Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
"Is there a way out of this arbitrariness?" Nope. Depends on context. – Michael Brown Apr 15 '13 at 11:30
I usually see $H(0) \equiv 1/2$ – DilithiumMatrix Apr 15 '13 at 13:01
Related: – Emilio Pisanty Apr 2 '15 at 18:12

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

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

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

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

or conversely

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

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

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

cf. the Sokhotski–Plemelj formula

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

share|cite|improve this answer

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 $H(0)=1/2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.