15
$\begingroup$

Denote $\square_m=\{\pmb{x}=(x_1,\dots,x_m)\in\mathbb{R}^m: 0\leq x_i\leq1,\,\,\forall i\}$ be an $m$-dimensional cube.

It is all too familiar that $\int_{\square_1}\frac{dx}{1+x^2}=\frac{\pi}4$.

QUESTION. If $\Vert\cdot\Vert$stands for the Euclidean norm, then is this true? $$\int_{\square_{2n-1}}\frac{d\pmb{x}}{(1+\Vert\pmb{x}\Vert^2)^n} =\frac{\pi^n}{4^nn!}.$$

$\endgroup$
  • $\begingroup$ Do you really mean $\square_{2n-1}$? $\endgroup$ – user64494 Dec 19 '18 at 17:56
  • 2
    $\begingroup$ Your formula also seems plausible for $n$ a half-integer (interpreting $n!$ as $\Gamma(n+1)$), at least it is true for $n=\frac{3}{2}$. This suggests an induction on $2n-1$. $\endgroup$ – Gro-Tsen Dec 19 '18 at 18:11
  • $\begingroup$ Alas, the obvious naive approach consisting of writing the integral over $[-1;1]^{2n-1}$ and decomposing the latter into the hypersurface of cubes of side $a$ themselves cut into their various $(2n-2)$-dimensional faces, does not seem to work (or maybe I tried it wrong). $\endgroup$ – Gro-Tsen Dec 19 '18 at 18:18
18
$\begingroup$

Yes, this is true, it follows from formulas in Higher-Dimensional Box Integrals, by Jonathan M. Borwein, O-Yeat Chan, and R. E. Crandall.

\begin{align} &\text{define}\;\;C_{m}(s)=\int_{[0,1]^m}(1+|\vec{r}|^2)^{s/2}\,d\vec{r},\;\;\text{we need}\;\;C_{2n-1}(-2n).\\ &\text{the box integral is}\;\;B_m(s)=\int_{[0,1]^m}|\vec{r}|^s\,d\vec{r},\\ &\text{related to our integral by}\;\;2n C_{2n-1}(-2n)=\lim_{\epsilon\rightarrow 0}\epsilon B_{2n}(-2n+\epsilon)\equiv {\rm Res}_{2n}. \end{align} The box integral $B_n(s)$ has a pole at $n=-s$, with "residue" ${\rm Res}_n$. Equation 3.1 in the cited paper shows that the residue is a piece (an $n$-orthant) of the surface area of the unit $n$- sphere, so it is readily evaluated, \begin{align} &{\rm Res}_{n}=\frac{1}{2^{n-1}}\frac{\pi^{n/2}}{\Gamma(n/2)},\\ &\text{hence}\;\;C_{2n-1}(-2n)=\frac{1}{2n}\frac{1}{2^{2n-1}}\frac{\pi^{n}}{\Gamma(n)}=\frac{1}{4^n}\frac{\pi^n}{n!}\;\;\text{as in the OP}. \end{align}

More generally, we can consider

$$C_{p-1}(-p)=\int_{[0,1]^{p-1}}(1+|\vec{r}|^2)^{-p/2}\,d\vec{r}=\frac{1}{p}\,{\rm Res}_p=\frac{1}{p} \frac{1}{2^{p-1}}\frac{\pi^{p/2}}{\Gamma(p/2)}. $$ This formula holds for even and odd $p\geq 2$, as surmised by Gro-Tsen in a comment.


There are similarly remarkable hypercube integrals where this came from, for example

$$\int_{[0,\pi/4]^k}\frac{d\theta_1 d\theta_2\cdots d\theta_k}{(1+1/\cos^2\theta_1+1/\cos^2\theta_2\cdots+1/\cos^2\theta_k)^{1/2}}=\frac{k!^2\pi^k}{(2k+1)!}.$$

$\endgroup$
  • $\begingroup$ Thanks for the reference and solution! As the first solver, it is accepted. $\endgroup$ – T. Amdeberhan Dec 19 '18 at 21:58
12
$\begingroup$

Here is an elementary proof. Using the formula \begin{equation} \frac1{a^n}=\frac1{\Gamma(n)}\,\int_0^\infty u^{n-1} e^{-u\,a}\,du \tag{1} \end{equation} with $a=1+\|x\|^2$, denoting your integral by $J_n$, letting $I:=[0,1]$, $\phi(z):=\frac1{\sqrt{2\pi}}\,e^{-z^2/2}$, and $\Phi(z):=\int_{-\infty}^z\phi(t)\,dt$, and making substitutions $x_1=z_1/\sqrt{2u}$ and then $u=z^2/2$, we have \begin{align} J_n&=\frac1{\Gamma(n)}\,\int_0^\infty u^{n-1} e^{-u}\,du \int_{I^{2n-1}}e^{-u\,\|x\|^2}\,dx \\ &=\frac1{\Gamma(n)}\,\int_0^\infty u^{n-1} e^{-u}\,du \Big(\int_I e^{-u\,x_1^2}\,dx_1\Big)^{2n-1} \\ &=\frac{\pi^{n-1/2}}{\Gamma(n)}\,\int_0^\infty u^{-1/2} e^{-u}\,du \Big(\Phi(\sqrt{2u})-\frac12\Big)^{2n-1} \\ &=\frac{2\pi^{n}}{\Gamma(n)}\,\int_0^\infty \Big(\Phi(z)-\frac12\Big)^{2n-1}\,\phi(z)\,dz \\ &=\frac{2\pi^{n}}{\Gamma(n)}\,\int_0^\infty \Big(\Phi(z)-\frac12\Big)^{2n-1}\,d\Phi(z) \\ &=\frac{\pi^{n}}{4^n \Gamma(n+1)}, \end{align} as desired.

This derivation obviously holds whenever $2n-1$ is a positive integer.


Further comments:

More generally, in view of Bernstein's theorem on completely monotone functions, one can quite similarly express the box integral \begin{equation} \int_{I^n} g(\|x\|^2)\,dx \end{equation} as an ordinary integral over $[0,\infty)$ -- for any completely monotone function $g$. Indeed, Bernstein's theorem states that any completely monotone function is a positive mixture of decreasing exponential functions; identity (1) is then such an explicit Bernstein representation of the completely monotone function $a\mapsto \frac1{a^n}$.

Now, furthermore, one does not have to insist that the mixture be positive, since we are dealing here with identities rather than inequalities. So, any function $g$ that can be represented as the mixture \begin{equation} g(a)=\int_0^\infty e^{-u\,a}\,\mu(du) \end{equation} of decreasing exponential functions $a\mapsto e^{-u\,a}$ for $a>0$, where $\mu$ is any finite, possibly signed ("mixing") measure, will do just as well.

$\endgroup$
  • $\begingroup$ Thanks for the nice elementary proof and commentaries. I'd have accepted this but your solution came a tad later. $\endgroup$ – T. Amdeberhan Dec 19 '18 at 22:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.