Show or prove that $\int_{-\infty}^{\infty}\frac{\sin(x)}{x} \mathrm{e}^{i \alpha x} \mathrm{d}x = \pi$ $$\int_{-\infty}^{\infty}\frac{\sin(x)}{x} \mathrm{e}^{i \alpha x} \mathrm{d}x = \pi$$
This particular integral rings a bell in our department(Mathematics). It has yet been solved and proved and keeps showing in every third year Complex Analysis Exam. 
An additional condition is that $\alpha$ is real. Assume $\alpha$ is 1, Matlab says the answer is $\pi$ but doesn't say how. Could anyone show me the proof? 
 A: $\sin(x)/x$ is roughly the Fourier transform of a box function, see http://en.wikipedia.org/wiki/Rectangular_function#Fourier_transform_of_the_rectangular_function
Then using the convolution theorem this integral is just the convolution of a box function (of width $\pi$ and height $1$) with the constant $1$ function, which is the result.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\bbox[10px,#ffd]{\int_{-\infty}^{\infty}{\sin\pars{x} \over x}\expo{\ic\alpha x}\,\dd x} =
\int_{-\infty}^{\infty}\bracks{%
\half\int_{-1}^{1}\expo{-\ic kx}
\,\dd k}\expo{\ic\alpha x}\,\dd x
\\[5mm] = &\
\pi\int_{-1}^{1}\dd k\int_{-\infty}^{\infty}\expo{-\ic\pars{k - \alpha}x}
\,{\dd x \over 2\pi} =
\pi\int_{-1}^{1}\delta\pars{k - \alpha}\,\dd k
= \pi\,\Theta\pars{1 - \verts{\alpha}}
\end{align}
However, when $\ds{\verts{\alpha} = 1}$ we can work it out directly from the initial integral as
$$
\int_{-\infty}^{\infty}{\sin\pars{x} \over x}\,\expo{\pm\ic x}\,\dd x=
\int_{-\infty}^{\infty}{\sin\pars{2x} \over 2x}\,\dd x = {\pi \over 2}
$$
Then,
$$\color{#00f}{\large%
\int_{-\infty}^{\infty}{\sin\pars{x} \over x}\expo{\ic\alpha x}\,\dd x
=\left\lbrace%
\begin{array}{ccl}
0 & \mbox{if} & \verts{\alpha} > 1
\\[2mm]
\pi  & \mbox{if} & \verts{\alpha} < 1
\\[2mm]
{\pi \over 2}  & \mbox{if} & \alpha = \pm 1
\end{array}\right.\,,\qquad\qquad\alpha\ \in\ {\mathbb R}}
$$
A: As $\dfrac{\sin x}{x}$ is even then
$$
\int_{-\infty}^{\infty}\frac{\sin x}{x}\mathrm{e}^{iax}\,dx=
\int_{-\infty}^{\infty}\frac{\sin x}{x}\cos ax\,dx=
\int_{-\infty}^{\infty}\frac{\sin (a+1)x-\sin(a-1)x}{2x}\,dx. \tag{1}
$$
Now for $b>0$,
$$
\int_{-\infty}^{\infty}\frac{\sin bx}{x}\,dx\,\,\stackrel{y=bx}=\,\, 
\int_{-\infty}^{\infty}\frac{\sin y}{y}\,dy,
$$
while for $b<0$
$$
\int_{-\infty}^{\infty}\frac{\sin bx}{x}\,dx=-\int_{-\infty}^{\infty}\frac{\sin x}{x}\,dx.
$$
Hence $(1)$ provides
$$
\int_{-\infty}^{\infty}\frac{\sin x}{x}\mathrm{e}^{iax}\,dx=\frac{1}{2}
\big(\sgn(a+1)-\sgn(a-1)\big)
\int_{-\infty}^{\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}
\big(\sgn(a+1)-\sgn(a-1)\big).
$$
