Complicated partition function emerges from simple constraints I'm working on a statistical model which involves many degrees of freedom $=1...$. Each degree of freedom is described by a gamma distribution with its own parameters, which we will assume to be all different $v_i \sim \Gamma(k_i,\theta_i)$ with $_i>0$, $_i>0$
I want to calculate the joint pdf of all the variables, given their sum:
$$ P(\vec{v}|V) = \frac{1}{Z_V} \Big [ \prod_{i=1}^{S} p(v_i) \Big ] \delta \Big ( \sum_{i=1}^{S}v_i - V \Big ).$$
Now, my degrees o freedom are gamma-distributed: to be rigorous, I have to use incomplete gamma pdfs, in order to make sense of the integral. In my setting $v_i << V$ allowing us, if needed, to send $V \to \infty$.
$$ Z_V = \frac{1}{\prod_{i=1}^{S} \gamma(V,k_i) \theta^{k_i}}\int_{0}^{V} dv_1 \dots \int_{0}^{V} dv_s \prod_{i=1}^{S}  v^{k_i-1}_i e^{-\frac{v_i}{\theta_i}}  \delta \Big ( \sum_{i=1}^{S}v_i - V \Big ) .$$
Using the Fourier representation of the delta function to split the integrals one gets :
$$ Z_V \propto \int_{0}^{V} dv_1 \dots \int_{0}^{V} dv_s \prod_{i=1}^{S}  v^{k_i-1}_i e^{-\frac{v_i}{\theta_i}} \int_{-\infty}^{\infty} dt \  e^{ it( \sum_{i=1}^{S}v_i - V} ).$$
Each term in the productory is the moment generating function of a Gamma, and I end up with this integral
$$ Z_V \propto \int_{-\infty}^{\infty} dt \  e^{ - it V} \frac{1}{\prod_{i=1}^{S} (1-it\theta_i)^{k_i}}. $$
I can not go further with the problem. I tried some tricks, like Feynman and Schwinger parameterization, but it did not helped. Alternatively, one may consider it as the Fourier transform of the product of the generating functions and try to apply the residues theorem, but the $k_i$ s are not integers and, at least to me, is not clear how to deal with "fractional" singularities. Another thing that one may try is to introduce the grand canonical formulation of the problem.
Anyone has an idea about how to tackle this integral?
 A: In the limit $V \to \infty$ you can apply the saddle-point method. We need the stationary points of the action
$S(t) = i t V + \sum_{j=1}^S k_j \log (1-i t \theta_j)$,
that is,
$0 = dS/dt = i V - i \sum_{j=1}^S k_j \theta_j (1-i t \theta_j)^{-1}$
In general this equation needs to be solved numerically, but we can look for a limit in which this becomes simple. The natural guess is that $t$ is small; we can later look for conditions that make this self-consistent. I will assume that the $k$ and $\theta$ are $O(1)$ and leave the scaling of $S$ for later. If $t \ll 1$ then we have
$ 0 = V - \sum_{j=1}^S k_j \theta_j (1 + i t \theta_j + O(t^2))$
$ = V - S \langle k \theta \rangle - i t S \langle k \theta^2 \rangle + O(S t^2)$
where $\langle k \theta^n \rangle \equiv \frac{1}{S} \sum_j k_j \theta_j^n$, so that
$i t = \frac{V - S \langle k \theta \rangle}{S \langle k \theta^2 \rangle} + \ldots$
Now if $V \sim S$ then $t = O(1)$ and in general the ansatz is not self consistent. Also, if $V \ll S$ or $S \ll V$ then again the ansatz is not self consistent. However, we see that if we tune $S$ or $V$ such that $V - S \langle k \theta \rangle \ll S$ then obviously $t \ll 1$, and then the ansatz is self consistent. Whether or not this gives a good approximation depends on how large $S$ and $V$ are for your problem.
In any case, once you have the saddle-point value $t^*$, the saddle point method (alias stationary phase) gives the asymptotics of the integral,
$Z_V \propto e^{-S(t^*)} (d^2 S/dt^2)^{-1/2}|_{t=t^*} + \ldots$
where I am neglecting some prefactors of $2 \pi$.
