Let $f(x)$ be an even function on $(-\infty , \infty)$ and let $g(x) = \sin(ax)$, $a >0$.
Show that $$(f \ast g)(x) = \int_{-\infty}^{\infty} f(t)\sin(a(x-t))\,dt= \sin(ax)\hat{f}(a).$$
I end up with $$\sin(ax)\hat{f}(a) - e^{-iax}\int_{-\infty}^{\infty}f(t)\sin(at)\,dt.$$ The last part goes two zero?
Also, I used angle expansion. Is there an easier way to do this?
Because $f$ is an even function you know that $f(-t) = f(t)$. Additionally, $\sin$ is an odd function, i.e. $\sin(-at) = -\sin(at)$. Therefore, $f(-t)\sin(-at) = -f(t)\sin(at)$, and so
$$\begin{align*} \int_{-\infty}^{\infty} f(t)\sin(at)\,dt &= \int_{-\infty}^{0}f(t)\sin(at)\,dt + \int_{0}^{\infty}f(t)\sin(at)\,dt\\ &= -\int_{0}^{\infty}f(t)\sin(at)\,dt + \int_{0}^{\infty}f(t)\sin(at)\,dt\\ &=0. \end{align*} $$