# Integral leading to step function of zero argument

Consider the following integral with real $a>0$

$$I(a) = \int_0^{\infty} \mathrm{d}x~\delta(a-x) \Theta(x-a) f(x)$$

with $f(x)$ a function such that $f(a) \ne 0$, $\delta$ the Dirac function and $\Theta$ the unit step function.

A naive solution appears to be

$$I(a) = \Theta(0) f(a)$$.

How to make sense of this result? What to take for $\Theta(0)$?

How to evaluate integral $I(a)$ in a more rigourous way?

• Do you agree with the style changes I have done on your text ? – Jean Marie May 10 '17 at 10:09
• What's the context of this? Without some additional constraints I'm suspicious of there being a well-defined answer. Also, while I wouldn't characterize it as a duplicate, this question is similar in spirit to these questions: math.stackexchange.com/q/1740355/137524, math.stackexchange.com/q/2267909/137524. (Full disclosure, I've got an answer on the second one.) – Semiclassical May 10 '17 at 16:30
• This short article by Griffiths and Walborn may also be helpful: physics.usyd.edu.au/~cross/PUBLICATIONS/… – Semiclassical May 10 '17 at 16:59
• @Semiclassical Thanks. Its context is the quantum mechanics. What additional constraints could be? – Nigel1 May 10 '17 at 19:06

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\mrm{J}:\mathbb{R}^{2}\setminus \braces{\pars{x,y}\ \mid\ x = 0\ \vee\ x = y} \to R}$.