# What does Dirac delta function of a constant mean?

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?

• The Dirac delta function of a constant (not equal to zero) is of course zero. – David G. Stork Sep 20 '17 at 16:20
• 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. – Erick Wong Sep 20 '17 at 16:23
• 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. – mathreadler Sep 20 '17 at 16:28
• 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. – JMoravitz Sep 20 '17 at 16:52
• 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. – 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.

$$H(x) := \int_{-\infty}^{x} \delta(\tau)~\mathrm d\tau$$ 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.
• 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? – Did Sep 20 '17 at 17:27
• How would you give meaning to $H(0)$? $\frac{1}{2}$ or 1 or no value at all? – ty. Sep 20 '17 at 17:51
• @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 $$\lim_{\alpha \to 0^+} \left( \frac{1}{2\sqrt{\alpha\pi}} \mathrm e^{-x^2/4\alpha} \right)$$ – Fly by Night Sep 20 '17 at 19:59
• 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. – Did Sep 20 '17 at 20:27