Mehta integral

Remarks

The formula follows from the Selberg integral by a limit. Put $\alpha = \beta = \lambda$ and $t_i = \frac12 + \frac{x_i}{2\sqrt{2\lambda}}$; then $(t_i(1 - t_i))^{\lambda - 1}$ tends, after rescaling, to $e^{-x_i^2/2}$ as $\lambda \to \infty$, and Stirling's formula evaluates the limit of the right side. For $\gamma = 0$ both sides equal $1$, by the Gaussian integral. Example: $n = 2$ and $\gamma = 1$. Then $x_1 - x_2$ is normal with mean $0$ and variance $2$ under the Gaussian weight, so the left side is $\mathbb{E}[(x_1 - x_2)^2] = 2$, and the right side is $\frac{\Gamma(2)}{\Gamma(2)} \cdot \frac{\Gamma(3)}{\Gamma(2)} = 2$. The bound $\operatorname{Re} \gamma > -\frac1n$ is needed: for smaller $\gamma$ the singularity where all $x_i$ coincide makes the integral diverge.

Going deeper

For a student

For $\gamma = \frac12, 1, 2$ the integrand is, up to the Gaussian factor, the joint density of the eigenvalues of the Gaussian orthogonal, unitary and symplectic ensembles of random matrices. So the formula gives the normalizing constants of these densities. It is the Gaussian member of a family of integrals over root systems, the Macdonald–Mehta integrals, of which this is the case of type $A_{n-1}$.