$$\int_{-\infty}^{\infty}\frac{\sin(x)}{x} \mathrm{e}^{i \alpha x} \mathrm{d}x = \pi$$ 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?
On
$\sin(x)/x$ 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.
On
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 $$ \int_{-\infty}^{\infty}\frac{\sin x}{x}\mathrm{e}^{iax}\,dx=\frac{1}{2} \big(\sgn(a+1)-\sgn(a-1)\big) \int_{-\infty}^{\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2} \big(\sgn(a+1)-\sgn(a-1)\big). $$
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\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, $$\color{#00f}{\large% \int_{-\infty}^{\infty}{\sin\pars{x} \over x}\expo{\ic\alpha x}\,\dd x =\left\lbrace% \begin{array}{ccl} 0 & \mbox{if} & \verts{\alpha} > 1 \\[2mm] \pi & \mbox{if} & \verts{\alpha} < 1 \\[2mm] {\pi \over 2} & \mbox{if} & \alpha = \pm 1 \end{array}\right.\,,\qquad\qquad\alpha\ \in\ {\mathbb R}} $$