Fourier transform of exponent? Mathematica fails to find a Fourier transform of exponent.
Yet according to this page
$$\mathcal{F}[e^{2\pi iat}]=\delta(t-a)$$
and via substitution,
$$\mathcal{F}[e^{at}]=\delta\left(t-\frac a{2\pi i}\right)$$
Yet this does not make much sense because inverse Fourier transform of this will give 0.
Thus my question is what one can do so to fix this and make Fourier transform of exponent useful and revertible.
Possibly some modification of Dirac Delta for complex argument? Or taking Fourier integral over complex plane? Or taking two integrals, one over reals the other over imaginary axis?
 A: Mathematica $10.0.2.0$ can find the Fourier transform of $f(t) = e^{2\pi i at}$.

Mathematica defines the Fourier transform as 
$$
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{i\omega t}dt
$$
Now we can derive the solution shown by Mathematica and the website.
\begin{align}
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{2\pi iat}e^{i\omega t}dt &=
\frac{1}{\sqrt{2\pi}}\lim_{x\to\infty}\int_{-x}^{x}e^{it(2\pi a+\omega)}dt\\
&= \sqrt{2\pi}\lim_{x\to\infty}\frac{\sin[(2\pi a+\omega)x]}{(2\pi a+\omega)\pi}
\end{align}
Let $y=2\pi a+\omega$ and $\epsilon=\frac{1}{x}$. Then
\begin{align}
\sqrt{2\pi}\lim_{\epsilon\to 0}\frac{\sin[y/\epsilon]}{y\pi} &= \sqrt{2\pi}\delta(y)\\
&= \sqrt{2\pi}\delta(\omega + 2\pi a)
\end{align}
Now, we can take the inverse Fourier transform $F(\omega) = \sqrt{2\pi}\delta[\omega - (-2\pi a)]$ so 
$$
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\sqrt{2\pi}\delta[\omega - (-2\pi a)]e^{-i\omega t}d\omega = e^{2\pi iat}\tag{1}
$$
which occurs by the sifting property of the Dirac Delta function.
