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$$\int_{a-\epsilon}^{a+\epsilon} \delta(x - a)dx = 1$$

I can see intuitively why this is so as $a$ is inside the the domain of integration and all other values in the domain contribute $0$ to the integral by the properties of the dirac delta function.

But how would you formally derive this? As in, taking the integral of the function, then evaluating it, resulting in an answer of $1$? (Aside: I'm not sure how to integrate a discontinuous function like this, ie. where $\delta = 1$ if $x = a$, and $\delta = 0$ otherwise).

sonicboom
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    The delta "function" is not actually a function but a distribution. There are several equivalent definitions of it. The wikipedia article is quite useful. – Loreno Heer Nov 06 '14 at 12:08
  • A distribution is a function. – sonicboom Nov 12 '14 at 15:03
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    yes, but not a function in the sense like $f:\mathbb{R}\to\mathbb{R}$ or similar. But in the sense of a functional $F:D(\mathbb{R})\to\mathbb{R}$ that takes function in a test-function space to $\mathbb{R}$. – Loreno Heer Nov 12 '14 at 15:09

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Let's consider the delta function centered at the origin for simplicity. Let $\mathcal{P}(r,\theta)$ be the Poisson kernel. Now the $\int_{-\pi}^{\pi}\mathcal{P}(r,\theta)d\theta = 1$ which can be seen here. $$ \delta(\theta) = \lim_{r\to 1}\mathcal{P}(r,\theta) = \begin{cases} \infty, & \theta = 0\\ 0, & \text{otherwise} \end{cases} $$ To show limit for $\theta = 0$, use the following form of the Poisson kernel $$ \frac{1}{\pi}\Bigl[\frac{1}{2}+\sum_{n = 1}^{\infty}r^n\cos(n\theta)\Bigr] $$ For $\theta\neq 0$, use the following form of the Poisson kernel $$ \frac{1}{2\pi}\frac{1-r^2}{1 - 2r\cos(\theta) + r^2} $$ Then we have $$ \int_{-\pi}^{\pi}\delta(\theta)d\theta = \int_{-\pi}^{\pi}\lim_{r\to 1}\mathcal{P}(r,\theta)d\theta = \lim_{r\to 1}\int_{-\pi}^{\pi}\mathcal{P}(r,\theta)d\theta = \lim_{r\to 1}(1) = 1 $$

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