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?
>  $$\int\_{\square\_{2n-1}}\frac{d\pmb{x}}{(1+\Vert\pmb{x}\Vert^2)^n}
> =\frac{\pi^n}{4^nn!}.$$
> 
> 
> 



Integrating over a hypercube, not a hypersphere
Yes, this is true, it follows from formulas in [Higher-Dimensional Box Integrals](https://projecteuclid.org/download/pdf_1/euclid.em/1317758103), 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.




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


$$\int\_{[0,\pi/4]^k}\frac{d\theta\_1 d\theta\_2\cdots d\theta\_k}{(1+1/\cos^2\theta\_1+1/\cos^2\theta\_2\cdots+1/\cos^2\theta\_k)^{1/2}}=\frac{k!^2\pi^k}{(2k+1)!}.$$


