I have a Fourier transform of a function here: $$\tilde{F}(k)=\frac{2}{k-k_0}\sin\left(\frac{k-k_0}{2}a\right),$$ where $a\in\mathbb{R}$. The question did not specify whether $k_0$ is real or complex, I'll take it to be real for now.
I am required to compute $\vert \tilde{F}(k) \vert^2$ and sketch the Fourier spectrum ( $\vert \tilde{F}(k) \vert^2$ versus $k$).
$$ \vert \tilde{F}(k) \vert^2 = \frac{4}{(k-k_0)^2}\sin^2\left(\frac{k-k_0}{2}a\right)$$
If I were to plot this spectrum, I am expecting a peak at $k=k_0$, but I am not sure how to compute the Full width half maximum. Any hints will be appreciated. Thanks!
Call $x = (k - k_0)a/2$, so that $F$ becomes
$$ \tilde{F}(x) = a\frac{\sin x}{x} = \tilde{F}(0)\frac{\sin x}{x} $$
and
$$ |\tilde{F}(x)|^2 = |\tilde{F}(0)|^2{\rm sinc}^2(x) $$
The FWHM of this is $2\sigma$ where $\sigma$ is the solution to
$$ |\tilde{F}(\sigma)|^2 = \frac{1}{2}|\tilde{F}(0)|^2 = |\tilde{F}(0)|^2{\rm sinc}^2(\sigma) $$
So basically the problem is reduced to solving
$$ {\rm sinc}(\sigma) = 1 / \sqrt{2} $$
So far I know this cannot be solved agebraically, but numerically you get $\sigma = 1.39156$ so that FWHM $=2\sigma = 2.78311$