Fourier transform of derivative of plane wave What is the Fourier transform $F(k)$ of:
$$ f(y) = A \, ik \, e^{iky} $$
If you calculate it with Wolfram Alpha, it says that there are no results found in terms of standard functions. 
 A: Wolfram Alpha is correct here. That Fourier transform doesn't exist as a standard function, but it does exist as a distribution - it's the distributional derivative of the Dirac delta.
Distributions are basically functionals, i.e. they're objects $\mathcal D$ which take a ("test") function $\varphi\in L_2(\mathbb R)$, say, and give a number $\mathcal D[\varphi] \in \mathbb C$. The Dirac delta is such one such distribution, and it is defined as
$$
\delta[\varphi] = \varphi[0].
$$
The distributional derivative is defined via a simple formula - you just defined as
$$
\delta'[\varphi] = -\varphi'(0),
$$
i.e., where the Dirac delta returns the value of what it's "integrated" against at the origin, the Dirac-delta's distributional derivative returns the value of the derivative of its argument at the origin. 
The origin for that identification comes from the heuristic that if you integrate by parts you should be able to get
\begin{align}
\delta'[\varphi]
& = \int_{-\infty}^\infty \delta'(x) \varphi(x) \mathrm dx
\\ & = \left[\delta'(x) \varphi(x) \right]_{-\infty}^\infty - \int_{-\infty}^\infty \delta(x) \varphi'(x) \mathrm dx
\\ & = -\varphi'(0),
\end{align}
with the boundary terms vanishing because $\varphi(x)$ should vanish at infinity. (This is obviously a shaky argument and it only works as a heuristic - if you want to make it work rigorously, you need to start with distribution theory from the beginning.)
