Im trying to understand a step in a certain proof, this is actually only easy calculus, but somehow I can't wrap my head around it. The equation is given as
$$ \frac{\partial}{\partial t} \int_0^t f(x+(s-t)b,s) \, \text{d}s = f(x,t) + \int_0^t -b \cdot \nabla f(x+(s-t)b,s) \text{d}s.$$
I don't get that, because Im thinking
\begin{align} \frac{\partial}{\partial t} \int_0^t f(x+(s-t)b,s) \, \text{d}s &= \int_0^t \frac{\partial}{\partial t} f(x+(s-t)b,s) \, \text{d}s \newline &= \int_0^t -b \cdot \nabla f(x+(s-t)b,s) \text{d}s. \end{align}
Obviously Im missing something easy here, anyone knows from top of their head?
Cheers
You can look at the integral as a multivariate function $$ I(A,B,t) := \int_A^B F(s,t)\ ds$$ Then if we set $\psi(t) := (0,t,t)$, $\psi'\equiv (0,1,1)$, and the multivariate chain rule gives $$ \frac{d}{dt} I\circ \psi(t) = \nabla I(t) \cdot \psi'(t)= 0\times (\partial_A I)(0,t,t)+1\times(\partial_B I)(0,t,t) + 1\times (\partial_t I)(0,t,t) $$ The second term $\partial_B I$ is just $F(s,t)$ by the Fundamental Theorem of Calculus, and the last term is the term you thought that should be there, $(\partial_t I)(0,t,t) = \int_0^t \partial_t F(s,t) \ ds$.
In your case $F(s,t) = f(x+(s-t)b,s)$, so we have $\partial_t F(s,t) = -b\cdot \nabla f(x + (s-t)b,s)$ as needed.