convergence of Fourier transform of $e^{-t}\sin(2\pi ft)u(t)$
Thanks to  $u(t)$,
 $$
\begin{align}
   \int\limits_{-\infty }^{+\infty }{{{e}^{-t}}\sin(2\pi ft)u(t){{e}^{-j2\pi ft}}\,dt}=\int\limits_{0}^{+\infty }{{{e}^{-t}}\sin(2\pi {{f}_{c}}t){{e}^{-j2\pi ft}}\,dt}
\end{align}
$$
Now 
$$0 \le | e^{-t}\sin(2\pi {{f}_{c}}t){{e}^{-j2\pi ft}}| \le | e^{-t}|$$
so your function is absolutely integrable.

The trouble comes for the exponential product rule:

$${{{e}^{-t}}{{e}^{j2\pi {\pm{f}_{c}}t}}{{e}^{-j2\pi ft}}} = e^{-t\left(1\mp j2\pi {{f}_{c}}+j2\pi f\right)}$$
which integrates quite well.