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: $$\int_0^a dx_1 \int_0^{a-x_1} dx_2 \ldots \int_0^{a-x_1 - \ldots - x_{a-2}} dx_{a-1}=\frac{a^{a-1} }{(a-1)!}$$
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.