Fourier transform of exp(-g(w)exp(-w))

Question

I am looking for inverse fourier transform of $$e^{(-g(\omega)e^{(-a|\omega|)})}$$ I am mainly interested in $g(\omega)=b$, $g(\omega)=b\omega ^2$ and $g(\omega)=bi\omega$ Let me know if you need further information. Thanks

Answer 1

lets $ g(\omega) =b$, then $$f(\omega) = e^{-b e^{-a|\omega|}} $$ Use the exponential series, $$e^{-x} = (-1)^n \sum_{n=0}^{\infty} \frac{x^n}{n!}$$ So $$f(\omega) = (-1)^n \sum_{n=0}^{\infty} \frac{ \left( b e^{-a|\omega|} \right)^n}{n!} = (-1)^n \sum_{n=0}^{\infty} \frac{ \left( b^n e^{-an|\omega|} \right)}{n!}$$ Now take the inverse Fourier transform

$$ f(t) = \sum_{n=0}^{\infty} \frac{(-1)^n b^n}{n!} \int_{-\infty}^{\infty} e^{-an|\omega|} e^{i \omega t} d \omega $$

I am bit confused by your absolute value $|\omega|$, but if its normal $\omega$ then its a simple Dirac delta function $$ f(t) = \sum_{n=0}^{\infty} \frac{(-1)^n b^n}{n!} \sqrt{2 π} \delta(t - i n a) $$

For $ g(\omega) =b \omega^2 $

$$ f(t) = \sum_{n=0}^{\infty} \frac{(-1)^n b^n}{n!} (i)^{2n} \sqrt{2 π} \delta^{(2n)}(t - i a n)$$

and for $ g(\omega) =ib\omega$

$$ f(t) = \sum_{n=0}^{\infty} \frac{(-1)^n b^n}{n!} (i)^{2n} (i)^{2n} \sqrt{2 π} \delta^{(2n)}(t - i a n)$$

$\delta^{(2n)}$ is the $2n^{th}$ derivative of Dirac delta function