Understanding a proof that $\delta(x-x') = \mathrm{d}/\mathrm{d}x\, \theta(x-x')$

Question

A solution to an exercise in Principles of Quantum Mechanics:

My question concerns the leap

This is apparently an application of Integration by Parts; however, it seems to me it is conflating $$ \int g(x)\,\mathrm{d}\theta(x-x') $$

with

$$ \int g(x) \theta'(x-x')\,\mathrm{d}x $$

which is the form required for us to able to apply Integration by Parts. Is this not the case?

Answer 1

Riemann-Stieltjes integral

Definition. The Riemann–Stieltjes integral of a real-valued function $f$ of a real variable on the interval $[a,b]$ with respect to another real-to-real function $\alpha$ is denoted by $$\int_a^b f(x) \, \mathrm{d}\alpha(x).$$ The definition of the Riemann–Stieltjes integral can be found here.

Lemma 1. Given an $\alpha(x)$ which is continuously differentiable over $\mathbb{R}$. Then $$\int_a^b f(x) \, \mathrm{d}\alpha(x) = \int_a^b f(x)\alpha'(x) \, \mathrm{d}x.$$

Lemma 2. Let $f$ be a Riemann-Stieltjes integrable function with respect to $\alpha$ on the interval $[a,b]$. Then $\alpha$ be a Riemann-Stieltjes integrable function with respect to $f$ on the interval $[a,b]$ and $$\int_a^b f(x) \, \mathrm{d}\alpha(x)=\left[\alpha(x)f(x)\right]_{a}^{b}-\int_a^b \alpha(x) \, \mathrm{d}f(x).$$

A proof of this theorem can be found here.

Analyzing the leap

Choose $f(x) =g(x)$ and $\alpha(x)=\Theta(x-x')$. By Lemma 2,

$$\begin{align}\int_{-\infty}^{\infty} g(x) \, \mathrm{d}\Theta(x-x') &= \left[\Theta(x-x')g(x)\right]_{-\infty}^{+\infty}-\int_{-\infty}^\infty \Theta(x-x') \, \mathrm{d}g(x). \end{align}$$

If we let $g(x)$ be continuously differentiable over $\mathbb{R}$, by Lemma 1 we get $$\begin{align}\int_{-\infty}^\infty g(x) \, \mathrm{d}\Theta(x-x') &= \left[\Theta(x-x')g(x)\right]_{-\infty}^{+\infty}-\int_{-\infty}^{\infty} \Theta(x-x')g'(x) \, \mathrm{d}x. \end{align}$$

Answer 2

Okay well in terms of understanding why this is the case, visualising it graphically, $\theta(x-x')$ is going to be a step from $0$ to $1$ at the point $x=x'$. Now clearly this is a discontinuity but we can visualise this as a vertical line i.e. infinite gradient at the point $x=x'$. Now if we are to plot the gradient of $\theta$ we would see that it is clearly $0$ everywhere other than at this point as it is a constant function over the two domains. As for the integral you mentioned, it is completely fine to use it in that form. the dirac delta is commonly defined as: $$\int_{-\infty}^\infty g(x)\delta(x-x')dx=g(x')$$ so bearing this in mind: $$\int_{-\infty}^\infty g(x)d\theta(x-x')=\int_{-\infty}^\infty g(x)\frac{d\theta(x-x')}{dx}dx$$ $$=\int_{-\infty}^\infty g(x)\delta(x-x')dx=g(x')$$ which is effectively what this is proving