# 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.

• It can certainly be estimated easily in polar coordinates. – T. Bongers Nov 29 '18 at 20:28
• @T.Bongers So no exact analytical formula? – becko Nov 29 '18 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 '18 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 '18 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 '18 at 21:05

## 2 Answers

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 '18 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 '18 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 '18 at 18:18
• 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 '18 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 '18 at 18:32

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:

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