The following is Problem 11.9 in "Mathematical Physics: A Modern Introduction to Its Foundations, Second Edition" by Sadri Hassani.
Given the following representation of the step function: $$\theta(x) = \lim_{\epsilon\to 0}\frac{1}{2\pi i} \int_{-\infty}^\infty \frac{e^{i t x}}{t-i\epsilon} \, dt, $$ show that $\theta'(x) = \delta(x)$.
This was a homework problem for an undergraduate physics class, but I (the grad student TA) am having trouble solving it rigorously. The derivative is clearly $$\theta'(x) = \frac{1}{2\pi} \int_{-\infty}^\infty \frac{t e^{itx}}{t-i\epsilon} \, dt,$$ which is essentially the usual Fourier representation of the Dirac delta function $$\delta(x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{itx} \, dt.$$ However, the textbook has not introduced the Fourier transform yet, so this problem should be doable without using this representation.
It is easy to show that $\theta'(x) = 0$ for $x>0$ and $x<0$ by coverting the integral over $t$ into a contour integral with a semicircular portion in the UHP for $x>0$ and LHP in the $x<0$. One can also see that $\theta'(0)$ is infinite. Lastly, one can show that $\theta'(x)$ integrates to 1 over any region containing the origin.
This is the answer that I will accept from the students. However, I would like to have a more rigorous derivation involving a test function. Here is my attempt.
Let $$\theta_\epsilon(x) = \frac{1}{2\pi i} \int_{-\infty}^\infty \frac{e^{i t x}}{t-i\epsilon} \, dt $$ We wish to show $\displaystyle{\lim_{\epsilon\to 0} \theta_{\epsilon}'(x) = \delta(x)}$. The derivative is: $$\theta_\epsilon'(x) = \frac{1}{2\pi i} \int_{-\infty}^\infty \frac{it e^{itx}}{t-i\epsilon} \, dt $$ Let $g(x) \in \mathcal{S}(\mathbb{R})$ be a smooth test function (Schwartz function). We wish to show that $$\lim_{\epsilon \to 0} A_\epsilon = g(0),$$ where $$ A_\epsilon \equiv \int_{-\infty}^\infty \theta_\epsilon'(x) g(x) \, dx.$$ We integrate and convert one of the integrals into a contour integral: \begin{align*} A_\epsilon &= \frac{1}{2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{g(x) t e^{itx}}{t-i\epsilon} \, dt \, dx \\ & = \frac{1}{2\pi} \int_{-\infty}^\infty \underbrace{\int_{R} \frac{g(x)z e^{ixz}}{z-i\epsilon} \ dz}_{I_R} \, dx \end{align*} where $R$ is the contour of real numbers in $\mathbb{C}$. By Jordan's lemma, the contour integral $I_R$ is: $$I_R = \left\{\begin{matrix} g(x) \cdot 2\pi i \cdot i\epsilon e^{-\epsilon x} & x>0 \\ 0 & x<0 \end{matrix}\right.,$$ where we have closed the contour in the UHP or LHP for $x>0$ and $x<0$, respectively. Thus: $$A_\epsilon = -\epsilon \int_0^\infty g(x) e^{-\epsilon x} \ dx $$ Integrating by parts: \begin{align*} A_\epsilon &= -\epsilon \left[ -\frac{1}{\epsilon} e^{-\epsilon x} g(x)\Big|_0^\infty + \frac{1}{\epsilon} \int_{0}^\infty e^{-\epsilon x} g'(x) dx \right] \\ & = -g(0) - \int_0^\infty e^{-\epsilon x} g'(x) \ dx \\ \end{align*} where we used the fact that $g\to 0$ as $x \to\infty$.
Examining the second term, we see that since $e^{-\epsilon x} g'(x)$ is dominated by $g'(x)$ on $[0,\infty)$, we can use the Dominated Convergence Theorem to bring the $\epsilon \to 0$ limit inside the integral and find $$\lim_{\epsilon \to 0} A_\epsilon = -g(0) - \int_0^\infty g'(x) \, dx = -g(0) - g(x)\Big\vert_0^\infty = g(0) - g(0) = 0.$$ This is clearly not the desired result. I don't see anything wrong with the manipulations after the contour integral, so that step must be wrong. I guess splitting up the integral for $x>0$ and $x<0$ somehow misses the $x=0$ piece, which at the end of the day is the only piece that matters?
Any ideas would be appreciated.
Here's an idea for a proof that seems to be semi-rigorous and within the limitations of the class material (I would give full points to it at least :P).
Using complex analysis arguments it is easy to prove that
$$\theta_{\epsilon}(x)=e^{-\epsilon x}\theta(x)~,~x\neq0$$
If we define $\theta'(x)\equiv\lim_{\epsilon\to 0}\theta'_{\epsilon}(x)$
then it is not hard at all to see that when g(x) is a test function:
$$\int_{-\infty}^{\infty}\theta'_{\epsilon}(x)g(x)dx=-\int_{-\infty}^{\infty}\theta_{\epsilon}(x)g'(x)dx=-\int_{0}^{\infty}g'(x)e^{-\epsilon x}dx=g(0)-\epsilon\int_{0}^{\infty}g(x)e^{-\epsilon x}dx$$
Then we conclude, taking the limit $\epsilon\to0$ and using the dominated convergence theorem where appropriate:
$$\int_{-\infty}^{\infty}\theta'(x)g(x)dx=\lim_{\epsilon\to 0}\int_{-\infty}^{\infty}\theta'_{\epsilon}(x)g(x)dx=g(0)$$
and thus $$\theta'(x)=\delta(x)$$