I want to solve this using Fourier theory and Plancharels theorem. This theorem states that:

Plancharels Theorem: For any $f\in L^2(\mathbb{R}), \ \widehat{f}\in L^2(\mathbb{R}).$ Moreover, $$\langle\widehat{f},\widehat{g}\rangle=2\pi\langle f,g \rangle,$$ and thus $$||\widehat{f}||^2_{L^2}=2\pi||f||^2,$$ $\forall \ f,g\in L^2(\mathbb{R}).$

The set of functions denoted by $L^2(\mathbb{R})$ is a Hilbert space with scalarprouduct

$$\langle f,g \rangle=\int_{\mathbb{R}}f(x)g(x) \ dx,\ \text{if} \ g,f\in\mathbb{R},$$

thus the first line in Plancharels can be rewritten as

$$\int_{\mathbb{R}}\hat{f}(x)\hat{g}(x) \ dx = 2\pi\int_{\mathbb{R}}f(x)g(x) \ dx.$$

But unfortunately our integral of concern is not over $\mathbb{R}$, to ammend this we can extend the integral to $\mathbb{R}$ by writing

$$\int_{0}^{\infty}=\frac{\sin(x)}{xe^{x}} \ dx = \frac{1}{2}\int_{\mathbb{R}}\frac{\sin(x)}{xe^{|x|}} \ dx = \frac{1}{2}\int_{\mathbb{R}}\frac{\sin(x)}{x} e^{-|x| \ dx}.$$

This is where I get stuck. In the notes it says the following in order to proceed:

Using the Plancharel trick, we can replace those functions (which I assume are $\sin(x)/x$ and $e^{-|x|})$ by the Fourier transforms:

$$\frac{1}{2}\int_{\mathbb{R}}\frac{\sin(x)}{x}e^{-|x|} \ dx =\frac{1}{4\pi}\int_{\mathbb{R}}\pi\chi_{(-1,1)}(x)\frac{2}{x^2+1} \ dx = \frac{1}{2}\int_{-1}^{1}\frac{1}{x^2+1} \ dx...$$

Can someone explain what has happened here? I don't see the $\chi-$ function mentioned anywhere in the notes and how to apply it. But reading online it seems to be some sort of indicator function which I'm not familiar with. Any help is greatly appreciated.

  • 1
    $\begingroup$ Yes - it's an indicator function. $\tilde f(k) = \int dx f(x)e^{ikx}=\int dx \frac{\sin x}{x}e^{ikx}=\int dx \frac{1}{2i}\frac{e^{ix}-e^{-ix}}{x}e^{ikx}=\int dx\int_{-1}^{1}dt \frac{1}{2}e^{itx}e^{ikx}=\int dx\int_{-1}^{1}dt \frac{1}{2}e^{i(t+k)x}=\int_{-1}^{1}dt \frac{2\pi}{2}\delta(t+k)=\pi\begin{cases}1,& k\in[-1,1]\\ 0,& \text{else}\end{cases}=\pi1_{[-1,1]}$. $\endgroup$ – DavidP Mar 4 at 15:59
  • $\begingroup$ Ok, so in my case I have that $\chi_{(-1,1)}=1?$ What happens to $\int \ dx?$ In your case you have $dx$ and $dt$ but in my case I only have $dx$. I still don't see how to apply this. $\endgroup$ – Parseval Mar 4 at 17:01
  • $\begingroup$ Have a look here (Post from Ron Gordon): math.stackexchange.com/questions/1226853/…. I calculated F(K). $\endgroup$ – DavidP Mar 4 at 18:24
  • $\begingroup$ @DavidP - Thank you, but that post doesn't explain why the limits of integration can change from $\mathbb{R}$ to $[-1,1].$ $\endgroup$ – Parseval Mar 4 at 19:24
  • 1
    $\begingroup$ The integrand is 0 else. $\chi_{[-1,1]}(x)=1$ for $x\in [-1,1]$ and 0 else. So $\int_{\mathbb{R}}=\int_{-\infty}^{-1}+\int_{-1}^1+\int_1^{\infty}$ and the first and third integral are zero. $\endgroup$ – DavidP Mar 4 at 19:38

Well, if you're trying to find out the integral (the question is too technical for me, I'm sorry and this is too long for a comment) we can do that. Let's define:


Using the 'evaluating integrals over the positive real axis' property of the Laplace transform, we can write:


Finding those Laplace transforms (they are not hard to show), we get:


Using the following property:




  • 1
    $\begingroup$ This was a nice method, I've never seen the property $(4)$ before. Thanks! $\endgroup$ – Parseval Mar 4 at 19:58
  • $\begingroup$ @Parseval You're welcome, I'm glad that I could help you! $\endgroup$ – Jan Mar 4 at 20:05

$$I=\int_0^\infty\frac{\sin(x)}{xe^x}dx=\int_0^\infty\sum_{n=0}^\infty\frac{(-1)^nx^{2n}e^{-x}}{(2n+1)!}dx=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\int_0^\infty x^{2n}e^{-x}dx=\sum_{n=0}^\infty\frac{(-1)^n(2n)!}{(2n+1)!}=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}$$


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{\sin\pars{x} \over x\expo{x}}\,\dd x & = \int_{0}^{\infty}\expo{-x}\pars{{1 \over 2}\int_{-1}^{1}\expo{\ic kx} \dd k}\,\dd x = {1 \over 2}\int_{-1}^{1}\int_{0}^{\infty}\expo{-\pars{1 - \ic k}x} \dd x\,\dd k \\[5mm] & = {1 \over 2}\int_{-1}^{1}{\dd k \over 1 - \ic k} = {1 \over 2}\int_{-1}^{1}{\dd k \over 1 + k^{2}} = \int_{0}^{1}{\dd k \over k^{2} + 1} = \arctan\pars{1} \\[5mm] & = \bbx{\pi \over 4} \end{align}


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