How can I compute the derivative of delta function using its Fourier definition?

Question

I am wondering if it's possible to compute the derivative of the Dirac Delta function using the definition obtained from Fourier transformation: $$\delta(x-x')=\frac{1}{2\pi}\int e^{-ik(x-x')}dk.$$ What I have attempted is the following (all the integrals are from -infinity to +infinity): $$\frac{d}{dx}\delta(x-x') = \frac{1}{2\pi} \int \frac{d}{dx}e^{-ik(x-x')}dk=\frac{-1}{2\pi} \int e^{-ik(x-x')} \cdot ik dk$$ $$=\frac{-1}{2\pi} \left\{\frac{-ke^{-ik(x-x')}}{x-x'}\bigg|_{-\infty}^{\infty} + \int_{-\infty}^{\infty}\frac{e^{-ik(x-x')}}{x-x'}dk\right\}$$ If the first term would to be zero, then the second term would become $\frac{-\delta(x-x')}{x-x'}$. Intuitively, this should be the derivative of the Delta function: when $x'$ is approached from the left, its derivative goes from 0 to infinity; from the right, the derivative goes from 0 to negative infinity. However, I have not been able to show that the first term is indeed zero. Can you prove/disprove this?

Answer 1

The first term is not zero in any direct sense, in fact the expression clearly diverges. The reason that in physics you can get away with pretending it is zero is that $\delta$ and its derivative $\delta'$ aren't actually functions with a converging Fourier expansion in the first place, but, as they are often called, distributions.

In my opinion the easiest way to understand it is that they're dual vectors of a function space. Specifically, $\delta$ is in the dual of the space $(\mathcal{C}^0(\mathbb{R}))^\ast$ of the space of continuous functions, and $\delta'$ is in $(\mathcal{C}^1(\mathbb{R}))^\ast$ i.e. continuously differentiable functions. An easy and rigorous way to define them is $$\begin{align} \delta_{x_0} \: f :=& f(x_0) \\ \delta'_{x_0} \: f :=& -f'(x_0) \end{align}$$ i.e. the argument of $\delta$ is actually a function, not a real number. Anything written in the $\delta(x-x')$ style is in fact just pseudo-notation, that only becomes well-defined when it appears in an integral: $$\begin{align} \int_\Omega\!\!\mathrm{d}x\ \delta(x-x_0) \cdot f(x) := \delta_{x_0}\:f =& f(x_0) & \text{if $x_0\in\Omega$} \end{align}$$ Equivalently, you can do all of this in Fourier space. The expansion $\delta(x) \propto \int\mathrm{d}k\ e^{-ikx}$ doesn't actually converge by itself, however it does converge when frequency-wise multiplied with the Fourier transform of a continuous function, because such an expansion has coeffients decaying with at least $O(k^{-1})$, so $$ \|e^{-ikx} \cdot\operatorname{FT}(f)(k)\| \leq O(k^{-1}) $$ and an oscillating function that decays in that way can be integrated.

Likewise, the Fourier expansion you derived for $\delta'$ does make sense after you multiply it frequency-wise with the expansion of a continuously differentiable function, because that decays in $O(k^{-2})$ and therefore $$ \bigl(\tfrac{\mathrm{d}}{\mathrm{d}x}\:\delta(x-x')\bigr)\:f(x) \propto \left.\left[\frac{-k\cdot e^{-ik(x-x')}}{x-x'}\cdot O(k^{-2})\right]\right|_{-\infty}^\infty + \ldots $$ and here $k\cdot O(k^{-2})$ gives something in $O(k^{-1})$, which therefore does vanish at infinity meaning your derivation is correct.