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.

  • 1
    $\begingroup$ I don't think you can apply Jordan's lemma in your calculation of $A_{\epsilon}$ since the function $g(z)=z/(z-i\epsilon)$ doesn't vanish as you push the radius of the semicircular contour to infinity. Jordan's lemma in this case states that indeed, the integral is bounded above, but it doesn't go to zero. $\endgroup$ Commented Nov 26, 2019 at 18:26

1 Answer 1


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 $$\theta'(x)=\delta(x)$$


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