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: $\int_{-\infty}^\infty \exp(-||\vec x||^2) ||\vec x||^{-m}\mathrm{d}\vec{x}$.

  • It can certainly be estimated easily in polar coordinates. – T. Bongers Nov 29 at 20:28
  • @T.Bongers So no exact analytical formula? – becko Nov 29 at 20:29
  • I wouldn't be surprised if there is one... but depending on the application, do you really need one, or is an estimate good enough? – T. Bongers Nov 29 at 20:29
  • @T.Bongers I need an analytical formula. But I will then take the limit $m,n \rightarrow\infty$ (with a fixed ratio $m/n$). So in reality I only require an asymptotic estimate. – becko Nov 29 at 20:31
  • @T.Bongers Specifically I'll be satisfied by a large deviation limit $\lim_{n\rightarrow\infty} (1/n) \ln I$, where $I$ is the above integral and the ratio $m/n$ is fixed. – becko Nov 29 at 21:05

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 $[0,1]^n$.

  • Your answer made me realize that I need to introduce a factor $\Gamma(n/2)$ to get simple exponential growth. Having done that, your technique produces lower and upper bounds that have different rates of exponential growth. Therefore these bounds are not tight enough to produce a large deviation principle. Or am I missing something? Thanks! – becko Dec 6 at 17:12
  • @becko No, I don't think you're missing anything. I haven't thought of a better approximation – Calvin Khor Dec 6 at 17:15
  • @becko While I don't think this resolves your question, this paper undoubtedly goes far further than the basic computation I made projecteuclid.org/download/pdf_1/euclid.em/1317758103 – Calvin Khor Dec 6 at 18:18
  • 1
    Thanks. According to davidhbailey.com/dhbpapers/boxintegrals.pdf, we have that $I(n,m) ~ (n/3)^{-m/2}$ for large $n$ and fixed $m$. Unfortunately they say nothing about the large $n,m$ limit with fixed ratio $m/n$. – becko Dec 6 at 18:27
  • @becko I'm very pleased to see divergence theorem useful, my small scribblings didn't bring me to that conclusion. And I would guess that a clever interchange of integrals doesn't help much either. Good luck – Calvin Khor Dec 6 at 18:32
up vote 1 down vote accepted

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:

$$1-\frac{2u\mathrm e^{-u^2}}{\sqrt{\pi} \mathrm{erf} (u)} = \alpha $$

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