# “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

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
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. $\it{Functional Analysis, 2nd Ed., Rudin}$.