TITLE: Integrate: $\int_0^1 \mathrm{d}x_1 \int_0^1 \mathrm{d}x_2 \ldots \int_0^1 \mathrm{d}x_n \delta\left( \sum_{i=1}^n k_i x_i \right)$
QUESTION [8 upvotes]: Is there a closed form formula for this integral:
$$\int_0^1 \mathrm{d}x_1 \int_0^1 \mathrm{d}x_2  \ldots \int_0^1 \mathrm{d}x_n \delta\left( \sum_{i=1}^n k_i x_i \right)$$
where $\delta(x)$ is the Dirac delta function, and the $k_i$ are real numbers.
Here's how far I've got. Doing the integral on $k_n$ gives:
$$\frac{1}{|k_n|}\int_0^1 \mathrm{d}x_1 \int_0^1 \mathrm{d}x_2  \ldots \int_0^1 \mathrm{d}x_{n-1} \left[0\leq -\frac{1}{k_n}\sum_{i=1}^{n-1} k_i x_i \leq 1 \right]$$
where the brackets are Iverson brackets:
$$\left[P\right]=
\begin{cases}
1 & \text{if }P\text{ is true }\\
0 & \text{otherwise}
\end{cases}$$
From here, I could calculate the appropriate limits of integration of $x_{n-1}$. But that seems like too much work, and I am not sure that I will get a closed form formula eventually as I keep going to $x_{n-2}$ and so on. I wonder if there is a simpler approach...

REPLY [4 votes]: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
&\color{#0000ff}{\large\int_{0}^{1}\dd x_{1}\int_{0}^{1}\dd x_{2}\ldots\int_{0}^{1}\dd x_{n}\,\delta\pars{\sum_{i = 1}^{n}k_{i}x_{i}}}
\\[3mm]&=\int_{0}^{1}\dd x_{1}\int_{0}^{1}\dd x_{2}\ldots\int_{0}^{1}\dd x_{n}
\int_{-\infty}^{\infty}\expo{\ic q\sum_{i = 1}^{n} k_{i}x_{i}}\,{\dd q \over 2\pi}
\\[3mm]&=
\int_{-\infty}^{\infty}{\dd q \over 2\pi}\prod_{i = 1}^{n}\int_{0}^{1}
\expo{\ic qk_{i}x_{i}}\,\dd x_{i}
=
\int_{-\infty}^{\infty}{\dd q \over 2\pi}\prod_{i = 1}^{n}
{\expo{\ic qk_{i}} - 1 \over \ic qk_{i}}
=
\int_{-\infty}^{\infty}{\dd q \over 2\pi}\prod_{i = 1}^{n}
\expo{\ic qk_{i}/2}\,{2\ic\sin\pars{qk_{i}/2} \over \ic qk_{i}}
\\[3mm]&=\color{#0000ff}{\large%
{2^{n - 1} \over \pi}\int_{-\infty}^{\infty}{\dd q \over q^{n}}\prod_{i = 1}^{n}
{\expo{\ic qk_{i}/2}\sin\pars{qk_{i}/2} \over k_{i}}}
\end{align}
I don't see any further reduction unless we know something else about the $\braces{k_{i}}$.