Dirac delta derivatives and discontinuities I have an integral of the form
$$
I = \int_{-\infty}^{x} f(x) \delta'(x) dx
$$
where
$$
f(x) =
\begin{cases}
a & x<0 \\
b & x>0
\end{cases}
$$
and $\delta'$ is the derivative of the Dirac delta. By the rules of using the Dirac delta, I expect that
$$
I = -f'(0) = (a-b) \delta(x)
$$
and, plotting graphs of functions that may be used to in the limit towards $\delta'$, this certainly looks plausible. Is this above correct, and how would I go about proving it?
 A: This integral is in general not well defined in the theory of distributions. Note that $f(x)$ is not a test function, and it has a singularity at $x=0$. If we try to give a meaning to it one runs into difficulties. Using the step function $H(x)$ one could try to give meaning to $I(x)$ as follows:
$$\begin{align}
I(x)&=\int_{-\infty}^{+\infty}H(x-t)f(t)\delta'(t)\,dt=-\left[\frac{d}{dt}H(x-t)f(t)\right]_{t=0}\\
&=\left[\delta(x-t)f(t)-H(x-t)(a-b)\delta(t)\right]_{t=0}\\
&=\delta(x)f(0)-H(x)(a-b)\delta(0) = \;???\,.
\end{align}$$
If $a\ne b$ you can not make sense of this; $f(0)$ and $\delta(0)$ do not make sense. So $I(x)$ is not a well defined distribution. To give meaning to it you may want to replace $f(t)$ by some smooth version of it (physicists will call doing that "regularization"), but the result will depend on the way you do this "regularization". This is a well known problem when you try to multiply distributions with common singularities.
A: If $f(t)=a+(b-a)H(t)$ and $x<0$, then $\int_{-\infty}^x \delta'(t)f(t)\,dt =\int_{-\infty}^x \delta'(t)a\,dt=0$ since $\delta'$ is of compact support on $\{0\}$.
Next note that there is no meaning to the square of the Dirac Delta as a distribution (See Here).  Consequenlty, if $f(t)=a+(b-a)H(t)$, then the expressions of this pseudo equalities, $\int_{-\infty}^x \delta'(t)f(t)\,dt=-\int_{-\infty}^x \delta(t)f'(t)\,dt=-\int_{-\infty}^x\delta(t)\,\left((b-a)\delta(t)\right)\,dt$, are meaningless for all $x\ge 0$.
