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.

  • 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

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


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