2

How can i calculate the Fourier transform of $e^{-|t|}\sin(t)$. I guess I need to do something with convolution, but I am not sure. Can somebody show me the way?

Matthew Conroy
  • 11,234
user1200276
  • 21
  • Changing the sin in (e^(iwt)-e^(-iwt))/2 and then split the integral in two integrals of e to the power of something. But the absolute is killing, because of e^(-abs(t)+i-iw)). – user1200276 May 14 '14 at 18:06
  • you can use latex in your comments as well, try to do that – mm-aops May 14 '14 at 18:09
  • Identical question, although no explicit answer http://math.stackexchange.com/questions/139050/fourier-transform-of-fx-sinx-cdot-e-x?rq=1 – glowsticc May 14 '14 at 19:31

2 Answers2

3

The function $f(t):=e^{-|t|}\sin t$ is odd. Therefore we can write $$\eqalign{\hat f(x)&=\int_{-\infty}^\infty f(t)\>e^{-ixt}\ dt=-2i\int_0^\infty e^{-t}\sin t\>\sin(xt)\ dt\cr &=i\int_0^\infty\bigl(\cos((1+x)t)-\cos((1-x)t)\bigr)e^{-t}\ dt\ .\cr &={i\over2}\int_0^\infty\bigl(e^{(-1+i(1+x))t}+e^{(-1-i(1+x))t}-e^{(-1+i(1-x))t}-e^{(-1-i(1-x))t}\bigr)\>dt\cr &={i\over2}\left({1\over1-i(1+x)}+{1\over1+i(1+x)}-{1\over1-i(1-x)} -{1\over1+i(1-x)}\right)\cr &={-4ix\over 4+x^4}\quad.\cr}$$

Christian Blatter
  • 226,825
0

Hope fully this will help, but I got stuck on the final integral:

$F(e^{-|t|}\sin(t))=\int_{-\infty}^\infty e^{-ixt}e^{-|t|}\sin(t)dt$

$=\int_0^\infty e^{-ixt}e^{-|t|}\sin(t)dt+\int_{-\infty}^0 e^{-ixt}e^{-|t|}\sin(t)dt$

$=\int_0^\infty e^{-t(ix+1)}\sin(t)dt+\int_{-\infty}^0 e^{-t(ix-1)}\sin(t)dt$

=$\int_0^\infty e^{-t}\sin(t)(e^{-itx}-e^{itx})dt$

$-2i\int_0^\infty e^{-t}\sin(t)\sin(tx)dt$

Ellya
  • 11,783