convergence of Fourier transform of $e^{-t}\sin(2\pi ft)u(t)$

Question

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

Answer 1

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.

Answer 2

Let $$x(t)={{e}^{-t}}\sin(2\pi f_ct)u(t)$$

Its Fourier transform denoted by $X(f)$ is calculated as follows $$ \begin{align} X(f)&=\int\limits_{-\infty }^{+\infty }{{{e}^{-t}}\sin(2\pi f_ct)u(t){{e}^{-j2\pi ft}}\,dt}\\ &=\int\limits_{0}^{+\infty }{{{e}^{-t}}\sin(2\pi {{f}_{c}}t){{e}^{-j2\pi ft}}\,dt}\\ &=\int\limits_{0}^{+\infty }{{{e}^{-t}}\left(\frac{{{e}^{j2\pi {{f}_{c}}t}}-{{e}^{-j2\pi {{f}_{c}}t}}}{2j}\right){{e}^{-j2\pi ft}}\,dt} \\ & =\frac{1}{2j}\int\limits_{0}^{+\infty }{{{e}^{-(1+j2\pi (f-f_c))t}}-{{e}^{-(1+j2\pi (f+f_c))t}}\,dt}\\ & = \frac{1}{2j}\left( \frac{-e^{-(1+j2\pi (f-f_c))t}}{1+j2\pi (f-f_c)}-\frac{-e^{-(1+j2\pi (f+f_c))t}}{1+j2\pi (f+f_c)} \right)\biggr\vert^{t=+\infty}_{t=0}\\ &=\frac{1}{2j}\left(\frac{1}{1+j2\pi (f-f_c)}-\frac{1}{1+j2\pi (f+f_c)} \right)\\ &=\frac{j}{2}\left(\frac{1}{1+j2\pi (f+f_c)}-\frac{1}{1+j2\pi (f-f_c)} \right) \end{align} $$

Notice that this result can also be found using the convolution theorem:

$$\mathcal{F}\{f(t)g(t)\}=F(\omega)*G(\omega)$$

where $f(t)=e^{-t}u(t)$ and $g(t)=\sin(2\pi f_c t)$