Integrating a Dirac delta of a sum 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.
 A: 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 $I_M=\frac{1}{(M-1)!}$.
A: Make the following change of variables:
$y_1 = f_1$
$y_2 = f_1 + f_2$
...
$y_k = \sum_{i=1}^k f_i$
...
$y_M  = \sum_{i=1}^M f_i$
and notice that $y_1<y_2<...<y_M$.
The Jacobian determinant of the above transformation is 1, therefore,
$\int \prod_{i=1}^M \ df_i\  \Theta(f_i) \ \delta\left(\sum_{i=1}^Mf_i-1\right)=  \int \prod_{i=1}^M \ dy_i\  \Theta(y_i) \ \delta\left(y_M-1\right)$
Since  $y_1<y_2<...<y_M$, the latter series of integrals  can be written as 
$\int^{1}_0 dy_M \ \delta\left(y_M-1\right)\int_0^{y_M} dy_{M-1} \int_0^{y_{M-1}} dy_{M-2} ... \int_0^{y_{k+1}} dy_{k} ... \int_0^{y_3} dy_2 \int_0^{y_2} dy_1 $
$= \int_0^1 dy_{M-1} \int_0^{y_{M-1}} dy_{M-2} ... \int_0^{y_{k+1}} dy_{k} ... 
\int_0^{y_3} dy_2 \ y_2$
$ =  \int_0^1 dy_{M-1} \int_0^{y_{M-1}} dy_{M-2} ... \int_0^{y_{k+1}} dy_{k} ... 
\int_0^{y_4} dy_3 \frac{y_3^2}{2}$ 
and integrating step by step we have
$\int_0^1 dy_{M-1} \int_0^{y_{M-1}} dy_{M-2} ... \int_0^{y_{k+1}} dy_k \ \frac{ y_{k}^{k-1}}{(k-1)!} = \int_0^1 dy_{M-1}\ \frac{ y_{M-1}^{M-2}}{(M-2)!} =\frac{1}{(M-1)!}  $
Therefore $C = (M-1)!$
and the normalized distribution is
$P({f}) = (M-1)!\int \prod_{i=1}^M \ df_i\  \Theta(f_i) \ \delta\left(\sum_{i=1}^Mf_i-1\right)$
Q.E.D.
