convergence of Fourier transform of $e^{-t}\sin(2\pi ft)u(t)$ As you see Fourier transform function is being divergent for the first statement but it seems to converge. What is my fault?
$$
\begin{align}
   \int\limits_{-\infty }^{+\infty }{{{e}^{-t}}\sin(2\pi ft)u(t){{e}^{-j2\pi ft}}\,dt}\\
&amp;=\int\limits_{0}^{+\infty }{{{e}^{-t}}\sin(2\pi {{f}_{c}}t){{e}^{-j2\pi ft}}\,dt}\\
&amp;=\int\limits_{0}^{+\infty }{{{e}^{-t}}\frac{{{e}^{j2\pi {{f}_{c}}t}}-{{e}^{-j2\pi {{f}_{c}}t}}}{2j}{{e}^{-j2\pi ft}}\,dt} \\ 
 &amp; =\int\limits_{0}^{+\infty }{\frac{{{e}^{j2\pi t({{f}_{c}}-f)-t}}-{{e}^{-j2\pi t({{f}_{c}}-f){-t}}}}{2j}\,dt}\\
&amp;=\int\limits_{0}^{+\infty }{\frac{{{e}^{t(j2\pi ({{f}_{c}}-f)-1)}}-{{e}^{-t(j2\pi ({{f}_{c}}-f){+1)}}}}{2j}\,dt} \\ 
 &amp; =\left(\frac{{{e}^{t(j2\pi ({{f}_{c}}-f)-1)}}}{2j(j2\pi ({{f}_{c}}-f)-1)}+\frac{{{e}^{-t(j2\pi ({{f}_{c}}-f){+1)}}}}{2j(j2\pi ({{f}_{c}}-f)+1)}\right)_{0}^{+\infty } \\ 
\end{align}
$$
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.