This is based on a previous post of mine: Question concerning the setup and process of solving an ODE.
Here, I am specifically looking for advice on how to compute \begin{equation} \tag{$*$} \mathcal{F}^{-1}\left(\sqrt{\frac{2}{\pi}}\frac{\sin(s)}{s(\mu^2+s^2)}\right), \end{equation} where $\mu$ is a positive constant. I know I can apply linearity to get $$\sqrt{\frac{2}{\pi}}\mathcal{F}^{-1}\left(\frac{\sin(s)}{s(\mu^2+s^2)}\right),$$ but I'm struggling to understand how to proceed from here. Using the convention that $$\mathcal{F}^{-1}(g(x))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty g(x)e^{isx}ds,$$ the constant coefficient becomes $\frac{1}{\pi}$, so $(*)$ becomes $$\frac{1}{\pi}\int_{-\infty}^\infty\frac{\sin(s)e^{isx}}{s(\mu^2+s^2)}ds.$$ Unfortunately, I'm a bit lost on how to actually evaluate this. My initial though was to expand $\sin(s)$ as a Taylor series, but this doesn't appear to be productive.
As discussed in the comments, my hint is as follows: just remember that the convolution theorem takes this form for the product of two functions
In our case the functions we're dealing with are:
$$ f(s) = \frac{1}{\mu^2+s^2}\\[10pt] g(s) = \frac{\sin(s)}{s} $$
which have very well known Fourier transforms, as mantiond by Jack D'Aurizio
$$ \begin{align} &\mathcal F^{-1}\left\{\frac{\mu}{\mu^2+s^2}\right\} = \sqrt{\frac{\pi}{2}}e^{-\mu|x|}&\textit{Laplace distribution}\\[10pt] &\mathcal F^{-1}\left\{\frac{\sin\left(as\right)}{as}\right\} =\frac{1}{a\sqrt{2\pi}}\operatorname{rect}\left(\frac{x}{2\pi a}\right)&\textit{Function of a symmetric interval} \end{align} $$
You can find more information on the rectangular function at this wikipedia link