In finance studies, maybe is necessary use Fourier Transform and convolution applications. In the book Introductory Mathematical Analysis for Quantitative Finance, at chapter 12 is proposed the following problem:
$$\mathrm{If} \ \ f(t) = \sin(t) \cdot \exp(-t) \ \ \mathrm{and} \ \ g(t) = \cos(t) \cdot \exp(-t) \ ,$$
show that:
$$(g \star f)(t) = \frac{1}{2}\cdot t \cdot\exp(-t)\cdot\sin(t) \ .$$
How can I do this?
HINT:
$$f(t)=\mathrm e^{-t}\sin t=\frac{\mathrm e^{\mathrm (\mathrm i-1)t}-\mathrm e^{-(\mathrm i+1)t}}{2\mathrm i} \\ g(t) =\mathrm e^{-t}\cos t= \frac{\mathrm e^{(\mathrm i-1)t}+\mathrm e^{-(\mathrm i+1)t}}{2}$$
The Fourier transform of $t\mapsto \mathrm e^{zt}$ is $$\int_{\mathbb R}\mathrm e^{zt}\mathrm e^{-\mathrm i\omega t}\mathrm dt=\int_{\mathbb R}\mathrm e^{(z-\mathrm i\omega)t}\mathrm dt$$ This is an easy integral. Apply it to your case and use linearity.
So, now, if you know the Fourier transforms of $f,g$, call them $F,G$, then you can calculate $g\ast f$ by taking an inverse transform,
$$g* f=\mathcal F^{-1}(F\cdot G)$$