How to calculate this complex definite integral? I'm studying a text and it has claimed that 
$$\int_{\infty}^{\infty}\left(\int_0^1e^{2\xi \pi y i} dy\right)^3 e^{-2\xi \pi i}d\xi = \frac{1}{2}$$
it doesn't seem so hard and i tried to compute it, but i failed!
Is there any hint?
Thanks
 A: Formally, if you write out the cube as a product of three integrals and reverse order of integration, what you really have is
$$\int_0^1 dy_1 \, \int_0^1 dy_2 \,\int_0^1 dy_3 \, \delta(y_1+y_2+y_3-1) $$
where $\delta$ is the Dirac delta function.  This delta confines the integration variables to the surface $y_1+y_2+y_3=1$ in the octant where all three variables are in the interval $[0,1]$.  Thus, the integral of the delta over $y_3$ is over a right triangle of side length $1$ in the $y_1-y_2$ plane, which has area $1/2$.
In general, if one replaces the three by, say, $m$, then the result is $1/(m-1)!$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{-\infty}^{\infty}
\pars{\int_{0}^{1}\expo{2\pi\ic y\xi}
\,\dd y}^{3}\expo{-2\pi\ic\xi}\,\dd\xi}}
\\[5mm] = &\
\int_{-\infty}^{\infty}
\pars{\int_{0}^{1}\expo{\ic y\xi}\dd y}^{3}\expo{-\ic\xi}
\,{\dd\xi \over 2\pi}
\\[5mm] = &\
\int_{-\infty}^{\infty}
\pars{\int_{0}^{1}\expo{\ic x\xi}\dd x}
\pars{\int_{0}^{1}\expo{\ic y\xi}\dd y}
\pars{\int_{0}^{1}\expo{\ic z\xi}\dd z}\ \times
\\[2mm] &\ \expo{-\ic\xi}\,{\dd\xi \over 2\pi}
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}
\delta\pars{x + y + z - 1}\dd x\,\dd y\,\dd z
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1}
\bracks{0 < 1 - y - z < 1}\dd y\,\dd z
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1}
\bracks{y < 1 - z}\dd y\,\dd z
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1 - z}\dd y\,\dd z =
\int_{0}^{1}\pars{1 - z}\dd z = \bbx{1 \over 2} \\ &
\end{align}
