Integrate: $\int_0^1 ||\vec x||^{-m}\mathrm{d}\vec{x}$ Let $m,n$ be two positive integers with $0 < m < n$. 
Can we integrate this:
$$I = \int_0^1 \mathrm{d}x_1 \dots \int_0^1 \mathrm{d}x_n  \left(\sum_{i=1}^n x_i^2\right)^{-m/2}$$
If a closed analytical expression is not possible, what I really need is to evaluate a large deviation limit of the form:
$$\lim_{n\rightarrow\infty} \frac{1}{n} \log [\Gamma(n/2) I]$$
where it is assumed that the ratio $m/n$ remains fixed. I introduce the factor $\Gamma(n/2)$ so that the argument to the log has simple exponential growth.
Related: Integrate: $\int_{-\infty}^\infty \exp(-||\vec x||^2) ||\vec x||^{-m}\mathrm{d}\vec{x}$.
 A: An upper bound for $n>m$,
$$ \int_{[0,1]^n} \frac1{|x|^m} dx \le \frac1{2^n}\int_{B(0,\sqrt2)} \frac1{|x|^m} dx = \frac {C_n} {2^n} \int_0^{\sqrt 2} r^{n-1-m} dr = \frac{C_n}{2^n(n-m)}\sqrt{2}^{n-m} $$
$C_n$ is the area of the unit sphere (coming from the integral in angles),
$ C_n = 2\pi^{n/2}/\Gamma (n/2)$. I think it should be clear from here how to get an upper bound on $\frac {\log I}n$. A very similar lower bound is also possible by similarly cutting up $B(0,1)$ into $2^n$ slices(one cut along each hyperplane $\{x_i=0\})$, one of which is contained in $[0,1]^n$.
A: According to https://www.davidhbailey.com/dhbpapers/boxintegrals.pdf, Eq. (33), we have the following exact relation:
$$\begin{aligned}
I(m,n) &= \int_0^1 \mathrm{d}x_1 \dots \int_0^1 \mathrm{d}x_n \left(\sum_{i=1}^n x_i^2\right)^{-m/2} \\ 
&= \frac{2^{1 - n} \pi^{n/2}}{\Gamma(m/2)} \int_0^{\infty}
\frac{[\mathrm{erf}(u)]^n}{u^{n-m+1}} \mathrm d u
\end{aligned}$$
For large $m,n$ and fixed $\alpha=m/n$, we can evaluate the integral by Laplace's method. 
$$\int_0^{\infty} \frac{[\mathrm {erf} (u)]^N}{u^{N - M + 1}} \mathrm d u \asymp \exp \{ N [\ln \mathrm{erf} (u^{\ast}) - (1 - \alpha) \ln u^{\ast}] \}$$
where $u^*$ maximizes $\ln \mathrm{erf}^{} (u) - (1 - \alpha) \ln u$ with respect to $u\ge0$. 
The notation $a_n\asymp b_n$ from large deviation theory means that $\lim (1/n)\ln a_n = \lim(1/n)\ln b_n$.
Differentiating we find that $u^{\ast}$ is the root of the equation:
$$1-\frac{2u\mathrm e^{-u^2}}{\sqrt{\pi} \mathrm{erf} (u)} = \alpha $$
