# Integral of product of Gamma densities over probability simplex

I want to find a simpler form or closed form for the following integral: $$\int_A \,\prod_{t=1}^T f_{\Gamma(a,\theta_t)}(x_t) \,d\mathbf{x}$$ where $A$ is the simplex $$A = \{\mathbf{x} \in \mathbb{R}^T : \sum_{t=1}^T x_t = 1 \hbox{ and } x_t \geq 0\}$$ and $f_{\Gamma(a,\theta_t)}$ is the Gamma density $$f_{\Gamma(a,\theta_t)}(x_t) = \theta_t^a x_t^{a-1} \exp(-x_t \theta_t) / \Gamma(a) .$$ We require that $a$ and $\theta_t$ are positive real numbers.

It seems that this shouldn't be too hard, but I've really been stumped. I have tried various methods from the paper by Wolpert and Wolf mentioned in this blog post - but when I follow the algorithm described in that paper, I end up running into even harder analytical difficulties at the end, when it is required to take an inverse Laplace transform of some product.

I feel that there must be a lower-tech way to do this! Thanks in advance for any help.

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Indeed, your integral is $g(1)$ where the Laplace trnsform $L(g)$ of $g$ is such that, for every positive $s$, $$L(g)(s)=\prod_{t=1}^T\left(\frac{\theta_t}{\theta_t+s}\right)^a.$$ For some simple values of $a$, one can decompose this rational fraction of $s$ into simple elements and inverse each term. If I am not mistaken, when $a=1$ and all the $\theta_t$ are distinct, the result is $$g(1)=\sum_t\theta_t\mathrm{e}^{-\theta_t}\prod_{i\ne t}\frac{\theta_i}{\theta_i-\theta_t}.$$ Unfortunately, I do not know a simple way of inverting $L(g)$ in the general case.