An integral over n-dimensional unit cube I'm trying to calculate the following $\text{n}$-dimensional integral:
$$ \int_{[0, 1]^n}  (1+ \sum_{k=1}^{n} x_k^{2}) ^ {- {{n+3}\over{2}}}  \ \mathrm dx_{1}\mathrm dx_2 ... dx_n  $$
where ${[0, 1]^n}$ is a $\text{n}$-dimensional unit cubic.
Is there a closed-form for this integral?
Thanks.
 A: This is solution for another integral ($\int_{[0, 1]^n}  (1+ \sum_{k=1}^{n} x_k^{2}) ^ {- {{n+1}\over{2}}}  \ \mathrm dx_{1}\mathrm dx_2 ... dx_n$), but it might be of interest as well.
We denote $\displaystyle I(n)=\int_{[0,1]^{n-1}}\frac{1}{(1+{x_1}^2+{x_2}^2+…+{x_{n-1}}^2)^{n/2}}dx_1…dx_{n-1}\tag*{}$
Let’s consider the space of $n$ dimensions; radius-vector $\vec R=x_1 \vec {e_1}+…+x_n \vec {e_n}$ ,
and $R=\sqrt{{x_1}^2+{x_2}^2+…+{x_{n}}^2}$ - its norm.
We also denote $\vec \eta=\frac{\vec R}{R}$, and the field $\vec F$
$\displaystyle\vec F(\vec R)=-\frac{1}{n-2}\vec\nabla\frac{1}{R^{n-2}}=\frac{\vec \eta}{R^{n-1}}\tag*{}$
Let’s consider the flow of the vector $\vec F(\vec R)$ through some closed hyper-surface surrounding the source (the origin):
$\displaystyle J_n=\int(\vec F(\vec R),d\vec S)=-\frac{1}{n-2}\int(\vec\nabla\frac{1}{R^{n-2}},d\vec S)\tag*{}$
According to Gauss theorem,
$\displaystyle J_n=-\frac{1}{n-2}\int\Delta\frac{1}{R^{n-2}}\,dV\tag*{}$ - the integral over the volume inside the surface ($\Delta$ denotes Laplacian).
But $\frac{1}{R^{n-2}}$ is the fundamental solution of the Laplace equation in $n$ dimensions (for example, look here computations problem with reverse Fourier transform).
$\displaystyle \Delta\frac{1}{R^{n-2}}=-\frac{2(n-2)\pi^{n/2}}{\Gamma\big(n/2\big)}\delta^n (\vec R)\tag*{}$
Integrating delta function
$\displaystyle J_n=\frac{2\pi^{n/2}}{\Gamma\big(n/2\big)}\tag*{}$
We see that the total flow through any closed hyper-surface is constant (does not depend on $R$).
Therefore, we can choose as a surface the hypercube surrounding the origin, with an edge length equal to 2. There are $2n$ equal faces, and the flow through one face of the cube is$ \int_{\text{one face}}(\vec F(\vec R),d\vec S)=\int_{\text{one face}}\frac{(\vec n, d\vec S)}{R^{n-1}}=\int_{\text{one face}}\frac{dS}{R^n}$. (Drawing a picture helps a lot).
The total volume of one $(n-1)$-dimensional face (with the edge length equal to 2) is $|[-1;1]|^{n-1}=2^{n-1}$, but we want to find only a portion of the flow - through the volume $|[0;1]|^{n-1}$.
Therefore, the desired integral $I(n)$ is the portion of total flow through the closed surface: $I(n)=\frac{1}{2n} \frac{1}{2^{n-1}}J_n$ (the total flow goes through $2n$ faces, and our integral is equal to the flow through $\frac{1}{2^{n-1}}$ part of one face).
Taking all together
$\displaystyle I(n)=\int_{[0,1]^{n-1}}\frac{1}{(1+{x_1}^2+{x_2}^2+…+{x_{n-1}}^2)^{n/2}}dx_1 … dx_{n-1}\tag*{}$
$\displaystyle=\frac{1}{2n\,2^{n-1}}\,\frac{2\pi^{n/2}}{\Gamma\big(n/2\big)}=\frac{(\sqrt\pi)^n}{2^n\Gamma\Big(\frac{n}{2}+1\Big)}\tag*{}$
For $n=3$ and $n=4$ we get
$\displaystyle I(3)=\int_{[0,1]^2}\frac{1}{(1+x^2+y^2)^{3/2}}dx\,dy=\frac{\pi}{6}\tag*{}$
$\displaystyle I(4)=\int_{[0,1]^3}\frac{1}{(1+x^2+y^2+z^2)^2}dx\,dy\,dz=\frac{\pi^2}{32}\tag*{}$
A: Start from $\int_0^\infty \lambda^{(n+1)\over 2} e^{-\lambda x}d\lambda= x^{-{(n+3)\over 2}} \int_0^\infty u^{(n+3)\over 2}e^{-u}du$
Then replace $e^{-\lambda x}$ with $e^{-\lambda (1+\sum x^2_k)}$
Can you take it from there?
