Multiple integral related to zeta function I am trying to calculate the following integral
$$\int_V \frac{d^d\vec{r}}{e^{x_1+...+x_d}-1},$$
where $V=[0,\infty)^d$ and $\vec{r}=(x_1,...,x_d)$. I know that the result should be related to the Riemann zeta function, but I do not see how to do it quickly and elementary (i.e. without the knowledge of all possible relations for zeta function). Any sugestion or hint ?
 A: Expand the integrand in a geometric series
\begin{equation*}
\dfrac{1}{e^{x_1+x_2+\ldots +x_d}-1} = \sum_{k=1}^{\infty}e^{-k(x_1+x_2+\ldots +x_d)} = \sum_{k=1}^{\infty}e^{-kx_1}\cdot e^{-kx_2}\cdot\ldots \cdot e^{-kx_{{d}}}.
\end{equation*}
For $d \ge 2$ the multiple integral over $V$  is given by
\begin{equation*}
\sum_{k=1}^{\infty}\left(\int_{0}^{\infty}e^{-kx_1}\, dx_1\right)^{d} =\sum_{k=1}^{\infty}\dfrac{1}{k^d} = \zeta(d).
\end{equation*}
A: By Fubini's theorem
$$\int_{(0,+\infty)^d}\frac{x_1}{e^{x_1+\ldots+x_d}-1}\,d\mu_d = \int_{(0,+\infty)^{d-1}}\int_{0}^{+\infty}\frac{x_1}{e^{x_1+\ldots+x_d}-1}\,dx_1\,d\mu_{d-1} $$
and the last integral equals
$$ \int_{(0,+\infty)^{d-1}}\text{Li}_2\left(e^{-(x_2+\ldots+x_d)}\right)\,d\mu_{d-1}=\int_{(0,1)^{d-1}}\frac{\text{Li}_2(v_2\cdots v_d)}{v_2\cdots v_d}\,dv_2\cdot dv_d$$
or
$$ \int_{(0,1)^{d-2}}\frac{\text{Li}_3(v_3\cdots v_d)}{v_3\cdots v_d}dv_3\cdots dv_d = \int_{0}^{1}\frac{\text{Li}_d(w)}{w}\,dw=\text{Li}_{d+1}(1)=\color{red}{\zeta(d+1)}.$$
With the same technique we have:
$$\int_{(0,+\infty)^d}\frac{d\mu_d}{e^{x_1+\ldots+x_d}-1} = \text{Li}_d(1)=\color{red}{\zeta(d)}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\left.\int_{\large\left[0,\infty\right)^{d}}
{\dd^{d}\vec{r} \over \expo{x_{1} + \cdots + x_{d}} - 1}\,
\right\vert_{\ d\ \geq\ 2}
\\[5mm] = &\
\int_{\large\left[0,\infty\right)^{d}}
\pars{\int_{0}^{\infty}
\braces{\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}\exp\pars{\bracks{z - x_{1} - \cdots - x_{d}}s}\,{\dd s \over 2\pi\ic}}
{\dd z \over \expo{z} - 1}}\dd^{d}\vec{r}
\\[5mm] = &\
\int_{0}^{\infty}{1 \over \expo{z} - 1}
\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}\expo{zs}
\pars{\int_{0}^{\infty}\expo{-xs}\,\dd x}^{d}{\dd s \over 2\pi\ic}\,\dd z =
\int_{0}^{\infty}{1 \over \expo{z} - 1}
\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}{\expo{zs} \over s^{d}}
{\dd s \over 2\pi\ic}\,\dd z
\\[5mm] = &\
\int_{0}^{\infty}{1 \over \expo{z} - 1}{z^{d - 1} \over \pars{d - 1}!}\,\dd z =
{1  \over \pars{d - 1}!}\sum_{n = 0}^{\infty}
\int_{0}^{\infty}z^{d - 1}\expo{-\pars{n + 1}z}\,\dd z
\\[5mm] = &\
{1  \over \pars{d - 1}!}
\sum_{n = 0}^{\infty}{\pars{d - 1}! \over \pars{n + 1}^{d}} =
\bbx{\zeta\pars{d}}
\end{align}
