Just wondering if there is a version of Leibniz integral rule for higher order derivatives. Specifically, I want to evaluate the differential $$ \frac{d^k}{dt^k}\int_0^t f(t - \tau) g(\tau) d\tau $$ at $t = 0$ for arbitrary integer $k$.
Edit
The Leibniz rule provides $$ \frac{d}{dt}\int_0^t f(t - \tau) g(\tau) d\tau = f(0) g(t) + \int_0^t f'(t - \tau) g(\tau) d\tau. $$ Then, $$ \frac{d^2}{dt^2}\int_0^t f(t - \tau) g(\tau) d\tau = f(0) g'(t) + f'(0) g(t) + \int_0^t f''(t - \tau) g(\tau) d\tau, $$ $$ \frac{d^3}{dt^3}\int_0^t f(t - \tau) g(\tau) d\tau = f(0) g''(t) + f'(0) g'(t) + f''(0) g(t) + \int_0^t f'''(t - \tau) g(\tau) d\tau, $$ etc. Eventually, $$ \frac{d^k}{dt^k}\int_0^t f(t - \tau) g(\tau) d\tau = f(0) g^{(k - 1)}(t) + f'(0) g^{(k - 2)}(t) + ... + f^{(k - 1)}(0)g(t) + \int_0^t f^{(k)}(t - \tau) g(\tau) d\tau. $$
Does $f(0) g^{(k - 1)}(t) + f'(0) g^{(k - 2)}(t) + ... + f^{(k - 1)}(0)g(t)$ stand for a simple expression. Apparently, $[f(t - \tau) g(\tau)]^{(k - 1)}_{t = \tau}$ is not what it should look like.
Edit 2
What I have encountered so far is as follows: $$ \frac{d^k}{dt^k}\int_0^t f(t - \tau) g(\tau) d\tau = \sum_{n = 1}^k f^{(n - 1)}(0) g^{(k - n)}(t) + \int_0^t f^{(k)}(t - \tau) g(\tau) d\tau. $$ Thus, evaluating both sides at $t = 0$, we get $$ \left[\frac{d^k}{dt^k}\int_0^t f(t - \tau) g(\tau) d\tau\right]\bigg|_{t = 0} = \sum_{n = 1}^k f^{(n - 1)}(0) g^{(k - n)}(0). $$ Can't encounter what the right hand side is.
THIS IS NOT AN ANSWER, but a long comment.
Using Dirac deltas and Heaviside functions, this can be seen as follows; $$ \int_0^t f(t-\tau)g(\tau)\, d\tau= fH\ast gH, $$ where $\ast$ denotes convolution and $$ H(t)=\begin{cases} 1, & t\ge 0, \\ 0, & t<0.\end{cases}$$ This can be useful, because convolution commutes with differentiation; $$ \frac{d^k}{dt^k} (fH\ast gH) = (fH)^{(k)}\ast gH = \sum_{h=0}^k \binom{k}{h} f^{(h-k)}\delta^{(h-1)}\ast gH.$$ Here we used that $H'=\delta$.