# Dirac Delta function at a point

From my understanding of the Dirac Delta function, it is infinitely thin and has a value of infinity at only a particular point. I also learned that $$\int_{-\infty}^{\infty} \delta(x-a) dx = 1$$ What if we have a case where we are faced with the integral: $$\int_{-\infty}^{a} \delta(x-a) dx$$ How does the fact that the function is infinitely thin at a affect the result?

• I think that's the problem if you learn about the Dirac Delta "function" this way. It is no function. Take it as a formal way to write it or you have to learn distribution theory.
– user301452
Commented Jan 10, 2016 at 14:15
• My guess is that a physicist's answer would be $$\int_{-\infty}^a\delta(x-a)dx=\frac12.$$ For a maths answer, see previous comment.
– Did
Commented Jan 10, 2016 at 14:31
• @user175257 the Dirac Delta is not a function at all. Commented Jan 10, 2016 at 14:39
• @user175257 True, as every mathematical object, and furthermore it is not a function.
– Did
Commented Jan 10, 2016 at 14:46
• @user175257: From the wiki page you quoted: "From a purely mathematical viewpoint, the Dirac delta is not strictly a function, because any extended-real function that is equal to zero everywhere but a single point must have total integral zero. The delta function only makes sense as a mathematical object when it appears inside an integral. While from this perspective the Dirac delta can usually be manipulated as though it were a function, formally it must be defined as a distribution that is also a measure." Commented Jan 10, 2016 at 14:58

Coming from a measure-theoretic point of view, one can write the following:

$\int_{-\infty}^{+\infty}\delta(x-a)dx = \int_{\mathbb R} 1 d \mu_a(x)$, where $\mu_a$ denotes the point measure with mass $1$ at $a$ and $0$ everywhere else.

Now, if we want to integrate over an interval $I$, we integrate the characteristic function $\chi_I(x)$ over $\mathbb R$:

$$\int_I \delta(x-a)dx= \int_{\mathbb R}\chi_I(x)d\mu_a(x).$$

And here comes a important distinction which we dont need to make when using the Riemann integral: If $a$ is a border point of the interval $I$, do we consider the open interval $I_1 =(-\infty,a)$, or the interval $I_2=(-\infty,a]$ ? Because the mass of $\mu_a$ lies at $a$, this is an important distinction:

$$\int_{(-\infty,a)} \delta(x-a)dx= \int_{\mathbb R}\chi_{(-\infty,a)}(x)d\mu_a(x) = 0$$

while

$$\int_{(-\infty,a]} \delta(x-a)dx= \int_{\mathbb R}\chi_{(-\infty,a]}(x)d\mu_a(x) = 1.$$

So we have to be very careful with what we mean by $\int_{-\infty}^a\delta(x-a)dx.$

Summing up: The fact that it's 'infinitely thin' translates to the measure having a point mass, and this means that the inclusion or the exclusion of single points will change the value of the integral.

• Although I don't have a background in measure theory, I think I understand where this can be useful - for example: in dealing with point masses or point charges in physics? Thank you so much! Commented Jan 10, 2016 at 22:30
• There are several applications for using the Dirac Delta in physics (the wikipedia page mentions some), also for the impulse response of dynamical systems this is of some use. Unfortunately, I cannot offer an explicit example where this kind of formula is used. Commented Jan 10, 2016 at 23:00

The Dirac Delta function is a symmetrical function.Thus $$\int_{-\infty}^{a} \delta(x-a) dx = \int_{a}^{+\infty} \delta(x-a) dx = \frac{1}{2}$$ Furthermore $$\int_{a^-}^{a^+}\delta(x-a) dx = 1$$ And $$\int_{a^-}^{a}\delta(x-a) dx = \int_{a}^{a^+}\delta(x-a) dx = \frac{1}{2}$$

• "The Dirac Delta function is a symmetrical function." Well, no, eventhough the sentence rings bizarrely, the Dirac Delta function is not a function.
– Did
Commented Jan 10, 2016 at 14:47
• What's the definition of the Dirac Delta as a function? What's its domain and what are its range? Commented Jan 10, 2016 at 14:53
• @Roland Thanks.I will be more careful about this.I learn the "Dirac Delta function" from a "signal and system" textbook not a professional mathematical textbook. Commented Jan 10, 2016 at 15:01
• @Roland, I guess the answer may be true in a certain context. Such bizarre things actually appear in physics where one takes a limit of a usual function to simplify calculations, e.g. A.O.Caldeira, A.J.Leggett used it in the formulas 3.28, 3.29, 3.31-3.35. The problem wouldn't emerge if they worked with an actual but unknown spectral density instead of a toy model 3.23. Perhaps, origins of OP's question are similar. Commented May 9, 2021 at 12:33