Is there a sense (perhaps distributionally) where the Fourier transform pair of: $$e^x\cos(x)$$ makes sense? i have seen online FT calculators such as https://www.dcode.fr/fourier-transform
give the following answer: $$(\frac{\pi}{2})^.5(\delta(w-(1+i))+(\frac{\pi}{2})^.5(\delta(w+(1-i))$$
wolfram alpha doesn't give an answer for the FT to the prior function, but does to $e^x(\cos(x)+\sin(x))$.
i would expect the exponential to cause issues for the FT, but perhaps there is some distributional interpretation?
thanks
For
$$f(x)=e^x \cos(x)\tag{1}$$
the Fourier transform
$$F(\omega)=\mathcal{F}_x[f(x)](\omega)=\frac{1}{\sqrt{2 \pi}}\int\limits_{-\infty}^{\infty} e^x \cos(x)\, e^{i \omega x}\,dx$$ $$=\sqrt{\frac{\pi}{2}} \delta(\omega-(1+i))+\sqrt{\frac{\pi}{2}} \delta(\omega+(1-i))\tag{2}$$
doesn't makes sense. This can be seen by trying to evaluate the inverse Fourier transform
$$f(x)=\mathcal{F}_{\omega }^{-1}\left[F(\omega)\right](x)=$$ $$\frac{1}{\sqrt{2 \pi}}\int_{-\infty }^{\infty } \left(\sqrt{\frac{\pi }{2}} \delta(\omega-(1+i))+\sqrt{\frac{\pi }{2}} \delta(\omega+(1-i))\right) e^{-i x \omega}\,d\omega\tag{3}$$
where the integral is evaluated over $\omega\in\mathbb{R}$.
Note for $\omega\in\mathbb{R}$ the arguments of the two Dirac delta functions in formula (3) above are both complex whereas the Dirac delta function $\delta(y)$ is only defined for $y\in\mathbb{R}$ (and when $y\ne 0$). Therefore the Fourier transform result illustrated in formula (2) above is incorrect.
Note that for
$$f(x)=\cos(x)\tag{4}$$
the Fourier transform
$$F(\omega)=\mathcal{F}_x[f(x)](\omega)=\frac{1}{\sqrt{2 \pi}}\int\limits_{-\infty}^{\infty} \cos(x) e^{i \omega x}\,dx=\sqrt{\frac{\pi}{2}} \delta(\omega -1)+\sqrt{\frac{\pi}{2}} \delta(\omega+1)\tag{5}$$
is consistent with the inverse Fourier transform
$$f(x)=\mathcal{F}_{\omega }^{-1}\left[F(\omega)\right](x)=\frac{1}{\sqrt{2 \pi}}\int\limits_{-\infty}^{\infty} \left(\sqrt{\frac{\pi}{2}} \delta(\omega-1)+\sqrt{\frac{\pi}{2}} \delta(\omega+1)\right) e^{-i x \omega}\,d\omega$$ $$=\frac{e^{-i x}+e^{i x}}{2}=\cos(x)\tag{6}$$