# Taking the derivative of the limit representation of the Heaviside function

(I should preface this by saying that my background is in physics-forgive me if I'm doing/asking something obvious!)

I am given this limit representation of the Heaviside step function

$$\theta(t) = \lim_{\epsilon \rightarrow 0^+} \frac{1}{2\pi i} \int_{-\infty}^{\infty} d\omega \frac{e^{i\omega t}}{\omega - i \epsilon}$$

to show that

$$\frac{d}{d t}\theta(t) = \delta(t)$$

My first instinct to simplify this problem was to move the limit inside the integral, like so:

\begin{equation*} \begin{aligned} \theta(t) &= \lim_{\epsilon \rightarrow 0^+} \frac{1}{2\pi i} \int_{-\infty}^{\infty} d\omega \frac{e^{i\omega t}}{\omega - i \epsilon} \\ &= \lim_{\epsilon \rightarrow 0^+} \int_{-\infty}^{\infty} \frac{1}{2\pi} \frac{e^{i\omega t}}{i\omega - i^2 \epsilon} d\omega \\ &= \int_{-\infty}^{\infty} \frac{1}{2\pi} \lim_{\epsilon \rightarrow 0^+} \frac{e^{i\omega t}}{i\omega - i^2 \underbrace{\epsilon}_\text{= 0}} d\omega \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} d\omega \frac{e^{i\omega t}}{i\omega} \end{aligned} \end{equation*}

However, when I googled moving limits inside integrals, I was told by the internet not to do so unless the integral is constant. I know that the Heaviside is not constant and has a discontinuity. But then I also thought "if we are doing the limit from the positive side of zero, wouldn't that be avoiding the discontinuity?" I am not sure this reasoning works, so that would be my question 1.

From this point I took a derivative with respect to time of the Heaviside

\begin{equation*} \begin{aligned} \frac{d}{dt}[\theta(t)] &= \frac{1}{2\pi} \int_{-\infty}^{\infty} d\omega \frac{\frac{d}{dt}[e^{i\omega t}]}{i\omega} \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} d\omega \frac{i\omega e^{i\omega t}}{i\omega} = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i\omega t} d\omega \\ &= \frac{1}{\sqrt{2\pi}}[\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{i\omega t} d\omega] \end{aligned} \end{equation*}

It was at this point I thought the integral reminded me of the Fourier transform. I know that the Fourier transform of the delta function is

$$\mathcal{F}_x[\delta(x-x_0)](k) = \int_{-\infty}^{\infty} e^{-2\pi i k x}\delta(x-x_0)dx = e^{-2\pi i k x_0}$$

and also

\begin{equation*} \begin{aligned} \mathcal{F}[\delta(x)](k) &= 1 \\ \mathcal{F}[\theta(x)](k) &= \frac{1}{ik} + \pi\delta(k) \end{aligned} \end{equation*}

I then took the Fourier transform of $\delta(x-0) = \delta(x)$:

\begin{equation*} \begin{aligned} \mathcal{F}[\delta(x)](\omega) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-i\omega x}\delta(x)dx \\ &= \frac{1}{\sqrt{2\pi}} e^{-i\omega (0)} = \frac{1}{\sqrt{2\pi}} \end{aligned} \end{equation*}

Since I know that $\mathcal{F}^{-1}[1](x) = \delta(x)$ I figured that could apply a similar method of performing the inverse Fourier transform on $\frac{1}{\sqrt{2\pi}}$.

\begin{equation*} \begin{aligned} \mathcal{F}^{-1}[\frac{1}{\sqrt{2\pi}}](x) &= \frac{1}{\sqrt{2\pi}}\delta(x) = \frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{i\omega x}dx \end{aligned} \end{equation*}

I returned to the time derivative of the Heaviside function

\begin{equation*} \begin{aligned} \frac{d}{dt}[\theta(t)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i\omega t} d\omega \end{aligned} \end{equation*}

The RHS of $\frac{d}{dt}[\theta(t)]$ looks incredibly similar to the RHS of $\mathcal{F}^{-1}[\frac{1}{\sqrt{2\pi}}](x)$, except for a change in variables. Thus I said that

\begin{equation*} \begin{aligned} \frac{d}{dt}[\theta(t)] = \mathcal{F}^{-1}[\frac{1}{\sqrt{2\pi}}](\omega) = \underbrace{\frac{1}{2\pi}\delta(\omega) = \frac{1}{2\pi}\delta(t)}_\text{change variables} \end{aligned} \end{equation*}

However! This is not the statement

$$\frac{d}{d t}\theta(t) = \delta(t)$$

$$\frac{d}{d t}\theta(t) = \frac{1}{\sqrt{2\pi}}\delta(t)$$

which means I obviously did something wrong while taking my Fourier transforms, but I'm not entirely sure what.

tl;dr: 1.) Can I take the limit inside of the integral in the expression $\theta(t) = \lim_{\epsilon \rightarrow 0^+} \frac{1}{2\pi i} \int_{-\infty}^{\infty} d\omega \frac{e^{i\omega t}}{\omega - i \epsilon}$? and 2.) what am I not understanding about Fourier transforms that resulted in this extra factor of $\frac{1}{\sqrt{2\pi}}$ in the final answer?

• you can't exchange the limit and the integral because otherwise the integral does not converge, and
• you mixed up your conventions for the Fourier transform.

The "right" way to show this equality invokes the theory of distributions.

For $\epsilon >0$, define the nascent step function by $$\theta_{\epsilon}(t)\stackrel{\text{def}}{=}\int_{-\infty}^{\infty}\frac{\mathrm{d}\omega}{2\pi\mathrm{i}}\frac{\mathrm{e}^{\mathrm{i}\omega t}}{\omega-\mathrm{i}\epsilon}\text{.}$$

We're tasked with showing $$\frac{\mathrm{d}}{\mathrm{d}t}\lim_{\epsilon\to 0^+}\theta_{\epsilon}(t)\stackrel{?}{=}\delta(t)\text{.}$$ This is an equality of distributions: by definition, it amounts to showing that for all test functions $\phi$, $$\langle(\lim_{\epsilon\to 0^+}\theta_{\epsilon})',\phi\rangle\stackrel{?}{=}\langle \delta,\phi\rangle\stackrel{\text{def}}{=}\phi(0)\text{.}$$

But by the definitions of $(\quad)'$ and $\lim$ for distributions we have $$\begin{split}\langle(\lim_{\epsilon\to 0^+}\theta_{\epsilon})',\phi\rangle &=-\langle\lim_{\epsilon\to 0^+}\theta_{\epsilon},\phi'\rangle \\ &=-\lim_{\epsilon\to 0^+}\langle \theta_{\epsilon},\phi'\rangle\text{.} \end{split}$$

Using the calculus of complex variables (or Fourier theory),

$$\theta_{\epsilon}(t)=\begin{cases}0 & t<0 \\ \tfrac{1}{2} & t=0 \\ \mathrm{e}^{-\epsilon t} & t>0\end{cases}$$ so $$\begin{split}-\langle \theta_{\epsilon},\phi'\rangle&=-\int_0^{\infty}\mathrm{e}^{-\epsilon t}\phi'(t)\mathrm{d}t\\ &=\phi(0)-\epsilon\int_0^{\infty}\mathrm{e}^{-\epsilon t}\phi(t)\mathrm{d}t\end{split}$$ for any test function $\phi$. Taking the limit of both sides as $\epsilon \to 0^+$ gives the result.