How to calculate the Integral $\int_0^t\delta'(t-t')d t'$ (where $\delta'$ is a derivative w.r.t $t$, and $t'\in(0,t)$)?

Question

Without choosing a specific representation for this distribution (by using something like a limit law on a Gaussian function):

How to calculate the Integral $\int_0^t\delta'(t-t')d t'$ (where $\delta'$ is a derivative of the Dirac Delta distribution, $\delta(t-t')$, with respect to $t$, and $t'\in(0,t)$)?

Is it correct to say that $\int_0^t\delta'(t-t')d t'=\delta(t)$? The latter is just some intuition I had, but have no real justification for it... maybe totally wrong..

Thanks!

Answer 1

We have $$ \int_0^t \delta'(t-t') \, dt' = \int_{-\infty}^{\infty} \chi_{[0,t]}(t') \, \delta'(t-t') \, dt' = (\chi_{[0,t]} * \delta')(t), $$ where $*$ means convolution.

Now, $$ \chi_{[0,t]} * \delta' = (\chi_{[0,t]} * \delta)' = (\chi_{[0,t]})' = \delta_0 - \delta_t, $$ i.e. $$ (\chi_{[0,t]} * \delta')(t) = \delta_0(t) - \delta_t(t) . $$

Unfortunately, the last term, $\delta_t(t)$ can only be given an infinite value.