Prove that the derivative of the step function is the delta function, i.e. show that \begin{equation} \frac{d}{dx}\,\Theta(x-a) \end{equation} satisfies the property of the delta function in equation (\ref{fdelta}) in Section \ref{deltaintro}.

We need to show that \begin{equation} \Int_{b}^{c} f(x)\,\delta(x-a) \,dx = f(a) \end{equation} for $b<a<c$, where \begin{equation} \delta(x-a)=\frac{d}{dx}\,\Theta(x-a) \end{equation} The main strategy is to use integration by parts, paying strict attention to the limits of integration. \begin{eqnarray} \Int_{b}^{c} f(x)\,\delta(x-a) \,dx &=& \Int_{b}^{c} f(x)\,\frac{d}{dx}\, \Theta(x-a) \,dx \nonumber\\ &=& \left. f(x)\, \Theta(x-a)\right|_b^c - \Int_{b}^{c} \frac{d}{dx}(f(x))\,\Theta(x-a) \,dx \nonumber\\ &=& \left\{ f( c )-0\right\} - \Int_{a}^{c} \frac{d}{dx}(f(x))\,dx \nonumber\\ &=& f( c )-\left\{\left. f(x)\right|_a^c\right\} \nonumber\\ &=& f( c )-\left\{f( c )-f(a)\right\} \nonumber\\ &=& f(a) \end{eqnarray}