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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?

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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

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

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  • $\begingroup$ Where can I find the full versions of the Fourier transforms? Wikipedia does not list the full form: en.wikipedia.org/wiki/Fourier_transform $\endgroup$ – Anixx May 6 '18 at 12:48
  • $\begingroup$ What do you mean? $\endgroup$ – Qmechanic May 6 '18 at 13:02
  • $\begingroup$ I mean I want the table of Fourier transforms that does not hide the infinite terms and works everywhere. For instance, I want $$\widehat H(x+a)=~PV\frac{-i}{k} +\pi\delta(k)+ a\operatorname {sinc} \left({\frac {k a }{2\pi}}\right)$$ instead of $$H(x+a)\to -\frac{i e^{i a k} }{k}+\pi \delta (k)$$ which they give in the tables. $\endgroup$ – Anixx May 6 '18 at 13:45
  • $\begingroup$ Where you got the (2)? $\endgroup$ – Anixx May 6 '18 at 15:23
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For analytical calculations, you just can't use $H(0)$ because it is not defined there. Also, you don't need it.

For numerical calculations, situations where a more continuous function might be of advantage; in physics, people often use the Fermi-Dirac distribution (which converges to a Heaviside function for $T=0$; you could call it a Heaviside sequence, analogous to a Delta sequence), also for situations which have no connection with statistical physics. In this case, you would also define $H(0)=1/2$.

There are other sequences/approximations at Wikipedia. I think that they all define $H(0)=1/2$.

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