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

  • $\begingroup$ Check the formula for Fourier transform..It should be -j2πft not j2πft.. $\endgroup$ Nov 20, 2016 at 14:21
  • 2
    $\begingroup$ What you wrote immediately after the third = sign is incorrect and the error propagates from there to give you nonsense at the end $\endgroup$ Nov 20, 2016 at 14:32
  • $\begingroup$ these are three totally unrealted questions. Don't ask three totally unrelated questions in one question! I'm removing the second and the third, please ask them separately (you can still find your original question in the edit history). $\endgroup$ Nov 20, 2016 at 14:38
  • $\begingroup$ Thanks for your response and Thank you Marcus for editing. I've edited the equation. $\endgroup$ Nov 20, 2016 at 15:26
  • $\begingroup$ Actually The $ e^t(...) $ converges when you set t to infinity $\endgroup$ Nov 20, 2016 at 15:59

2 Answers 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:


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

  • $\begingroup$ Dear msm , I appreciate you for your help and spend much time to solve this problem.Thanks $\endgroup$ Nov 21, 2016 at 6:35
  • $\begingroup$ My pleasure. I'm glad it helped. $\endgroup$
    – msm
    Nov 21, 2016 at 8:51

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.

  • 1
    $\begingroup$ Very true, my mistake. I changed the $\pm$ into $\mp$ too. $\endgroup$ Nov 21, 2016 at 9:44

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