I'm trying to understand the following derivative, where $r\geq0$:
$$ \psi(r):=\int_r^\infty e^{-(y-r)}\Bigl(\int_0^y \psi(y-x)dx\Bigr)dy $$ $$ \Longrightarrow\\ $$ $$ \psi'(r)=-\int_0^r \psi(r-x)dx +\int_r^\infty -e^{-(y-r)}\Bigl(\int_0^y \psi(y-x)dx\Bigr)dy $$
More specifically, I was under the impression that, by the fundamental theorem of calculus, the derivative of $\psi$ was simply the function inside the integral, $$ e^{-(y-r)}\Bigl(\int_0^y \psi(y-x)dx\Bigr) $$ evaluated at $r$. (Or rather, the negative thereof, since the $r$ was the lower bound.)
However, that would just be the first summand in the above result. The second summand I can't explain. Where does it come from? And where does my application of the FtoC fail?
This should follow directly from the Leibniz integral rule. With
$$\psi(r)=\int_r^\infty \underbrace{e^{-(y-r)} \left( \int_0^y \psi(y-x)\,\mathrm dx\right)}_{f(r,y)} \,\mathrm dy$$
the derivative would be
$$\begin{align} \frac{\mathrm d\psi}{\mathrm dr}&=-f(r,r)\frac{\mathrm dr}{\mathrm dr}+\int_r^\infty \frac{\partial f(r,y)}{\partial r}\,\mathrm dy\\[1ex] &=-e^{-(r-r)}\int_0^r \psi(r-x)\,\mathrm dx + \int_r^\infty e^{-(y-r)}\left(\int_0^y\psi(y-x)\,\mathrm dx\right)\,\mathrm dy \end{align}$$
though there seems to be a sign error either here or in the result you've provided.