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

Question

Let $m,n$ be two positive integers with $0 < m < n$.

Can we integrate this:

$$I = \int_{-\infty}^\infty \mathrm{d}x_1 \dots \int_{-\infty}^\infty \mathrm{d}x_n \left(\sum_{i=1}^n x_i^2\right)^{-m/2} \exp\left(-\sum_{i=1}^n x_i^2\right)$$

If a closed analytical expression is not possible, what I really need is to evaluate a large deviation limit of the form:

$$\lim_{n\rightarrow\infty} \frac{1}{n} \log I$$

where it is assumed that the ratio $m/n$ remains fixed.

Note that $I$ resembles a negative moment of the radious of a multivariate Gaussian.

I asked a very similar question here: Integrate: $\int_0^1 ||\vec x||^{-m}\mathrm{d}\vec{x}$. But note that this integral has different limits and a decaying quadratic exponential. Maybe I have better luck with this variant ;)

Answer 1

Consider the homomorphism $x \mapsto \left( \|x\|, \dfrac{x}{\|x\|} \right)$ between $\mathbf{R}^d \setminus \{0\}$ and $(0, \infty) \times \mathbf{S}_{d-1}.$ By the change of measures formula, the integral becomes $$\begin{align*} \int\limits_0^\infty dr \int\limits_{\mathbf{S}_{d-1}} d\sigma_d\ e^{-r^2}r^{-m} r^{d-1} &= \sigma_{d-1}(\mathbf{S}_{d-1}) \int\limits_{0}^\infty dr\ e^{-r^2} r^{d-m-1} \\ &= \dfrac{\sigma_{d-1}(\mathbf{S}_{d-1})}{2} \int\limits_0^\infty dr\ e^{-r} r^{\frac{d-m-2}{2}} \\ &= \dfrac{\sigma_{d-1}(\mathbf{S}_{d-1})}{2} \Gamma\left(\dfrac{d-m}{2}\right). \square \end{align*}$$

Answer 2

Introduce the spherical coordinates $r, \theta, \xi_1, \ldots, \xi_{N - 2}$, $$ x_1 = r \cos (\xi_1), \quad x_2 = r \sin (\xi_1) \cos (\xi_2), \quad \ldots, \quad x_N = r \sin (\xi_1) \ldots \sin (\xi_{N - 2}) \sin (\theta) $$ where $0 \leqslant \xi_i \leqslant \pi$, $0 \leqslant \theta \leqslant 2 \pi$ and $0 \leqslant r < \infty$. The volume element is given by $\mathrm d x_1 \ldots \mathrm d x_N = r^{N - 1} \mathrm d r \mathrm d \sigma$, where $\mathrm d \sigma$ is the unit-sphere surface element,

$$\mathrm d \sigma = \sin^{N - 2} (\xi_1) \ldots \sin (\xi_{n - 2}) \mathrm d \theta \mathrm d \xi_1 \ldots \mathrm d \xi_{N - 2}$$

It follows that: $$\begin{eqnarray*} I & = & \int_{- \infty}^{\infty} \| {\vec x} \|^{- M} \mathrm e^{- \| {\vec x} \|^2 / 2} \mathrm d \vec x\\ & = & \left( \int_0^{\infty} r^{N - M - 1} \mathrm e^{- r^2/2} \mathrm d r \right) \left( \int_{\mathcal{S}} \mathrm d \sigma \right)\\ & = & 2^{(N - M) / 2 - 1} \Gamma \left( \frac{N - M}{2} \right) \frac{2 \pi^{N / 2}}{\Gamma (N / 2)} \end{eqnarray*}$$

where $\mathcal S$ is the unit sphere.