I'm looking to find the derivative of a convolution integral of the following form:
\begin{equation} \frac{d}{dr}((G(r,t)*f(t)) = \frac{d}{dr} (\int_{-\infty}^{\infty} G(r,t-\tau)f(\tau) d\tau) \end{equation}.
Here, $G(r,t)$ is the 2D free space Greens function for the wave equation defined as: $ G(r,t) =\frac{1}{2\pi} \frac{H(t-r/c)}{\sqrt{t^2 - r^2/c^2}}$, where $H$ is a Heaviside and c is the wave speed constant. And $f(t)$ is a smooth function, I currently have let $f(t) = e^{-\alpha(t-t0)^2}$, ($\alpha$ is constant) and have computed the convolution integral numerically,
\begin{equation} G(r,t)*f(t) = \int_{0}^{max(0,tn-r/c)} \frac{e^{-\alpha(\tau-t0)^2}}{\sqrt{(t-\tau)^2 - r^2/c^2}}d\tau. \end{equation}
I was wondering if i could compute the derivative of the above convolution integral as $\frac{dG(r,t)}{dr}*f(t)$, then complete the convolution integral numerically as before but with $\frac{dG}{dr}$ or if there are some complications im not taking into account.
Thanks in advance.