I have seen a formula that unit step function is the integration of Dirac delta function.

$$H(x) = \int_{-\infty}^{x} \delta(t)~\mathrm dt $$

In evaluating the integral if we take the integral as sum of infinite terms ,what does delta function of a constant represent?

  • 1
    $\begingroup$ The Dirac delta function of a constant (not equal to zero) is of course zero. $\endgroup$ – David G. Stork Sep 20 '17 at 16:20
  • $\begingroup$ Keep in mind that Dirac delta is actually a distribution, not a function. So it is not something that can be evaluated in the sense of Riemann integrals. $\endgroup$ – Erick Wong Sep 20 '17 at 16:23
  • $\begingroup$ If you integrate it multiplied together with a function it "picks out" the value at t=0 of that function. Well, assuming 0 is in the interval of integration of course. $\endgroup$ – mathreadler Sep 20 '17 at 16:28
  • $\begingroup$ That image hurts to look at... so pixelated. We have the ability to type with $\LaTeX$ here. Visit this page to learn more about how to properly typeset mathematics on this site. $\endgroup$ – JMoravitz Sep 20 '17 at 16:52
  • 1
    $\begingroup$ I think OP wants to know how to use the classical idea of Riemann integration, where integration is a limit of more and more, thiner and thiner rectangles. $\endgroup$ – Fly by Night Sep 20 '17 at 17:03

The $\delta$-function is not actually a function - it's a distribution. The idea of finding an integral by using lots of thin rectangles doesn't work here. This is a more abstract form of integration. In fact, lots of integrals from Quantum Mechanics do not converge in the classical sense.

The $\delta$-function has the property that $\delta(x) = 0$ for all $x \neq 0$.

So, for example, $\delta(2) = 0$ and $\delta(-3.4)=0$.

The value of $\delta(0)$ is not well-defined, but we do know that $$\int_{-\infty}^{\infty}\delta(x)~\mathrm dx = 1$$

Since $\delta(x) = 0$ for all $x \neq 0$, the values of $x$ away from zero contribute nothing to the integral:

$$\int_{-\varepsilon}^{\varepsilon} \delta(x)~\mathrm dx = 1$$

for any $\varepsilon > 0$, as small as you like!

If $S$ is some open subset of the real numbers then $$\int_S \delta(x)~\mathrm dx \ \ = \ \ \left\{ \begin{array}{ccc} 1 & : & 0 \in S \\ 0 & : & 0 \notin S \end{array}\right.$$

In fact, you can make even stronger statements, e.g. $S$ doesn't need to be open, but you need to be careful how you word it.

In your example

$$H(x) := \int_{-\infty}^{x} \delta(\tau)~\mathrm d\tau$$ the set $S$ is the interval $(-\infty,x)$. If $x < 0$ then $0 \notin S$ and so $H(x) = 0$ for all $x < 0$. If $x>0$ then $0 \in S$ and so $H(x) = 1$ for all $x > 0$. What happens when $x=0$ depends on whom you speak to.

  • 1
    $\begingroup$ Why do you write things such as $$\delta(x)$$ after a first paragraph which explains (correctly) that, whatever $\delta$ is, $\delta$ is not a function? $\endgroup$ – Did Sep 20 '17 at 17:27
  • $\begingroup$ How would you give meaning to $H(0)$? $\frac{1}{2}$ or 1 or no value at all? $\endgroup$ – ty. Sep 20 '17 at 17:51
  • $\begingroup$ @Did It's a commonly used abuse of notation. It kind of makes sense if you think of the $\delta$-function as the limit of the Gaussian functions $$\lim_{\alpha \to 0^+} \left( \frac{1}{2\sqrt{\alpha\pi}} \mathrm e^{-x^2/4\alpha} \right)$$ $\endgroup$ – Fly by Night Sep 20 '17 at 19:59
  • $\begingroup$ Sorry but I am not the one you should explain what $\delta$ is, to. The OP asks for the details of $\delta$ hence every slip of language, even one which is used by the experts, is detrimental here. $\endgroup$ – Did Sep 20 '17 at 20:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.