Calculate $$f(w) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) e^{-iwx} {\rm d} x$$ where $$f(x) = \begin{cases} e^{-x} \sin(x), & x \geq 0 \\ 0, & x < 0\end{cases}$$
My solution was, of course, to compute the integral, and it's easy to do so but then I am stuck with the $\sin(\infty)$. So I thought of another way by replacing $$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$ but I couldn't continue. Can someone please give me any hints?
Well, you can, as the exponentials are multplied, take the exponentials together $$ \big ( \frac{e^{ix}-e^{-ix}}{2i} \big ) e^{-x}e^{-iwx} = \frac{1}{2i} \big(e^{x(i-1-iw)}-e^{-x(i+iw+1)} \big ). $$ Hence, you get for the integral $$ \int_0^\infty \frac{1}{2i} \big(e^{x(i-1-iw)}-e^{-x(i+iw+1)} \big ) dx $$ From this you be able to compute the integral.