Calculating value of integral of convolution using Fourier transform 
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?
 A: 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 $$I=\int_{-\infty}^\infty\frac{\sin a\omega\sin b\omega}{\omega\cdot \omega}d\omega=\frac{1}{2} \pi  (|a+b|-|a-b|)$$ 
A: No convolution here. All you need is the Parseval's identity

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

A: 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.
A: 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
$$
     \int_{-\infty}^{\infty}\frac{\sin^{2}(as)}{s^{2}}ds = \frac{\pi}{2}\int_{-\infty}^{\infty}|\chi_{[-a,a]}^{\wedge}(s)|^{2}=\frac{\pi}{2}\int_{-a}^{a}dx=\pi a.
$$
