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Calcuate the integral $$I=\int_{-\infty}^\infty\frac{\sin a\omega\sin b\omega}{\omega\cdot \omega}d\omega.$$

First I noticed that $$\mathcal{F}(\mathbb{1}_{[-h,h]})(\omega)=\frac{\sin h \omega}{\omega}$$ and by the definition of fourier transform $$I=\int_{-\infty}^\infty\widehat{1_{[-a,a]}\cdot1_{[-b,b]}}\cdot e^{i\omega\cdot 0} d\omega=({1_{[-a,a]}\ast 1_{[-b,b]}})(0)$$If we denote $c:=\min\{a,b\}$, then the convolution is simply $$({1_{[-a,a]}\ast 1_{[-b,b]}})(0)=\int_{-c}^c1dt=2c$$but when I use Mathematica I don't get these values.

What is my mistake in the calculation?

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Maybe you have missed factors of $\sqrt{2\pi}$. For example, $$ \chi_{[-a,a]}^{\wedge}(s)=\frac{1}{\sqrt{2\pi}}\int_{-a}^{a}e^{-isx}dx = \frac{1}{\sqrt{2\pi}}\frac{e^{-isa}-e^{+isa}}{-is}= \sqrt{\frac{2}{\pi}}\frac{\sin(as)}{s}. $$ And don't forget: $(f\star g)^{\wedge}=\sqrt{2\pi}f^{\wedge}g^{\wedge}$. Therefore, $$ \begin{align} \int_{-\infty}^{\infty}\frac{\sin(as)\sin(bs)}{s^{2}}e^{isx}ds & = \frac{\pi}{2}\int_{-\infty}^{\infty}\chi_{[-a,a]}^{\wedge}(s)\chi_{[-b,b]}^{\wedge}(s)e^{isx}ds \\ & = \frac{1}{\sqrt{2\pi}}\frac{\pi}{2}\int_{-\infty}^{\infty}(\chi_{[-a,a]}\star \chi_{[-b,b]})^{\wedge}(s)e^{isx}ds \\ & = \frac{\pi}{2}(\chi_{[-a,a]}\star\chi_{[-b,b]})(x) \end{align}. $$ Hence, $$ \int_{-\infty}^{\infty}\frac{\sin(as)\sin(bs)}{s^{2}}ds = 2\frac{\pi}{2}\min\{b,a\} = \pi\min\{b,a\}. $$ As a quick check, Parseval's identity gives $$ \int_{-\infty}^{\infty}\frac{\sin^{2}(as)}{s^{2}}ds = \frac{\pi}{2}\int_{-\infty}^{\infty}|\chi_{[-a,a]}^{\wedge}(s)|^{2}=\frac{\pi}{2}\int_{-a}^{a}dx=\pi a. $$

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  • $\begingroup$ Hi, regarding the convolution of indicator functions, do we have in general that $1_{[-1,1]}*1_{[-t,t]}=min\{1,t\}?$ I've tried reaching this here, but not really gotten there yet. It would be definitely interesting to have your input on this post, if time allows: math.stackexchange.com/questions/1509579/… Best. $\endgroup$ – user186225 Nov 2 '15 at 16:29
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I do not know if this is what you are waiting but, since I did not work Fourier transforms for more than $50$ years, I shall just try to explain what happens with the integral.

$$J=\int\frac{\sin a\omega\sin b\omega}{\omega\cdot \omega}d\omega=\frac{\omega (b-a) \text{Si}((a-b) \omega )+\omega (a+b) \text{Si}((a+b) \omega )-2 \sin (a \omega ) \sin (b \omega )}{2 \omega }$$ where appear the since integral function and so, using bounds $$I=\int_{-\infty}^\infty\frac{\sin a\omega\sin b\omega}{\omega\cdot \omega}d\omega=\frac{1}{2} \pi (|a+b|-|a-b|)$$

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  • $\begingroup$ What do you mean by $Si$? $\endgroup$ – user65985 Mar 26 '15 at 4:43
  • $\begingroup$ $Si(z)=\int_0^z \frac{\sin(t)}{t}~dt$ is the sine integral function. $\endgroup$ – Claude Leibovici Mar 26 '15 at 4:46
  • $\begingroup$ $\dfrac{|~|a+b|-|a-b|~|}2~=~\min(|a|,~|b|)~$ and $~\dfrac{|a+b|+|a-b|}2~=~\max(|a|,~|b|)$. $\endgroup$ – Lucian Mar 26 '15 at 11:10
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No convolution here. All you need is the Parseval's identity

$$ \int_{-\infty}^{\infty} f(x)g(x)dx = \int_{-\infty}^{\infty} F(w)G(w)dw $$

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  • $\begingroup$ But why my solution is not correct? p.s.: you get the same result as I get. $\endgroup$ – user65985 Mar 26 '15 at 4:43
  • $\begingroup$ @DanisFischer: If you use this identity you get the correct answer. $\endgroup$ – science Mar 26 '15 at 4:44
  • $\begingroup$ I get $\int_{-c}^c1dx=2c$ again. $\endgroup$ – user65985 Mar 26 '15 at 4:46
  • $\begingroup$ @DanisFischer: You need to see the valid domain of the product of the two different heaviside functions. $\endgroup$ – science Mar 26 '15 at 4:50
  • $\begingroup$ So what is the domain of integration? why it includes $\pi$ suddenly? $\endgroup$ – user65985 Mar 26 '15 at 4:53
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With your convention for the Fourier transform (which you should state), we have $$ \begin{align*} \int_{-\infty}^\infty \hat{f}(\omega) \overline{\hat{g}(\omega)} \, d\omega &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x)\overline{g(y)} e^{i\omega(x-y)} \, dx \, dy \, d\omega \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x)\overline{g(y)} \int_{-\infty}^{\infty} e^{i\omega(x-y)} \, d\omega \, dx \, dy \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x)\overline{g(y)} \int_{-\infty}^{\infty} e^{i\omega(x-y)} \, d\omega \, dx \, dy, \end{align*}$$ and then you interpret the inside integral as the Fourier transform of $1$, which is $2\pi \delta(x)$ because $$ \mathcal{F}^{-1}(\delta)(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \delta(\omega) e^{i\omega x} \, d\omega = \frac{1}{2\pi}. $$ You then get Plancherel's theorem as $$ \begin{align*} \int_{-\infty}^\infty \hat{f}(\omega) \overline{\hat{g}(\omega)} \, d\omega &= \int_{-\infty}^{\infty} f(x) \overline{g(y)} 2\pi \delta(x-y) \, dx \, dy \\ &= 2\pi\int_{-\infty}^{\infty} f(x)\overline{g(x)} \, dx, \end{align*}$$ and exactly the same style of reasoning would give you the convolution theorem. So the point is that you're missing a $\frac{1}{2\pi}$ because of your Fourier transform convention.

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