# A slight generalization of Mehta's integral.

I am trying to find the value of following integral

$$\int_{-\infty}^{\infty}\dots\int_{-\infty}^{\infty}\prod_{i=1}^ne^{-\frac{t_i^2}{2}+\alpha_i t_i}\prod_{1\le i<j\le n}\left|t_i-t_j\right|^{2\gamma}dt_1\dots dt_n$$ where, $\alpha_i$'s are complex numbers and $\gamma$ is a positive real number. When $\alpha_i$'s are all zero then the integral reduces to Mehta's integral and its value is $(2\pi)^{n/2}\displaystyle\prod_{j=1}^n\frac {\Gamma(1+j\gamma)}{\Gamma (1+\gamma)}$. Can anyone suggest some approach to compute the value of above integral for nonzero $\alpha_i$'s ?

I am also very interested in this type of integrals. For $\gamma$ positive integer, and $\alpha_i\in\mathbb{R}$ (but perhaps the latter condition may be relaxed), I feel that Heine's theorem for (hyper)determinants might be helpful (see e.g. eq. (13) in http://arxiv.org/pdf/0912.1228.pdf). If you call your integral $I(\alpha_1,\ldots,\alpha_n)$, this is symmetric under the exchange $\alpha_i\to\alpha_j$, so it should be equal to $$I(\alpha_1,\ldots,\alpha_n)=\frac{1}{n!}\sum_{\sigma}\int_{-\infty}^\infty\cdots\int_{-\infty}^\infty \prod_{i=1}^n d t_i\ e^{-\frac{t_i^2}{2}+a_{\sigma(i)}t_i}\prod_{i<j}|t_i-t_j|^{2\gamma}\ ,$$ where $\sigma$ is a permutation of the first $n$ integers. Then you can rewrite your integral introducing a permanent $$I(\alpha_1,\ldots,\alpha_n)=\frac{1}{n!}\int_{-\infty}^\infty\cdots\int_{-\infty}^\infty \prod_{i=1}^n d t_i\ e^{-\frac{t_i^2}{2}}\mathrm{perm}(e^{a_i t_j})\prod_{i<j}|t_i-t_j|^{2\gamma}\ .$$ This should be then written as a hyperdeterminant (again, only for $\gamma$ positive integer), or as a sum over permutations of conventional determinants. For example, for $\gamma=1$ this should read $$I(\alpha_1,\ldots,\alpha_n)=\sum_{\sigma}\det\left(\int_{-\infty}^\infty dx\ e^{-x^2/2+a_{\sigma(i)}x}x^{i+j-2}\right),$$ (if I am not mistaken). The integral can be computed, and perhaps some progress on the determinant evaluation can be achieved. It is admittedly not a very efficient nor complete solution, but I hope it might help.