-1
$\begingroup$

I need help to calculate the Fourier transform of this function:

$f(t)=\sin (e^{-at})$

Where $a$ is a constant.

I tried to use the definition of the Fourier transform:

$f(\omega)=\int_{-\infty}^\infty f(t) * e^{-i\omega t}dt$

Substituting $f(t)$:

$f(\omega)=\int_{-\infty}^\infty \sin (e^{-at}) * e^{-i\omega t}dt$

I used the Euler's identity: $2i\sin \theta = {e^{i \theta}-e^{-i \theta}} $

Substituting: $\sin (e^{-at})$ and $\theta=e^{-at}$

$f(\omega)=\frac{1}{2i}\int_{-\infty}^\infty ({e^{i e^{-at} }-e^{-i e^{-at}}}) * e^{-i\omega t}dt$

And I can't solve that integral. I don't know if the way i'm taking is the correct but i would like some help to solve it. Thanks.

$\endgroup$
  • 2
    $\begingroup$ What makes you sure about the assumption that there is a solution to this transform in terms of standard functions? $\endgroup$ – mrtaurho Aug 7 '18 at 23:02
  • $\begingroup$ I think the best you can get is a series expansion (coming from the power series expansion of $\sin$). $\endgroup$ – metamorphy Aug 8 '18 at 2:04
0
$\begingroup$

With the change of variables $t = -\ln(\xi)/a$, $$\int_{-\infty}^\infty \sin(e^{-a t}) e^{-i \omega t} dt = \frac 1 {|a|} \int_0^\infty \xi^{i \omega/a - 1} \sin \xi \,d\xi = \frac {i \Gamma \!\left( \frac {i \omega} a \right) \sinh \frac {\pi \omega} {2 a}} {|a|}, \\ a \in \mathbb R \backslash \{0\}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.