Find the Fourier transform for this function
$$f(x)=e^{x-e^x}$$
My Solution:-
$T[f(x)]=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx} f(x)dx$ $=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx}e^{x-e^x} dx$ $=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} (e^{x})^{-ik}e^{x}e^{-e^x} dx$
let $u=e^x$ then $du=e^x dx $ , $u$ from $ 0 \to \infty $
$T[f(x)]=\frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} (u)^{-ik}e^{-u} du$ $=\frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} (u)^{(1-ik)-1}e^{-u} du=\frac{\Gamma(1-ik)}{\sqrt{2\pi}}$
Is this true solution?
Is there a simplification of the final answer
Please, help me
Thank you for your participation