Find the Fourier transform of $\dfrac{\sin(x)}{x}$ "Let $\chi : \mathbb{R}\to \{0, 1\}$ be the characteristic function of the interval $[−1, 1]$ and let $f(x)=\sin(x)/x$.


*

*find the Fourier transform of $\chi$,

*find the Fourier transform of $f$,

*compute the $\int_{-\infty}^\infty f^2$ "


I have no solutions for this problem, this is what I got so far: 
1: Simply using my definition of the Fourier transform gives:
\begin{align*} 
(2\pi)^{1/2} \int_{-\infty}^\infty \chi(x) \exp(ikx) dx 
&= (2\pi)^{1/2} \int_{-1}^1 \exp(ikx) dx\\ 
&= \frac{(2\pi)^{1/2}}{ik} (\exp(ik)-\exp(-ik))
\end{align*}
2: using the definition again gives:
$$f^{(k)}=(2\pi)^{1/2} \int_{-\infty}^\infty \frac{\sin(x)}{x} \exp(ikx) dx $$
then I try to rewrite the sine with Euler so that $\sin(x)=(\exp(ix)-\exp(-ix))/2i$.
using this in the integral gives:
$$
f^{(k)} = \frac{(2\pi)^{1/2}}{(2i)}
          \int_{-\infty}^\infty \frac{\exp(ix)-\exp(-ix)}{x} \exp(ikx) dx $$
Then I can divide the integral into two integrals but I am stuck here anyway, I don't know how to calculate the Integral.
3: maybe I can do this one if I can do the second.
 A: I write the Fourier transform as
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: \frac{\sin{x}}{x} e^{i k x} $$
Consider, rather, the integral
$$ \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{i x}-e^{-i x}}{x} e^{i k x} $$
$$ = \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{i (1+k) x}}{x} - \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{-i (1-k) x}}{x} $$
Consider the following integral corresponding to the first integral:
$$\oint_C dz \: \frac{e^{i (1+k) z}}{z} $$
where $C$ is the contour defined in the illustration below:

This integral is zero because there are no poles contained within the contour.  Write the integral over the various pieces of the contour:
$$\int_{C_R} dz \: \frac{e^{i (1+k)z}}{z}  + \int_{C_r} dz \: \frac{e^{i (1+k) z}}{z}  + \int_{-R}^{-r} dx \: \frac{e^{i (1+k) x}}{x} + \int_{r}^{R} dx \: \frac{e^{i (1+k) x}}{x} $$
Consider the first part of this integral about $C_R$, the large semicircle of radius $R$:
$$\int_{C_R} dz \: \frac{e^{i (1+k)z}}{z} = i \int_0^{\pi} d \theta e^{i (1+k) R (\cos{\theta} + i \sin{\theta})}  $$
$$  = i \int_0^{\pi} d \theta e^{i (1+k) R \cos{\theta}} e^{-(1+k) R \sin{\theta}} $$
By Jordan's lemma, this integral vanishes as $R \rightarrow \infty$ when $1+k > 0$.  On the other hand,
$$ \int_{C_r} dz \: \frac{e^{i (1+k) z}}{z} = i \int_{\pi}^0 d \phi \: e^{i (1+k) r e^{i \phi}} $$
This integral takes the value $-i \pi$ as $r \rightarrow 0$.  We may then say that
$$\begin{align} & \int_{-\infty}^{\infty} dx \: \frac{e^{i (1+k) x}}{x} = i \pi & 1+k > 0\\ \end{align}$$
When $1+k < 0$, Jordan's lemma does not apply, and we need to use another contour.  A contour for which Jordan's lemma does apply is one flipped about the $\Re{z}=x$ axis.  By using similar steps as above, it is straightforward to show that
$$\begin{align} & \int_{-\infty}^{\infty} dx \: \frac{e^{i (1+k) x}}{x} = -i \pi & 1+k < 0\\ \end{align}$$
Using a similar analysis as above, we find that
$$\int_{-\infty}^{\infty} dx \: \frac{e^{-i (1-k) x}}{x} = \begin{cases} -i \pi & 1-k < 0 \\ i \pi & 1-k >0 \\ \end{cases} $$
We may now say that
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: \frac{\sin{x}}{x} e^{i k x} = \begin{cases} \pi & |k| < 1 \\ 0 & |k| > 1 \\ \end{cases} $$
To translate to your definition of the FT, divide the RHS by $\sqrt{2 \pi}$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{}$
\begin{align}
{\sin\pars{x} \over x}
&=\half\int_{-1}^{1}\expo{\ic kx}\,\dd k
=\int_{-\infty}^{\infty}
\bracks{\color{#c00000}{\pi\,\Theta\pars{1 - \verts{k}}}}
\expo{\ic kx}\,{\dd k \over 2\pi}
\end{align}
where $\ds{\Theta\pars{t}}$ is the
Heaviside Step Function.

Then, $\ds{\color{#c00000}{\pi\,\Theta\pars{1 - \verts{k}}}}$ is the
Fourier Transform of $\ds{\sin\pars{x} \over x}$:
$$
\color{#c00000}{\large\pi\,\Theta\pars{1 - \verts{k}}}
=\color{#66f}{\large\left\lbrace\begin{array}{lcl}
\pi & \mbox{if} & \verts{k} < 1
\\[2mm]
0 & \mbox{if} & \verts{k} > 1
\end{array}\right.}
$$

A: Hint: Notice that your expression for the transform of $\chi$ and the integrand in your transform for $f$ look eerily similar.
A: $f(x)=\frac{\sin\pi x}{\pi x}$ is the inverse Fourier Transform of the so called low pass filter, which is a $rect$ function (a normed pulse symmetric to zero ordinate):
$$\int_{-\infty}^\infty \frac{\sin(\pi x)}{\pi x} \, dx = \mathrm{rect}(0) = 1$$
which is a special case of the continuous
$$\int_{-\infty}^\infty \mathrm{sinc}(t) \, e^{-i 2 \pi f t}\,dt = \mathrm{rect}(f)$$
You may also try it via the Euler relation:
$$\frac{\sin(x)}{x} = \prod_{n=1}^\infty \cos\left(\frac{x}{2^n}\right)$$
see for instance >>> here page 96 or other literature (quite straight forward). Another example here >>>
I think the problem is that you were missing the term sinc when searching so just google this against Fourier and rect function.
By the way a nice vid over here >>>
I just saw you asked also for the Fourier of the square. The procedure is likely and you will get instead of a $rect$ function a triangular $tri$ function see also here >>> Again this is standard literature and just google sinc squared against trig function and you will endless reference how to calculate it.
Resume $sinc$ and $rect$ are paired and $sinc^2$ and $trig$ are paired.
