The Heaviside step function at zero and the integral of the Dirac delta function

Question

Just for fun I was reading about the Heaviside step function on Wikipedia. The definition in terms of the Dirac delta function makes sense:

$$ H(x) = \int_{-\infty}^x \delta(s)\ ds $$

$\delta(s) = 0$ for all $s \in [-\infty, 0)$. For other values of $s$, $\delta(s) \not= 0$ only at one point, so the integral (area) equals the value at that point, $1$.

However, later on in the article it reads (emphasis mine):

Since $H$ is usually used in integration, and the value of a function at a single point does not affect its integral, it rarely matters what particular value is chosen of $H(0)$.

If the value at a single point does not affect the integral for $H$, why does it for $\delta$? In other words, why does the following equation not hold, assuming the above quote?

$$ \int_{-\infty}^\infty \delta(s)\ ds = 0 $$

I must be misunderstanding something about either integrals or the Dirac delta function.

Answer 1

Formally, the dirac delta is not a function in the same sense as the heaviside function.

Rather, it is a distribution.

Answer 2

The best way to think of the dirac delta function is: $$\delta_n(x)=\begin{cases} 0 & x<0 \\ n & 0<x<\frac1n \\ 0 & x>\frac1n \end{cases}$$ $$\delta(x)=\lim_{n\to\infty}\delta_n(x)$$ Now try and integrate $\delta_n(x)$ first then take the limit: $$\int_{-\infty}^\infty\delta_n(x)dx=\int_{-\infty}^0\delta_n(x)dx+\int_0^{\frac1n}\delta_n(x)dx+\int_{\frac1n}^\infty\delta_n(x)dx$$ $$=\int_{-\infty}^00.dx+\int_0^{\frac1n}ndx+\int_{\frac1n}^\infty0.dx$$ $$=n\left[x\right]_0^{1/n}=1$$ and now: $$\int_{-\infty}^\infty\delta(x)dx=\lim_{n\to\infty}\int_{-\infty}^\infty\delta_n(x)dx=1$$