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We know that the Fourier transform of the sinc function is the rectangular function (or top hat). However, I’m at a loss as to how to prove it. Most textbooks and online sources start with the rectangular function, show that

$$\int_{-\infty}^\infty \text{rect}(x)e^{i\omega x}dx=\int_{-1/2}^{1/2}e^{i\omega x}dx=\left.\frac{e^{i\omega x}}{i\omega}\right\vert_{-1/2}^{1/2}=\text{sinc}(\omega/2)$$

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- Integral of Bessel function multiplied with sine

and then just invoke duality and claim that the Fourier transform of the sinc function is the rectangular function. Is there any way of deriving this directly? i.e., starting with the sinc function?

I’ve tried, but I’m not sure as to how to proceed. I know that the sinc is not Lebesgue integrable and only improper Riemann integrable. Some vague recollection of these being important in the Fourier transform hinders my thought process. Can someone clear things up for me?

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Since sinc is an entire function and decays with $1/\omega$, we can slightly shift the contour of integration in the inverse transform, and since there’s no longer a singularity then, we can split the integral in two:

$$\begin{eqnarray}\int_{-\infty}^\infty e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega

&=&

\int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\\\

&=&

\int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}}{i\omega}\mathrm{d}\omega –

\int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\;.

\end{eqnarray}$$

Now we can apply different substitutions to the two parts, $\omega'=\omega(x+\frac{1}{2})$ in the first part and $\omega'=\omega(x-\frac{1}{2})$ in the second part. That transforms the integrand into $e^{i\omega'}/(i\omega')$ in both cases. Now if $x$ lies outside the rectangle, the signs of the factors in the substitutions are the same, so the two integrals stay on the same side of the origin and go in the same direction, and hence yield the same value and cancel to $0$. But if $x$ lies inside the rectangle, then there’s a sign change due to $x-\frac{1}{2}$ but not due to $x+\frac{1}{2}$, so we get

$$\int_{-\infty+\epsilon i}^{\infty+\epsilon i} \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega' +

\int_{\infty-\epsilon i}^{-\infty-\epsilon i} \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega'\;,$$

which is (again using the sufficient decay at infinity)

$$\oint \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega'$$

on a contour that encloses the pole at the origin, and hence the value is $2\pi$.

I think you can have a look at this post:

http://eagle.lamost.org/?p=38679

Every steps for the Fourier Transform is clearly shown.

I think for function $f(x)$ the fourier transform is defined by $$f(t)=\int f(x)e^{-2\pi ixt}dx$$

Now the original function is $$\frac{sin(\pi x)}{\pi x}=\frac{e^{i\pi x}-e^{-i\pi x}}{2i\pi x}$$ hence the transformed one become $$(2i\pi)^{-1}\int \frac{(e^{i\pi(1-2t) x}-e^{-i\pi(1+2t) x})}{ x}dx$$

Hence we need to evaluate $$\int \frac{e^{kix}}{x}dx, k=\pi(\pm 1-2t)$$ and we need to consider function $$g(z)=\frac{e^{ikz}}{z}$$

Consider the indented semi circle on $x$-axis. with $x\rightarrow \infty$ the upper part just evaporated. Hence evaluating the inner circle we have $$\int \frac{e^{kix}}{x}dx=\int^{0}_{\pi}ie^{ikre^{i\phi}}d\phi=?$$

and $$\int \frac{e^{-kix}}{x}dx=\int^{0}_{\pi}ie^{-ikre^{i\phi}}d\phi=?$$

In the case $|t|=1/2$, either left or the right side becomes $1$. Now note $$\frac{1}{2i}\int \frac{e^{ix}-1}{x}=\frac{\pi}{2}$$ hence $$\frac{1}{2i\pi}\int \frac{e^{ikx}-1}{x}=\frac{1}{2}$$

Otherwise the result should be ideally 0. I examined my computation several times, could not find the problem when $|t|<1/2$. If it does transform to the rectangular function then when $|t|<1/2$ it should be 1.

I believe you can see Brad Osgood do it in this video or its successor.

