0

It is well known that $$\int_{\mathbb{R}^N} e^{-\Vert \mathbf{x}\Vert^2}\Vert \mathbf{x}\Vert^{-k} \mathrm{d}\mathbf{x}$$ is convergent for $k<N.$ But what exactly is its value? Or at least an upper and lower bound? I don't know which way to go to try to find out, whether polar coordinates, or a complex variable. Any suggestion?

Math Attack
  • 1
Ilovemath
  • 2,921

1 Answers1

1

Radial integration: Given $f:[a,b]\to \mathbb{R}$ a continuous function, then $$ \int_{\{a\le \lVert {\mathbf{x}} \rVert\le b\}} f(\lVert \mathbf{x} \rVert)\, \mathrm{d}^nx = \frac{2 \pi^{\frac{n}{2}} }{\Gamma\left(\frac{n}{2}\right)} \int_{a}^{b}f(t) t^{n-1}\, \mathrm{d}t $$

Proof: We use the $n$-dimensional spherical coordinates substitution suggested by @GEdgar to get \begin{align*} \int_{a\le \lVert {\mathbf{x}} \rVert\le b} f(\lVert \mathbf{x} \rVert)\, \mathrm{d}^n x& = \int_{0}^{2\pi}\int_{0}^{\pi} \dots \int_{0}^{\pi}\int_{a}^{b} f(r) r^{n-1} \prod_{k=1}^{n-2}\sin^{n-k-1}(\varphi_k)\, \mathrm{d}r\mathrm{d}\varphi_1\dots\varphi_{n-2}\varphi_{n-1}\\ & \overset{\color{purple}{n-k\to m}}{=} 2\pi\left(\prod_{m=2}^{n-1}\int_{0}^{\pi} \sin^{m-1}(\varphi_{n-m})\, \mathrm{d}\varphi_{n-m}\right) \int_{a}^{b} f(r) r^{n-1}\mathrm{d}t \end{align*} so the problem reduces to evaluating $\int_{0}^{\pi} \sin^{m-1}(t)\, \mathrm{d}t = B\left(\frac{m}{2},\frac12\right)$. To evaluate the resulting product we use that $B(\alpha, \beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$, so we get the following \begin{align*} \prod_{m=2}^{n-1} B\left(\frac{m}{2}\right) &= \prod_{m=2}^{n-1} \frac{\Gamma\left(\frac{m}{2}\right)\sqrt{\pi}}{\Gamma\left(\frac{m}{2} + \frac12\right)}\\ &=\pi^{\frac{n-2}{2}}\left(\prod_{\substack{2\le m\le n-1\\ m \text{ even}}} \frac{\Gamma\left(\frac{m}{2}\right)}{\Gamma\left(\frac{m+1}{2} \right)}\right)\left(\prod_{\substack{2\le m\le n-1\\ m \text{ odd}}} \frac{\Gamma\left(\frac{m}{2}\right)}{\Gamma\left(\frac{m+1}{2} \right)}\right)\\ &=\pi^{\frac{n}{2}-1}\left(\prod_{\substack{1\le k\le \frac{n}{2}-\frac12}} \frac{\Gamma\left(k\right)}{\Gamma\left(k + \frac12 \right)}\right)\left(\prod_{\substack{1\le k\le \frac{n}{2}-1}} \frac{\Gamma\left(k + \frac12\right)}{k\Gamma\left(k\right)}\right)\\ &=\pi^{\frac{n}{2}-1}\left(\prod_{\substack{1\le k\le \frac{n}{2}-1}} \frac{1}{k}\right)\left(\prod_{\substack{\frac{n}{2}-1< k\le \frac{n}{2}-\frac12}} \frac{\Gamma\left(k\right)}{\Gamma\left(k + \frac12 \right)}\right) \end{align*} $\require{cancel}$ Lastly, we consider the cases for $n$ even/odd to simplify the resulting product. For even $n$: $$ \left(\prod_{\substack{1\le k\le \frac{n}{2}-1}} \frac{1}{k}\right)\left(\prod_{\substack{\frac{n}{2}-1< k\le \frac{n}{2}-\frac12}} \frac{\Gamma\left(k\right)}{\Gamma\left(k + \frac12 \right)}\right) = \frac{1}{\left(\frac{n}{2}-1\right)!}\cancel{\left(\prod_{k \in \emptyset} \frac{\Gamma\left(k\right)}{\Gamma\left(k + \frac12 \right)}\right)} = \frac{1}{\Gamma\left(\frac{n}{2}\right)} $$ using that $\xi ! = \Gamma(\xi +1)$ for $\xi \in \mathbb{N}$. Now analyzing the case for odd $n$: $$ \left(\prod_{\substack{1\le k\le \frac{n}{2}-1}} \frac{1}{k}\right)\left(\prod_{\substack{\frac{n}{2}-1< k\le \frac{n}{2}-\frac12}} \frac{\Gamma\left(k\right)}{\Gamma\left(k + \frac12 \right)}\right) = \frac{1}{\left(\frac{n-3}{2}\right)!}\frac{\Gamma\left(\frac{n-1}{2}\right)}{\Gamma\left(\frac{n-1}{2} + \frac12 \right)} = \frac{1}{\cancel{\Gamma\left(\frac{n-1}{2}\right)}}\frac{\cancel{\Gamma\left(\frac{n-1}{2}\right)}}{\Gamma\left(\frac{n}{2}\right)} $$ again using that $\xi ! = \Gamma(\xi +1)$ for positive integer inputs.

Applying the formula to your problem we get $$ \int_{\mathbb{R}^n}e^{-\lVert \mathbf{x} \rVert^2} \lVert \mathbf{x} \rVert^{-k}\, \mathrm{d}^n x = \frac{2 \pi^{\frac{n}{2}} }{\Gamma\left(\frac{n}{2}\right)}\int_{0}^{\infty}e^{-t^2} t^{n-k-1}\, \mathrm{d} t \overset{\color{purple}{u=t^2}}{=}\frac{\pi^{\frac{n}{2}} }{\Gamma\left(\frac{n}{2}\right)}\int_{0}^{\infty}e^{-u} u^{\frac{n-k}{2}-1}\, \mathrm{d}u = \boxed{\frac{\pi^{\frac{n}{2}}\Gamma\left(\frac{n-k}{2}\right) }{\Gamma\left(\frac{n}{2}\right)}} $$ where in the last step we used the definition of the Gamma function, noticing that the convergence of the integral requires $\frac{n-k}{2}>0 \iff n>k$. Lastly, since the dimension $n$ is a positive integer, $\Gamma\left(\frac{n}{2}\right)$ can be expressed in terms of factorials. This gives an equivalent (but arguably simpler) answer as $$ \int_{\mathbb{R}^n}e^{-\lVert \mathbf{x} \rVert^2} \lVert \mathbf{x} \rVert^{-k}\, \mathrm{d}^n x = \begin{cases} \frac{\pi^{\frac{n}{2}}}{\left(\frac{n}{2}-1\right)!}\Gamma\left(\frac{n-k}{2}\right) , & n \text{ even}\\ \frac{\pi^{\frac{n-1}{2}}2^{n-1}\left(\frac{n-1}{2}\right)!}{(n-1)!}\Gamma\left(\frac{n-k}{2}\right) , & n \text{ odd}\end{cases} $$ and if $k$ is also chosen to be an integer then the last gamma function can analogously be simplified into factorials and powers of $\pi$.

Robert Lee
  • 7,233