Convolution between the derivative Dirac delta function and other function

Question

I'm having trouble with the following convolution: \begin{equation}\label{eq:1} \int_{-\infty}^{+\infty} \delta'(a-t) g(t) dt \textrm{.} \end{equation}

I know that (I prove this) \begin{equation}\label{eq:2} \int_{-\infty}^{+\infty} \delta'(t) g(t) dt \textrm{.}=-g'(0), \end{equation}

but I can not calculate the other. Can someone tell me how to solve this: \begin{equation}\label{eq:3} \int_{-\infty}^{+\infty} \delta'(a-t) g(t) dt \textrm{.} \end{equation}

Answer 1

Let $s = a-t$, so that $ds=-dt$. Then $$ \int_{-\infty}^\infty \delta'(a-t)g(t)\,dt =\int_{-\infty}^\infty \delta'(s)g(a-s)\,ds $$ If we set $\tilde g(s) = g(a-s)$, then according your your previous result, the integral above is $-\tilde g'(0)$. What is that in terms of the original $g$?

Answer 2

this result you showed me agree with this??: \begin{equation}\label{eq:5} \int_{-\infty}^{+\infty} \delta'(a-t) g(t) dt = g'(a) \textrm{.} \end{equation} and \begin{equation}\label{eq:6} \int_{-\infty}^{+\infty} \delta'(t-a) g(t) dt = -g'(a) \textrm{.} \end{equation} See the equatios (19) and (20) in: http://mathworld.wolfram.com/DeltaFunction.html