+|\mathbf{x}|^2}\Big)^Nd\mathbf{x}=\int d\Omega_N\int_0^\infty\Big(\frac{\lambda}{\lambda^2+r^2}\Big)^N r^{N-1}dr$$ where $\int d\Omega_N$ denotes the integration over all angles in the spheric system of coordinates.

Making substitution $r=\lambda x$ $$I=\int d\Omega_N\int_0^\infty\Big(\frac{1}{1+x^2}\Big)^N x^{N-1}dx=I_1\,I_2$$ We got the factorisation of our integral. Making substitution $t=\frac{1}{1+x^2}$in the second integral $$I_2=\int_0^\infty\Big(\frac{1}{1+x^2}\Big)^N x^{N-1}dx=\frac{1}{2}\int_0^1(1-t)^{\frac{N}{2}-1}t^{\frac{N}{2}-1}dt=\frac{1}{2}B\Big(\frac{N}{2};\frac{N}{2}\Big)=\frac{1}{2}\frac{\Gamma^2\Big(\frac{N}{2}\Big)}{\Gamma(N)}$$ To evaluate $\int d\Omega_N$, we consider $$J=\int_{R^N}e^{-\mathbf{x}^2}d\mathbf{x}=\int_{-\infty}^\infty...\int_{-\infty}^\infty e^{-x_1^2...-x_N^2}dx_1...dx_N=(\sqrt\pi)^N$$ On the other hand, $$J=\int d\Omega_N\int_0^\infty e^{-r^2}r^{N-1}dr=\frac{1}{2}\Gamma\Big(\frac{N}{2}\Big)\int d\Omega_N$$ $$\Rightarrow\,\,\,I_1=\int d\Omega_N=2 \frac{(\sqrt\pi)^N}{\Gamma\Big(\frac{N}{2}\Big)}$$ Therefore, $$I=\pi^{\frac{N}{2}}\frac{\Gamma\Big(\frac{N}{2}\Big)}{\Gamma(N)}$$ Using the duplication formula for Gamma-function (https://en.wikipedia.org/wiki/Gamma_function) $$\Gamma\Big(\frac{N}{2}\Big)\Gamma\Big(\frac{N}{2}+\frac{1}{2}\Big)=2^{1-N}\sqrt\pi\,\Gamma(N)$$ we get the desired result.