I'm interested in calculating:
$$ \mathscr{F}\{\theta \operatorname{sinc}(x)\}$$
Where $\theta$ is the Heaviside function. I attempted to solve that by transforming it into:
$$ \mathscr{F}\{\theta\}*\mathscr{F}\{\operatorname{sinc}(x)\}$$
where both Fourier transforms in the expression above are known. Of course, we have $$\mathscr{F}\{\theta\}=\sqrt{\frac{\pi}{2}}\delta(\omega) + \frac{i}{\sqrt{2\pi}\omega}$$
And then we get a convolution of $1/\omega$ with a $\operatorname{rect}$ function. Of course we have:
$$ \left(\frac{1}{\omega} * \operatorname{rect}(\omega)\right)(x) = \int_{x – 1/2}^{x+1/2} \frac{1}{\omega}$$
And in the locations where this integral is defined the answer is obvious. Of course, where this integral is undefined one could use a symmetry argument in order to give a probable solution, but I'm looking for a more formal approach to this.
How would one go about calculating this convolution formally?
Best Answer
The Fourier transform of the Heaviside function $\theta(t)$ is given by
$$\mathscr{F}\{\theta\}(\omega)=\sqrt{\frac\pi2}\delta(\omega)+\frac i{\sqrt{2\pi}}\text{PV}\left(\frac1\omega\right)\tag1$$
Here, $\psi(x)=\text{PV}\left(\frac1x\right)$ denotes the Principal Value distribution defined by as
$$\begin{align} \langle \psi, \phi\rangle &=\text{PV}\int_{-\infty}^\infty \frac{\phi(x)}{x}\,dx\tag2\\\\ &=\lim_{\varepsilon\to 0^+}\int_{|x|\ge \varepsilon}\frac{\phi(x)}{x}\,dx \end{align}$$
for $\phi$ continuous in a neighborhood of $0$ and having compact support.
The Fourier transform of the sinc function, $\text{sinc}(t)=\frac{\sin(t)}{t}$, is given by
$$\mathscr{F}\{\text{sinc}\}(\omega)=\sqrt{\frac\pi2}(H(\omega+1)-H(\omega-1))\tag3$$
Using $(1)$-$(3)$ along with the Convolution Theorem, we have
$$\begin{align} \left(\mathscr{F}\{\theta\}*\mathscr{F}\{\text{sinc}\}\right)(\omega_0)&=\frac12\langle \delta_{\omega_0},\mathscr{F}\{\text{sinc}\}\rangle +\frac i{2\pi}\text{PV}\int_{-\infty}^\infty \frac{\mathscr{F}\{\text{sinc}\}(\omega)}{\omega_0-\omega}\,d\omega\\\\ &=\sqrt{\frac\pi8} (H(\omega_0+1)-H(\omega_0-1))+\frac i{\sqrt{8\pi}}\text{PV}\int_{-1}^1 \frac{1}{\omega_0-\omega}\,d\omega\\\\ &=\sqrt{\frac\pi8} (H(\omega_0+1)-H(\omega_0-1))+\frac i{\sqrt{8\pi}} \log\left(\left|\frac{\omega_0+1}{\omega_0-1}\right|\right) \end{align}$$
And we are done!