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

  1. find the Fourier transform of $\chi$,
  2. find the Fourier transform of $f$,
  3. 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.

  • $\begingroup$ I have typeset your math. Please verify that I've done so correctly. $\endgroup$ – Neal Jul 12 '13 at 17:09
  • $\begingroup$ @Neal looks like this edit is problematic, lots of errors - formula in (2) particularly $\endgroup$ – gt6989b Jul 12 '13 at 17:10
  • $\begingroup$ What is your comfort level? Are you familiar with integration in the complex plane? $\endgroup$ – Ron Gordon Jul 12 '13 at 17:24
  • $\begingroup$ I think what makes things confusing here is that the original problem was set by someone probably who wanted to motivate the comprehension of "Fourier pairs". Here Fourier Forward and Inverse pair.The expectation is to run Fourier of $f$ and the Characteristic function and show that Fourier of each is the other. $\endgroup$ – al-Hwarizmi Jul 12 '13 at 18:11
  • $\begingroup$ It should not be f^(k), it should be a "hat" over f and k is the argument of f (my notation for Fourier transform), the rest looks alright. Thank you. $\endgroup$ – Ingvar Jul 13 '13 at 11:02

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:

integration contour

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 $\sqrt{2 \pi}$.

  • $\begingroup$ @Ingvar: if you found this useful, you might consider accepting it by clicking the check ark to the left. $\endgroup$ – Ron Gordon Jul 13 '13 at 11:41

Hint: Notice that your expression for the transform of $\chi$ and the integrand in your transform for $f$ look eerily similar.


$f(x)=\frac{\sin\pi x}{\pi x}$ 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.


$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{}$ \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 $\ds{\sin\pars{x} \over x}$: $$ \color{#c00000}{\large\pi\,\Theta\pars{1 - \verts{k}}} =\color{#66f}{\large\left\lbrace\begin{array}{lcl} \pi & \mbox{if} & \verts{k} < 1 \\[2mm] 0 & \mbox{if} & \verts{k} > 1 \end{array}\right.} $$


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