# “Line integral” of delta distribution

I have seen engineers write something to the effect : $\int_{-\infty}^x \delta_{0}(t) dt = H(x)$. Here $\delta_0(x)$ is the dirac delta distribution concentrated at origin and $H(x)$ is the step function. Even though the distributional derivative of $H(x)$ is the delta distribution, taking an integral of the delta distribution seems morally wrong, especially since $\delta(x)$ is not a lebesgue measurable function even. Besides an "intuitive" justification for this, is there a rigorous way to justify this?

• The dirac delta is a measure concentrated on a point. That is, if $E\subset \mathbb{R}$ then $\delta_0(E) = 1$ if $0\in E$ and $\delta_0(E) = 0$ otherwise. When you integrate is like integrating the 1 function respect to the $\delta$ measure. – I.C. Sep 17 '17 at 2:47
• I see, so $\int \delta dx := \int d(\mu)$ where $\mu$ is the dirac measure? – Abhi. A Sep 17 '17 at 2:54
• @Abhi, Yes, so if $x>0$ in your integral then $H(x)=\mu((-\infty, x)) = \delta_0((-\infty, x))=\int_{-\infty}^x 1 d(\delta_0)=1$ and if $x<0$ then $H(x)=0$. – I.C. Sep 17 '17 at 3:30
• @reuns I was trying to make sense of the integral of the distribution as evaluating it on the characteristic function but wasn't sure because they act on $C_c^{\infty}$ functions. I preferred to let my comment like that because maybe there is a way to see the integral but I wasn't sure. In any case if you agree just tell and I'll delete this comment also, was just because I supposed you were addressing your last comment to my edit. – I.C. Sep 17 '17 at 3:39
• $S(x)= \int_{-\infty}^x T(y)dy$ always is a well-defined distribution when $T$ is a compactly-supported distribution. When $T$ is a distribution of order $0$ (ie. a measure, as $\delta$) then $S$ is (represented by) a function. Note if $T$ is a compactly supported distribution then $\langle \int_a^x T(y)dy, \varphi \rangle = - \langle T, \int_a^x \varphi(t)dt \rangle$ perfectly makes sense as $\int_a^x \varphi(t)dt \in C^\infty$. And that's why distributions are useful : they make sense in general in analysis, and not only for $\varphi \in C^\infty_c$. – reuns Sep 17 '17 at 3:40

## 2 Answers

I tend to disagree with the approach that has been proposed.

In this case, I think that it would be better (or simpler, if you prefer) to think of $\delta_0$ simply as a measure, instead of a distribution. Namely, you have $$\int_{\mathbb R} \chi_{(-\infty,x]}d\delta_0=H(x),$$ where the integral is the Lebesgue integral of the function $\chi_{(-\infty,x]}$ with respect to the measure $\delta_0$.

As you see there is no need for convolutions or really distribution theory altogether. Note also that $\delta_0$ is a Radon measure and so lives in the subset $(C^0_c)'$ of $(C^\infty_c)'$, no need for $C^\infty$ functions.

• I agree that one can interpret $\delta$ as a measure and do what you have proposed, see also @I.C comment above. I was just curious, since $\delta$ is a distribution and so is $H(x)$ and engineers tend to write $\int_{-\infty}^{x}\delta(t)dt = H(x)$, can we prove equality of the left and the right hand sides as distributions. Hence , the approach with convolutions. – Abhi. A Sep 18 '17 at 1:32
• Sure, as @reuns noted it (I liked the comment!). But distributions are of very little use, if at all. They failed to solve the initial motivating problem of nonlinear PDE's besides the well known problem of multiplication, which is needed in particular in physics. From my point of view, one should avoid distributions whenever possible. – John B Sep 18 '17 at 1:42

Here is my attempt to justify this rigorously: Note that $\int_{-\infty}^{x}\delta(t)dt = \int_{-\infty}^{\infty}\delta(t)H(x-t)dt = H*\delta = H(x)$. Here $*$ represents the operation of convolution. Note that the convolution of two distributions is justified as long as one of the distributions is compactly supported, see e.g. $\it{Functional Analysis, 2nd Ed., Rudin}$.