Multivariable integral, probably related to the gamma function Let $x=\left[ x_1,x_2,..,x_n \right]^{T}$ represent the vector of all variables and $D$ be a diagonal matrix, the question is to integrate or give an approximate answer:
$\idotsint_{[0,\infty]^{n}} (x^{T}Dx)^{\alpha-1}exp(-x^{T}Dx) \,dx_1 \dots dx_n$
 A: The integral is easy, let me show you why:
$$
\idotsint_{[0,\infty]^{n}} (x^{T}Dx)^{\alpha-1}exp(-x^{T}Dx) \,dx_1 \dots dx_n
$$
let's first express:
$$
x^TDx=\sum_{i=1}^n d_i\,x_i^2
$$
where $d_i = D_{i,i}$.
To have a convergent integral every $d_i$ must be non negative.
hence we can use an alternate coordinate system:
$$
\eta_i=\sqrt{d_i}x_i
$$
and $x^T D x$ becomes:
$$
x^TDx=\sum_{i=1}^n d_i\,x_i^2=\sum_{i=1}^N \eta_i^2=\eta^T\eta
$$
and the integral using substitution obviously becomes:
$$
I_k\equiv \idotsint_{[0,\infty]^{n}} (\eta^T\eta)^k exp(-\eta^T\eta) \,\frac{d\eta_1 \dots d\eta_n}{\sqrt{\prod_{i=1}^n d_i}}
$$
Now look at this technique : Differentiation by  a parameter
$$
T(\beta)\equiv \idotsint_{[0,\infty]^{n}} exp(-\beta\eta^T\eta) \,d\eta_1 \dots d\eta_n
$$
$$
\frac{\partial^k}{{\partial \beta}^k}T(\beta)=\idotsint_{[0,\infty]^{n}}(-\eta^T\eta)^k exp(-\beta\eta^T\eta) \,d\eta_1 \dots d\eta_n
$$
so
$$
I_k=\frac{(-1)^k}{\sqrt{\prod_{i=1}^n d_i}}\lim_{\beta\to1}\frac{\partial^k}{{\partial \beta}^k}T(\beta)
$$
now all that there is to it is calculating $T(\beta)$
$$
T(\beta)\equiv \idotsint_{[0,\infty]^{n}} exp(-\beta\eta^T\eta) \,d\eta_1 \dots d\eta_n=\left(\int_0^{\infty} exp(-\beta z^2) \,dz\right)^n=\left(\frac{1}{2}\sqrt{\frac{\pi}{\beta}}\right)^n
$$
Try and complete the work, else i'll continue to help :)
