Can someone help me to compute this integral with a delta function I don't know how to compute this integral:
$$\int_{0}^{\infty}  \prod_{i=1}^a dx_i \,\delta \left(\sum_{i=1}^a x_i - a\right)$$
The result should be:
$$\frac{a^{a-1} }{(a-1)!}$$
Thanks very much for helping!
edit:
Thanks to the link I am one step further:
$$\int_0^\infty dx_a\delta \left(\sum_{i=1}^a x_i - a\right)=1$$
if
$$x_a=a-\sum_{i=1}^{a-1}x_i\geq 0\\
\Leftrightarrow \quad \sum_{i=1}^{a-1}x_i \leq a$$
So:
$$\int_{0}^{\infty}  \prod_{i=1}^a dx_i \,\delta (\sum_{i=1}^a x_i - a)=\int_0^a dx_1 \int_0^{a-x_1} dx_2 \ldots \int_0^{a-x_1 - \ldots - x_{a-2}} dx_{a-1}$$
But how do I now show:
$$\int_0^a dx_1 \int_0^{a-x_1} dx_2 \ldots \int_0^{a-x_1 - \ldots - x_{a-2}} dx_{a-1}=\frac{a^{a-1} }{(a-1)!}$$
 A: Let $n\in\mathbb{N}$ and $a>0$. Then, taking $n+1$ integrals and evaluating the innermost (the one over $x_{n+1}$), we get
$$
\int_{0}^{\infty} dx_1
\cdots
\int_{0}^{\infty} dx_n
\int_{0}^{\infty} dx_{n+1}
\,\delta(x_1+x_2+\cdots+x_n+x_{n+1}-a)
\\=
\int_{0}^{\infty} dx_1
\cdots
\int_{0}^{\infty} dx_n
\int_{-\infty}^{\infty} dx_{n+1} \, H(x_{n+1})
\,\delta(x_1+x_2+\cdots+x_n+x_{n+1}-a)
\\=
\int_{0}^{\infty} dx_1
\cdots
\int_{0}^{\infty} dx_n
\,
H(a-(x_1+x_2+\cdots+x_n))
,
$$
where $H$ is the Heaviside step function.
Now, set
$$
V_n(a) :=
\int_{0}^{\infty} dx_1
\cdots
\int_{0}^{\infty} dx_n
\,
H(a-(x_1+x_2+\cdots+x_n)) 
$$
Then we can create a recursive formula:
$$
V_n(a) = 
\int_{0}^{\infty} dx_1
\left(
\int_{0}^{\infty} dx_2
\cdots
\int_{0}^{\infty} dx_n
\,
H((a-x_1)-(x_2+\cdots+x_n)) 
\right)
\\= 
\int_{0}^{a} dx_1 \, V_{n-1}(a-x_1)
$$
where the upper limit was changed from $\infty$ to $a$ since we should have $a-x_1>0$.
We have
$$
V_1(a) 
= \int_0^\infty dx_1 \, H(a-x_1)
= \int_0^a dx_1
= a \\
V_2(a)
= \int_0^a dx_1 \, V_1(a-x_1)
= \int_0^a dx_1 \, (a-x_1)
= \frac12 a^2 \\
V_3(a)
= \int_0^a dx_1 \, V_2(a-x_1)
= \int_0^a dx_1 \, \frac12(a-x_1)^2
= \frac16 a^3 \\
$$
and so on.
I leave it to you to turn the "and so on" into an induction proof.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{\pars{0,\infty}^{\,\,a}}\
\delta\pars{%
\sum_{i = 1}^{a}x_{i} - a}\prod_{j = 1}^{a}\dd x_{j}}
\\[5mm] = &
\int_{\pars{0,\infty}^{\,\,a}}
\braces{\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}
\exp\pars{\bracks{a - \sum_{i = 1}^{a}x_{i}}s}\,{\dd s \over 2\pi\ic}}
\prod_{j = 1}^{a}\dd x_{j}
\\[5mm] = &
\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}\expo{as}\pars{\int_{0}^{\infty}\expo{-sx}\dd x}^{a}\
{\dd s \over 2\pi\ic} =\
\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}
{\expo{as} \over s^{a}}\,{\dd s \over 2\pi\ic}
\\[5mm] = & \bbx{a^{a - 1} \over \pars{a - 1}!} \\ &
\end{align}
