Is the fourier transform F of $\cosh (ax)$: $ F(\cosh (ax))=\sqrt \frac{\pi}{2} \delta(\omega-ia)+\sqrt \frac{\pi}{2} \delta(\omega+ia)$
This is the answer wolfram alpha gives. I am looking up the meaning of a complex delta and can't find anything. So I think wolfram is just spitting out nonsense. Is there any meaning to a complex valued delta function?
$\cosh$ is a divergent integral. So normally you would say it has no fourier transform. However it is possible to assign consistent transformations to divergent integrals (like $1/t$ $1/t^2$)
$\delta(\omega-ia)$ means you need to insert $ia$ where you see $\omega$ in the integral, which indeed will result in $\cosh x$ if you were to apply the inverse transform integral.
This representation clearly does not make sense, however it is consistent and if you make calculations with it, it will work out in the end.