For what it’s worth, I enjoy his lecture style. Very playful, like “I’m about to tell you a **major secret of the universe**.”

$\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{\dd}{{\rm d}}%

\newcommand{\ds}[1]{\displaystyle{#1}}%

\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%

\newcommand{\expo}[1]{{\rm e}^{#1}}%

\newcommand{\ic}{{\rm i}}%

\newcommand{\imp}{\Longrightarrow}%

\newcommand{\pars}[1]{\left( #1 \right)}%

\newcommand{\pp}{{\cal P}}%

\newcommand{\sgn}{{\rm sgn}}%

\newcommand{\ul}[1]{\underline{#1}}%

\newcommand{\verts}[1]{\left\vert #1 \right\vert}$

Following @Juha-Matti Vihtanen definition of ${\rm sinc}\pars{t}$:

\begin{align}

\int_{-\infty}^{\infty}{\rm sinc}\left(t\right)\,\expo{\ic\omega t}\,\dd t

&=

\int_{-\infty}^{\infty}\pars{%

{1 \over 2}\int_{-1}^{1}\expo{\ic\nu\pars{\pi t}}\,\dd\nu}

\expo{\ic\omega t}\,\,\dd t

=

\pi\int_{-1}^{1}\dd\nu\int_{-\infty}^{\infty}

\expo{\ic\pars{\omega + \nu\pi}t}\,{\dd t \over 2\pi}

\\[3mm]&=

\pi\int_{-1}^{1}\delta\pars{\omega + \nu\pi}\,\dd\nu

=

\pi\int_{-1}^{1}{\delta\pars{\nu – \bracks{-\omega/\pi}} \over \pi}\,\dd\nu

=

\Theta\pars{1 – \verts{\,-\,{\omega \over \pi}}}

\\[1cm]&

\end{align}

$$\color{#ff0000}{\Large%

\int_{-\infty}^{\infty}{\rm sinc}\left(t\right)\,\expo{\ic\omega t}\,\dd t

\color{#000000}{\Large\ =\ }

\Theta\pars{\pi – \verts{\omega}}}

$$

Write down the expression for the fourier transform g( x ). differentiate with respect to x . you will get a hitherto meaningless integral. use the -epsilon x^2 trick ( pretty standard, if not known look into the derivation of the fourier transform of the gaussian ). You will see some gaussians appearing. Integrate and see ( by a change of variable ) that the it is zero. Use the fundamental theorem of calculus to do all this business. Now take an interval where it is non-zero say (-1,1) and zero in the rest( because it has to vanish at infinity ). evaluate the constant by the inversion formula. taking nice intervals will give you nice constants.

You can use the Fourier inversion theorem. Recall that $\textrm{sinc}(t)$ is defined by

\begin{eqnarray}

\textrm{sinc}(t) = \begin{cases} \frac{\sin(\pi t)}{\pi t} & , \ t \neq 0 \\

1 & , \ t = 0

\end{cases} .

\end{eqnarray}

I use here the normed version of $\textrm{sinc}(t)$ because there wasn’t other definitions in my studybooks. Recall that Fourier transform is usually defined by improper Riemann-integral of the first kind by

\begin{equation}

\mathscr{F}f(\omega) = \frac{1}{2\pi} \int_{-\infty}^\infty f(t) e^{-i\omega t} dt

\end{equation}

This is the version in my complex analysis book “Mathews & Howell: Complex Analysis for Mathematics and Engineering, 2nd edition”. However, it was assumed that the function $f$ is in $L^1$. Hence we have to calculate the transform an other way. The first thing is to define the transform s.t. it exists for every $\omega \in \mathbb{R}$. It turns out that one such a definition is

\begin{eqnarray}

\mathscr{F}f(\omega) = \frac{1}{2\pi} \int_0^\infty (f(t)e^{-i\omega t} + f(-t)e^{i\omega t}) dt \ ,

\end{eqnarray}

that is an improper Riemann-integral of the first kind. It has the representation

\begin{eqnarray}

\mathscr{F}f(\omega) = \frac{1}{2\pi} PV \int_{-\infty}^\infty f(t) e^{-i\omega t} dt \ ,

\end{eqnarray}

that is seen easily. Now define

\begin{eqnarray}

g(\omega) = \frac{1}{2}(\chi_{[-\pi,\pi]}(\omega) + \chi_{(-\pi,\pi)}(\omega)) \ .

