As part of an inference project, I'm normalising a prior distribution which vanishes unless the set of $M$ data points $f_1,\dots, f_M$ satisfies
$$ \sum_{i=1}^M f_i = 1. $$
Accordingly this is encoded with a Delta function in the prior, along with the condition that $0 \leq f_i \leq 1$, as such:
$$ \mathrm{Pr}(f_i) \propto \delta_\mathrm{D}\left(\sum_{i=1}^M f_i - 1\right)\prod_{i=1}^M \Theta(f_i), $$
where $\Theta$ is the Heaviside step function. To normalise we integrate and solve for $C$:
$$ C\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty} \delta_\mathrm{D}\left(\sum_{i=1}^M f_i - 1\right)\prod_{i=1}^M \Theta(f_i) \,\mathrm{d}f_i = 1. $$
Trying to integrate this with a few values of $M$ in Mathematica shows that the integral itself is $(M-1)!$.
What I'm looking for is some insight into why this is. Is the first integral over $f_M$ just the integral of a delta function of $f_M$ plus some constants? So is it $1$? Then the rest of the integrals are just over constants. I can see that you do this $M-1$ times but then I don't see how the $(M-1)!$ comes out.
The integral can be calculated using Laplace transform.
The original integral is $$ I_M=\int_0^\infty\int_0^\infty \delta\left(\sum_{i=1}^M f_i - 1\right) \prod_{i=1}^M \,\mathrm{d}f_i\ . $$ Define an auxiliary integral as $$ I_M(T)=\int_0^\infty\int_0^\infty \delta\left(\sum_{i=1}^M f_i - T\right) \prod_{i=1}^M \,\mathrm{d}f_i\ . $$ Clearly, $I_M=I_M(1)$.
Take the Laplace transform $$ \hat{I}(s)=\int_0^\infty \mathrm{d}T~e^{-sT}I_M(T)=\left[\int_0^\infty\mathrm{d}f~e^{-sf}\right]^M=\frac{1}{s^M}\ , $$ whose inverse Laplace transform is $$ I_M(T)=\frac{T^{M-1}}{(M-1)!} $$ as can be verified by direct substitution. Setting $T=1$, we get $I_M=\frac{1}{(M-1)!}$.