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I was trying to figure out what is the integral of $$\int_{-\infty}^{+\infty} H(t)\delta(t)dt,$$

where $H(t)$ is the Heaviside step and $\delta(t)$ is the Dirac delta.

A first approach: We observe that $\delta(t) = H'(t)$, and hence:

$$\int_{-\infty}^x H(t)\delta(t)dt = \int_{-\infty}^{+\infty} H(t)H'(t)dt = \left.\frac{1}{2}H^2(t)\right|_{t=-\infty}^{t = +\infty} = \frac{1}{2}(1-0) = \frac{1}{2}.$$

A second approach: Since $\int_{-\infty}^{+\infty} f(t)\delta(t)dt = f(0)$, then:

$$\int_{-\infty}^x H(t)\delta(t)dt = H(0) = 1.$$

What's wrong?

the_candyman
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    Your second approach is wrong, because the identity $\langle \delta, f\rangle = f(0)$ holds for smooth functions, whereas $H$ is a distribution. –  Jun 14 '18 at 19:32
  • @T.Bongers Thanks! Then, can I rely on the first approach? – the_candyman Jun 14 '18 at 19:33
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    Also note that $H(0)$ is often defined as being $1/2$ ( http://mathworld.wolfram.com/HeavisideStepFunction.html ) – Winther Jun 14 '18 at 19:35
  • Dirac delta is Distribution, you can only define it properly on test functions like functions with compact support.. – john Jun 14 '18 at 19:35
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    @john Thus, both approaches are wrong? – the_candyman Jun 14 '18 at 19:39
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    I'm not sure that the integral is even well defined. I suspect that depending on what you do, you can get any value between $0$ and $1$. – celtschk Jun 14 '18 at 19:41
  • First, we have to know how to multiply distributions, like $H(t) \delta(t)$ – GEdgar Jun 14 '18 at 19:45
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    @the_candyman not wrong but you have to be aware of what you use as a definition. I think the first approach is a bit more problematic, as you are differentiating something that is not continuous. This approach still works if the functions you are looking at are well behaved, but here it breaks down. – john Jun 14 '18 at 20:09
  • If one uses a non-even regularization of the Dirac Delta, say a pulse function around $0$ that is not centered at $0$, then the distribution $\langle H,\delta \rangle$ can be any value between $0$ and $1$. In that respect, the distribution is undefined. – Mark Viola Jun 14 '18 at 20:10

1 Answers1

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Let $a\in (0,1)$ and $\delta_n(x)$ be the regularized Dirac Delta given by

$$\delta_n(x)=\begin{cases}n&,x\in[-\frac an,\frac{1-a}n]\\\\0&,\text{elsewhere}\tag1\end{cases}$$

For all smooth functions with compact support $\phi$, we have

$$\lim_{n\to \infty}\int_{-\infty}^\infty \phi(x)\delta_n(x)\,dx=\phi(0)$$

Now, let's analyze the integral of $\delta_n(x)H(x)$. Proceeding, we have

$$\begin{align} \lim_{n\to\infty}\int_{-\infty}^\infty H(x)\delta_n(x)\,dx&=\lim_{n\to\infty}\int_{0}^{(1-a)/n}n\,dx \\\\&=1-a\tag2 \end{align}$$

Inasmuch as the value the integral in $(2)$ depends on the regularization of the Dirac Delta, $\delta(x)$, we assert that the distribution $H \delta$ fails to exist.

It is of interest to note that the regularization $\delta_n(x)$ as given in $(1)$ is consistent with defining the Heaviside function as

$$H(x)=\begin{cases}1&,x>0\\\\ a&,x=0\\\\0&,x<0\end{cases}$$

And naively evaluating $\langle H,\delta\rangle$ as $H(0)$ would give $\int_{-\infty}^\infty H(x)\delta(x)\,dx=a$, which doesn't agree with the result in $(2)$ unless $a=1/2$.

Mark Viola
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  • Thanks for this, Mark! I never had the need to study distributions beyond the chapter of Rudin's Functional Analysis where they were introduced. I came here from this thread in a review queue, felt that something is not right in splitting that integral at zero, then saw copper.hat's comment, then yours. I learned something today :-) – Jyrki Lahtonen Aug 05 '18 at 05:51
  • Hi Jyrki. Your comment just made my day. You're welcome of course. And thank you. – Mark Viola Aug 05 '18 at 12:40