I want to find the Fourier transform of $ g(t)=(2t+\sin t)e^{e^{-t^2}}$.
Take the formula $V(\omega)=\int_{-\infty}^\infty v(t)e^{-i\omega t}$, and we get:
\begin{equation} V(\omega)=\int_{-\infty}^\infty (2t+\sin t)e^{-t^2}e^{-i\omega t}dt=\int_{-\infty}^\infty 2te^{-t^2}e^{-i\omega t}dt+\int_{-\infty}^\infty \sin te^{-t^2}e^{-i\omega t}dt \end{equation}
here, one gets three integrals, if replacing $\sin t= \frac{e^{it}-e^{-it}}{2i}$:
\begin{equation} V(\omega)=\int_{-\infty}^\infty 2te^{-t^2}e^{-i\omega t}dt+\frac{1}{2i}\int_{-\infty}^\infty e^{it}te^{-t^2}e^{-i\omega t}dt-\frac{1}{2i}\int_{-\infty}^\infty e^{-it}te^{-t^2}e^{-i\omega t}dt \end{equation}
From here, I am not able to see any known form from the Fourier transform table, except $f(t)sin(\Omega t)=\frac{\hat{f}(\omega-\Omega)-\hat{f}(\omega+\Omega}{2i}$, which would be used before changing the sine term. But either forms, have the $t^2+it-i\omega t$ factor in the exponents, which I can't really amend with the Fourier transform of $e^{-at^2}=\frac{\sqrt{\pi}}{\sqrt{a}}e^{-\omega^2/4a}$ either, since $t^2+it-i\omega t$ is very different from $at^2$.
What would be the right solution here?
Thanks