Calculate $\int_{\mathbb{R}}e^{ix\omega}(\frac{\sin(\omega)}{\omega})^2 d\omega$ using convolution theorem.

Question

I have study Fourier transformation and found this exercise: Calculate $\int_{\mathbb{R}}e^{ix\omega}(\frac{\sin(\omega)}{\omega})^2 d\omega$ using convolution theorem. I have notice that integral looks like inverse Fourier transformation, where $(\frac{\sin(\omega)}{\omega})^2$ is some function's Fourier transformation. Is there any hint how I can use convolution theorem to solve this?

Answer 1

Let the Fourier Transform of $f(\omega)$ be given by

$$F(x)=\int_{-\infty}^\infty f(\omega)e^{i\omega x}\,d\omega$$

We have the Fourier Transform pairs

$$\begin{align} f(\omega) &\leftrightarrow F(x)\\\\ f^2(\omega) &\leftrightarrow \frac{1}{2\pi}F(x)*F(x)\tag1\\\\ \frac{\sin(\omega)}{\omega}&\leftrightarrow \pi\,\text{rect}(x/2)\tag2\\\\ \frac{\sin^2(\omega)}{\omega^2}&\leftrightarrow \frac{1}{2\pi}\left(\pi\,\text{rect}(x/2)\right)*\left(\pi\,\text{rect}(x/2)\right)=\pi\, \text{tri}(x/2)\tag3 \end{align}$$

where $\text{rect}(t)$ and $\text{tri}(t)$ are the Rectangle Function and Triangle Function, respectively.

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NOTES:

The pair in $(1)$ expresses the result of the Convolution Theorem.

To arrive at $(1)$, we use the result $\int_{-\infty}^\infty \frac{\sin(ax)}{x}\,dx=\pi \text{sgn}(a)$. Proceeding we have

$$\begin{align} \int_{-\infty}^\infty \frac{\sin(\omega)}{\omega}e^{i\omega x}\,d\omega&=\frac1{2i}\int_{-\infty}^\infty \frac{e^{i\omega(x+1)}-e^{i\omega(x-1)}}{\omega}\,d\omega\\\\ &=\frac\pi2 \text{sgn}(x+1)-\frac\pi2 \text{sgn}(x-1)\\\\ &=\pi\,\text{rect}(x/2) \end{align}$$

And finally, using $(1)$ and $(2)$ we have

$$\begin{align} \int_{-\infty}^\infty \left(\frac{\sin(\omega)}{\omega}\right)^2 e^{i\omega x}\,d\omega&=\frac1{2\pi}\int_{-\infty}^\infty \left(\pi\,\text{rect}((x-x')/2)\right)\left(\pi\,\text{rect}(x'/2)\right)\,dx'\\\\ &=\pi\,\text{tri}(x/2) \end{align}$$