Consider the following integral with real $a>0$
$$ I(a) = \int_0^{\infty} \mathrm{d}x~\delta(a-x) \Theta(x-a) f(x) $$
with $f(x)$ a function such that $f(a) \ne 0$, $\delta$ the Dirac function and $\Theta$ the unit step function.
A naive solution appears to be
$$I(a) = \Theta(0) f(a) $$.
How to make sense of this result? What to take for $\Theta(0)$?
How to evaluate integral $I(a)$ in a more rigourous way?
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\begin{align} \mrm{J}\pars{a,b} & \equiv \int_{0}^{\infty}\delta\pars{a - x}\Theta\pars{x - b}\,\mrm{f}\pars{x}\,\dd x = \bracks{\int_{0}^{\infty}\delta\pars{a - x}\,\dd x} \Theta\pars{a - b}\,\mrm{f}\pars{a} \\[5mm] & = \Theta\pars{a}\Theta\pars{a - b}\,\mrm{f}\pars{a}\implies \left\{\begin{array}{rcl} \ds{\mrm{J}\pars{a,a^{-}}} & \ds{=} & \ds{\Theta\pars{a}\,\mrm{f}\pars{a}} \\[1mm] \ds{\mrm{J}\pars{a,a^{+}}} & \ds{=} & \ds{0} \end{array}\right. \end{align}