$\DeclareMathOperator{\sech}{sech}$ $\DeclareMathOperator{\csch}{csch}$
I was curious why Wolfram Alpha does not give a Fourier transform for $\tan(t)$ when it does provide one for $\tanh(t)$. According to Wolfram Alpha $\mathcal{F}[\tanh(t)] = i\sqrt{\frac{\pi}{2}} \csch(\frac{\pi \omega}{2}) $. This is all well and good, however using the fact that $\mathcal{F}[g(a t)] = \frac{1}{|a|}\hat{g}(\frac{t}{a})$, we could easily show that $\mathcal{F}[\tanh(t)] = \sqrt{\frac{\pi}{2}} \csc(\frac{\pi \omega}{2})$. Note that, $-i \tanh(i t) = \tan(t)$ and $i \csch(i \omega) =\csc(\omega)$. Therefore, setting $a = i$ the following formula holds.
$-i\sqrt{\frac{\pi}{2 |a|}} \csch(\frac{\pi \omega}{2 a}) =-i\sqrt{\frac{\pi}{2|i|}} \csch(\frac{\pi \omega}{2 i}) =-\sqrt{\frac{\pi}{2}}\csc( \frac{-\pi \omega}{2}) = \csc(\frac{\pi \omega}{2}) $
This question implies that this sort of manipulation is allowed. Have I voilated some underlying integrability assumption where this is not a valid operation? Or does Wolfram Alpha have some oversight.
As context I was working a derivation of the fact $\sech(\pi t)$ is its own Fourier transform, and the method works for $\tan(t)$ as well. I was a bit surprised that Wolfram Alpha does not have $\tan(t)$, leading me to question my method.
$$\tan(t)=\frac{i \left(e^{-i t}-e^{i t}\right)}{e^{i t}+e^{-i t}}\tag{1}$$
so
$$\mathcal{F}_t[\tan(t)](\omega)=i \left(\mathcal{F}_t\left[\frac{e^{-i t}}{e^{i t}+e^{-i t}}\right](\omega)-\mathcal{F}_t\left[\frac{e^{i t}}{e^{i t}+e^{-i t}}\right](\omega)\right)=-\sqrt{\frac{\pi}{2}} \csc\left(\frac{\pi \omega}{2}\right)\tag{2}$$
(see Wolfram Alpha evaluation).