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I'm trying to show that $\int_{-\infty}^x\delta(a)da= \theta(x)$ for $x\neq0$, where $\delta(x)$ is the dirac delta function, and $\theta(x)$ is the step function , which equal to $0$ for $x\leq0$ and $1$ when $x>0$.

By intuition, this integral makes sense to me. However, to evaluate this integral, we will need to find the anti-derivative of $\delta(x)$, and I saw in some definitions this is just $\theta(x)$. How can I evaluate this integral? Also, how can we take care of the case where $x = 0$? Thanks!

IGY
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    It is confusing to use the same symbol $x$ for both the upper end of the range of integration and also the dummy variable in the integral. To evaluate an integral with $\delta$ in there, you need to apply what the $\delta$ means as a distribution (generalized function). In general, distributions need not make sense when evaluated at a point. – GEdgar Oct 09 '22 at 20:39
  • @GEdgar Thanks for the comment! Should I take $\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}x^2/\sigma^2}$ and make $\sigma \rightarrow 0$? – IGY Oct 09 '22 at 20:52
  • No, I mean take a test function and see how this behaves when applied to it. – GEdgar Oct 09 '22 at 21:06
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    $\delta$ is not really a function. It might be understood as a measure on the sets of $\mathbb{R}$ that assigns $1$ to all sets containing $0$ and $0$ otherwise, or as a distribution with acts on $\mathcal{C}^\infty_c(\mathbb{R})$ by assigning $\phi$ tp $\phi(0)$. As a measure, you have $\int\mathbb{1}{(-\infty,x]}(t)\delta(dt)=\mathbb{1}{(-\infty,x]}(0)=\mathbb{1}_[0,\infty)(x)$. – Mittens Oct 09 '22 at 23:48

2 Answers2

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By definition, for any $\varphi$ continuous at $y=0$, $\int_{\Bbb R} \varphi(y)\,\delta_0(\mathrm d y) = \varphi(0)$. In this case, notice that since $x\neq 0$, $\theta(x-y) = \mathbf{1}_{(-\infty,x]}(y)$ is indeed continuous at $y=0$, hence $$ \int_{-\infty}^x \delta_0(\mathrm d y) = \int_{\Bbb R} \theta(x-y)\, \delta_0(\mathrm d y) = \theta(x) $$ since $\theta(x-y) = \theta(x)$ when $y=0$.

Edit: About your edit about what happens at $x=0$, there is no clear meaning to this. See e.g. What is the value of the integral $\int_{-\infty}^a \delta(x - a) dx$ and related integrals? if you want my opinion on this.

LL 3.14
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  • Thanks so much for the answer! I might be a bit confused with the notation. Is $\delta_0(dy)$ the delta function centered at 0? Why dy is in the patrenthesis? – IGY Oct 10 '22 at 20:15
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    Yes, $\delta_0$ indicates that it is centered at $0$. The notation is here to underline the fact that the Dirac delta is not a function, but either a distribution or a measure. As a measure, it leads to the integration with respect to this measure. If you do not like it, you can think of it as being "$\delta_0(x),\mathrm d x$" if $\delta_0$ was a function. I explained this notation here https://math.stackexchange.com/questions/3801916/working-with-infinitesimals-of-the-form-dfx-for-example-dax-and-relating/3807217#3807217 for example. – LL 3.14 Oct 10 '22 at 21:18
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    But the idea is that in general a measure $\mu$ is defined on sets $\mu(A)$, and $\int f(x),\mu(\mathrm d x)$ is a common notation for integration of a function $f$ with respect to a measure $\mu$, while the notation $\int f,\mathrm d g$ should be rather used when $g' = \mu$ in the sense of distribution to be compatible with the the Stieltjes integral (see e.g. https://en.wikipedia.org/wiki/Lebesgue%E2%80%93Stieltjes_integration). – LL 3.14 Oct 10 '22 at 21:19
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    I could also have written it with the duality pairing as in Mark Viola's answer. To summarize $$ \langle \delta_0,\varphi\rangle = \langle \theta',\varphi\rangle = \int \varphi(x),\mathrm d \theta(x) = \int \varphi(x),\delta_0(\mathrm d x) = \varphi(0) $$ The first is the expression of the Dirac delta as a linear form on continuous functions, the second comes from the fact $\theta' = \delta_0$ in the sense of distributions, the third is a Stieltjes integral, the fourth is the integral with respect to the Dirac measure defined by $\delta_0(A) = 1$ if $0\in A$ and $0$ else. – LL 3.14 Oct 10 '22 at 21:30
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As a measure, and using the Riemann-Stieltjes integral, we can write for $x\ne0$

$$\begin{align} \int_{-\infty}^x \,d\{\delta(x)\}&=\int_{-\infty}^x \,d\theta(a)\\\\ &=\theta(x) \end{align}$$

And we are done!

As a distribution, $\delta$ has compact support on $\{0\}$. As such, we can write for any compactly supported $\phi$ that is continuous at $0$

$$\langle \delta_0,\phi \rangle=\phi(0)$$

So, for $\phi(a)=\theta(x-a)$ we find that for $x\ne0$

$$\langle \delta_0, \theta_x\rangle=\theta(x)$$

And we are done!

Response to "How can we take care of the case where $x=0$?

Note that the distribution $\langle \delta_0,\theta_0\rangle$, sometimes written $\int_{-\infty}^\infty \delta(x)\theta(x)\,dx=\int_0^\infty \delta(x)\,dx$, is meaningless as shown in THIS ANSWER.

Mark Viola
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  • Strictly speaking, distributions should only have $C^\infty_c$ test functions, so your second "As a distribution" example should rather be called a dual element of $C^0$ (which is the way the Bourbaky group actually decided to define measures for example). This is why from my point of view, measures are both measures on sets and elements of the dual of $C^0$. Hence I would call the two approaches "as a measure" (of course the theorem behind this is the fact that $\int \varphi\mathrm d\theta = \int \varphi \delta_0 = \varphi(0)$ if $\delta_0$ is defined as a measure on sets. – LL 3.14 Oct 10 '22 at 19:06
  • On another side, to prove it from distributions, I would rather argue that the measure $\delta_0\in(C^0)'$ is the unique extension of the distribution $\delta_0$ and Riesz representation theorem, or prove that for any family of test functions $\varphi_n$ in $C^\infty_c$ approaching $\mathbf{1}_{(-\infty,x)}$, it holds $\langle\delta_0,\varphi_n\rangle \to \theta(x)$. That is a completely different type of theorems behind, and what I would call the proof "as a distribution". – LL 3.14 Oct 10 '22 at 19:10
  • @LL3.14 Indeed. Distributions operate on test functions (i.e., function that are $C_C^\infty$). Inasmuch as $\delta_0$ is a distribution with compact support ${0}$, then $\delta_0$ can be extended to compactly supported functions that are continuous at $0$. I had included this in the posted solution. And you are correct that we could have approached the problem using a sequence of test functions that approach $\mathbf{1}_{(-\infty,x)}$, which requires a different apparatus. – Mark Viola Oct 11 '22 at 17:38
  • Many thanks for the answer and comment! Some definitions, like Riemann-Stieltjes integral and 'measure', are new to me, so I'm trying to read a bit more about how that works. Could you explain a bit more about the first equality (The two integrals are equal)? – IGY Oct 11 '22 at 22:58
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    @IGY I've embedded links to reference the terms "As a measure," "Riemann-Stieltjes integral," and "As a distribution." Please let me know how I can improve my answer further. I really want to give you the best answer I can. – Mark Viola Oct 11 '22 at 23:33