Differentiating under the integral sign with variable limit

The formula

\frac{d}{dt}\int_0^t \phi(t,s) u(s)\, ds = \phi(t,t)u(t) + \int_0^t \phi_t(t,s) u(s)\, ds

is not new but it doesn’t seem so easy to track down a simple derivation. A simple (formal) derivation of this runs as follows. First define,

F(t,\xi) := \int_0^t f(\xi, s)\, ds = \digamma(\xi, t) - \digamma(\xi,0)

where \digamma is a primitive such that \digamma_s(\cdot,s) = f(\cdot,s). Then

\frac{dF}{dt}= \frac{\partial F}{\partial t} + \frac{\partial F}{\partial\xi}\frac{d\xi}{d t}

But on substituting

\frac{\partial F}{\partial t}(t,\xi) = \frac{\partial\digamma}{\partial t}(\xi,t) = f(\xi,t)


\frac{\partial F}{\partial\xi}(t,\xi) = \int_0^t f_\xi(\xi,s)\, ds

this becomes

\frac{dF}{dt}= f(\xi,t) + \frac{d\xi}{d t} \int_0^t f_\xi(\xi,s)\, ds.

That’s it really. As a corollary take \xi = t and f(\xi,s) = \phi(t,s)u(s) and then we see that

\frac{d}{dt}\int_0^t \phi(t,s)u(s)\, ds = \phi(t,t)u(t) + \int_0^t(\phi(t,s)u(s))_t\, ds

which is what we set out to demonstrate.