Integral of Dirac Delta Function

Question

If the Dirac Delta function has an infinitely tall "spike" at $x=0$, then how is it that its integral over the entire real number line is one, instead of infinity?

Answer 1

To talk about the Delta function rigorously, you need to introduce the idea of distributions. In this case you never look at $\delta(x)$ outside an integral, and you define it to have the property

$$\int\delta(x-x_0)f(x)dx=f(x_0),$$ which then implies it having unit integral if we take $f(x)=1$.

In fact, if we actually define $\delta(x)$ to be $0$ everywhere and $\infty$ at $0$, then its Lebesgue integral should be zero.

We can however look at $\delta(x)$ as the limit of a series of Gaussians:

$$\delta_a(x)=\frac{1}{a\sqrt{\pi}}e^{-x^2/a^2},$$

as on the Wikipedia page. Each Gaussian $\delta_a$ has unit integral, and we can show that as $a\rightarrow 0$, we start to arrive at the properties of $\delta(x)$. You can also see in the animation that it approaches the shape of being zero everywhere, but highly peaked at the origin.