dirac delta distributions - What is the most natural value of Heaviside step function at zero argument?

In many physical applications, the Heaviside step fuction is defined as $$H(x) = \left\{\begin{eqnarray} 1, \quad x>0 \\ 0, \quad x<0 \end{eqnarray}\right.$$ The value $H(0)$ is left undefined. Is there a physically prefered value of $H(0)$ or does it depend on the problem at hand?

For example, due to relation $H'(x) = \delta(x)$, it would be nice to have $H(0)=\frac{1}{2}$ if we think of $\delta(x)$ as an even function. However, when considering signal processing, we would like our functions to be (left/right) continuous at $x=0$.

Is there a way out of this arbitrariness?

Answer

Continuous Fourier analysis, which contains both the Fourier transform and the Fourier series, and which is used in e.g. signal processing, naturally picks the average value of the left and right limits, cf. the Dini-Dirichlet criterion. For the Heaviside step function, this means that

$$ H(0)~=~\frac{1}{2} \left(\lim_{x\to 0^-} H(x)+ \lim_{x\to 0^+} H(x)\right) ~=~\color{red}{\frac{1}{2}}.\tag{1}$$

The Fourier transform of the Heaviside step function $H(x)$ is the distribution

$$\widehat{H}(k)~=~\int_{\mathbb{R}}\! dx e^{-ikx} H(x) ~=~\frac{-i}{k-i0^+},\tag{2} $$

or conversely

$$ H(x)~=~ \int_{\mathbb{R}}\! \frac{dk}{2\pi} e^{ikx} \widehat{H}(k) ~=~ \int_{\mathbb{R}}\! \frac{dk}{2\pi i} \frac{e^{ikx}}{k-i0^+}.\tag{3} $$

As a check, setting $x=0$ in eq. (3) yields precisely eq. (1),

$$ H(x=0) ~\stackrel{(3)}{=}~\int_{\mathbb{R}}\!\frac{dk}{2\pi i} \frac{1}{k-i0^+}~\stackrel{(5)}{=}~ \int_{\mathbb{R}}\! \frac{dk}{2\pi i}\left( P\frac{1}{k} + i\pi\delta(k)\right) ~=~\color{red}{\frac{1}{2}}, \tag{4} $$

cf. the Sokhotski–Plemelj formula

$$ \frac{1}{k-i0^+}~=~P\frac{1}{k} +i\pi\delta(k). \tag{5}$$