About integrating product of two sinc function using Fourier transform

Question

So the problem is

which I think is pretty straight-foward by using Fourier transform and convolution property of two sinc functions and evaluating the convolution at 5. However, I got sinc(t) for the convolution result(So the answer is sinc(5)?). But when I use matlab to check, it says that the result is 100sinc(t)(again 100sinc(5))? I'm really confused and got stuck here.

Answer 1

The convolution theorem states that a convolution of two functions is equal to the inverse FT of the product of the FTs of each of the functions. That is,

$$\int_{-\infty}^{\infty} dx' \, f(x') g(x-x') = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G(k) e^{-i k x}$$

where

$$F(k) = \int_{-\infty}^{\infty} dx \, f(x) e^{i k x}$$ $$G(k) = \int_{-\infty}^{\infty} dx \, g(x) e^{i k x}$$

So when each of $f$ and $g$ are $\operatorname{sinc}{x} = \sin{x}/x$, then $F(k) = G(k) = \pi 1_{[-1,1]} $.

Then

$$\int_{-\infty}^{\infty} dx' \operatorname{sinc}{x'} \operatorname{sinc}{(x-x')} = \frac{\pi}{2 \pi} \int_{-1}^1 dk \, e^{i k x} = \operatorname{sinc}(x)$$

So, yes, the answer to your question is $\operatorname{sinc}(5)$. No idea how Matlab could return what it did.

Answer 2

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\begin{align} &\bbox[5px,#ffd]{% \int_{-\infty}^{\infty}\on{sinc}\pars{t}\on{sinc}\pars{t - 5}\, \dd t} \\[5mm] = &\ \int_{-\infty}^{\infty} \bracks{{1 \over 2}\int_{-1}^{1}\expo{\ic\omega t}\dd\omega} \bracks{{1 \over 2}\int_{-1}^{1} \expo{-\ic\nu\pars{t - 5}}\,\,\dd\nu}\dd t \\[5mm] = &\ {\pi \over 2}\int_{-1}^{1}\expo{5\nu\ic}\ \overbrace{\int_{-1}^{1}\ \underbrace{\int_{-\infty}^{\infty}\expo{\ic\pars{\omega - \nu}t} \,\,{\dd t \over 2\pi}}_{\ds{\delta\pars{\omega - \nu}}}\ \,\dd\omega}^{\ds{=\ 1}}\ \,\dd\nu \\[5mm] &\ {\pi \over 2}\int_{-1}^{1}\expo{5\nu\ic}\,\dd\nu = \pi\int_{0}^{1}\cos\pars{5\nu}\,\dd\nu = \pi\,{\sin\pars{5} \over 5} \\[5mm] = &\ \bbx{\pi\,\on{sinc}\pars{5}} \approx -0.6025 \\ & \end{align}