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The Heaviside step function implicitly expands to a piecewise function:

UnitStep[t - 3] // PiecewiseExpand

$$ \begin{cases} 1 & t\geq 3 \\ 0 & \text{True} \\ \end{cases} $$

But the Heaviside step with strict inequalities does not have an implicit expansion:

HeavisideTheta[t - 3] // PiecewiseExpand

$$ \theta (t-3) $$

and the expansion must be given explicitly (Solution from this answer):

% /. {
   HeavisideTheta[x_] :> Piecewise[{{1, x > 0}, {0, x < 0}}]
} // PiecewiseExpand

$$ \begin{cases} 1 & t>3 \\ 0 & \text{True} \\ \end{cases} $$

Is this just an oversight in the implementation or is there a more nuanced distinction?

Related:

Converting HeavisideTheta[]s and Sign[]s functions to a single Piecewise[]

UnitStep vs HeavisideTheta; KroneckerDelta vs DiscreteDelta

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    $\begingroup$ D[HeavisideTheta[x], x] evaluates to DiracDelta[x]. If HeavisideTheta converted to a Piecewise expression this relation to DiracDelta would be obscured. $\endgroup$
    – Bob Hanlon
    May 3, 2020 at 21:55
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    $\begingroup$ @Bob I saw that you converted your answer into a comment. I think it’s insightful enough to stand as an answer though. Would you consider undeleting? $\endgroup$
    – MarcoB
    May 5, 2020 at 3:48

1 Answer 1

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D[HeavisideTheta[x], x] evaluates to DiracDelta[x]. If HeavisideTheta converted to a Piecewise expression this relation to DiracDelta would be obscured.

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    $\begingroup$ Thank you! (+1) $\endgroup$
    – MarcoB
    May 5, 2020 at 3:52

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