Q:

Finding the peak and width of Fourier spectrum

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!

A:

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 $=2\sigma = 2.78311$ 