\end{eqnarray}

Then $g\in L^1$, it is piecewise differentiable with finite number of pieces in every finite interval and satisfies

\begin{eqnarray}

g(\omega) = \frac{1}{2}(g(\omega+)+g(\omega-)) \ ,

\end{eqnarray}

where

\begin{eqnarray}

f(x+) & = & \lim_{t \rightarrow x^+} f(t) \ , \\

f(x-) & = & \lim_{t \rightarrow x^-} f(t) \ .

\end{eqnarray}

We now have for $t \neq 0$

\begin{eqnarray}

\mathscr{F}g(t) & = & \frac{1}{2\pi} PV \int_{-\infty}^\infty \frac{1}{2} (\chi_{[-\pi,\pi]}(\omega) + \chi_{(-\pi,\pi)}(\omega)) e^{-i t \omega} d\omega = \frac{1}{2\pi} \int_{-\pi}^\pi e^{-i\omega t} d\omega \\

& = & \frac{1}{2\pi} \bigg|_{-\pi}^\pi \frac{1}{-it} e^{-i\omega t} = \frac{1}{2\pi} \frac{1}{-it} (e^{-i\pi t} – e^{i\pi t}) = \frac{\sin(\pi t)}{\pi t} \ .

\end{eqnarray}

For $t = 0$ we have

\begin{eqnarray}

\mathscr{F}g(t) & = & \frac{1}{2\pi} PV \int_{-\infty}^\infty \frac{1}{2} (\chi_{[-\pi,\pi)}(\omega) + \chi_{(-\pi,\pi)}(\omega)) e^{-i 0 \omega} d\omega = \frac{1}{2\pi} \int_{-\pi}^\pi d\omega = 1 \ .

\end{eqnarray}

Hence

\begin{eqnarray}

\mathscr{F}g(t) = \textrm{sinc}(t) \ .

\end{eqnarray}

We can now calculate

\begin{eqnarray}

\mathscr{F}\textrm{sinc}(\omega) & = & \mathscr{F}\mathscr{F}g(\omega) = \frac{1}{2\pi} PV \int_{-\infty}^\infty \frac{1}{2\pi} PV \int_{-\infty}^\infty g(\omega’) e^{-it \omega’} d\omega’ e^{-i\omega t} dt \\

& = & \frac{1}{(2\pi)^2} \lim_{M \rightarrow \infty} \int_{-M}^M \int_{-\infty}^\infty g(\omega’) e^{-it \omega’} d\omega’ e^{-i\omega t} dt = \frac{1}{(2\pi)^2} 2\pi g(-\omega) \\

& = & \frac{1}{2\pi} g(\omega) = \frac{1}{2\pi} \frac{1}{2} (\chi_{[-\pi,\pi]}(\omega) + \chi_{(-\pi,\pi)}(\omega)) \\

& = & \frac{1}{2\pi} \frac{1}{2} (\textrm{sgn}(\pi+\omega) + \textrm{sgn}(\pi-\omega)) \ .

\end{eqnarray}

The transform of transform identity applied in the fourth equation is shown here. Note, that transform of transform differs from the form of Fourier inversion theorem only by one minus-sign. Note also that $\mathscr{F}g = \textrm{sinc} \notin L^1$. Hence we cannot apply the inversion theorem in “W. Rudin: Functional Analysis” directly. The transform of transform identity mentioned above is applied instead. It is also shown behind the link with a special definition of Fourier transform that implies convergence of transform of transform for every element of $\mathbb{R}$.

The answer is quite close to the Mathematica-answer. The difference is due to the coefficient $\frac{1}{2\pi}$ instead of $\frac{1}{\sqrt{2\pi}}$ in front of the transform.

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