How to compute the following differentiation? Is there a general rule that can be applied?
$$\frac{d}{dt}\int_0^t e^{x(s)}ds$$
in the case of $x=W$ where $W$ is a standard brownian motion, is there some particular facts to take into account?
if one applies the fundamental theorem of calculus :
$$\frac{d}{dx}\left( \int_{a(x)}^{b(x)}f(x,t)dt \right)=f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int_{a(x)}^{b(x)}\frac{d}{dx}f(x,t)dt$$
letting in our case $x=t$, $b(x)=t$, $a(x)=0$ and $f(W,t)=e^{W_t}$ we have :
$$\frac{d}{dt}\left( \int_{0}^{t}e^{W_s}ds \right)=e^{W_t}+\int_{0}^{t}\frac{d}{ds}e^{W_s}ds$$
the problem is that $\frac{d}{ds}e^{W_s}$ cannot be computed as a borwian motion is not derivable.
any help?
Note that
$$\int_0^t e^{W_s} \, ds$$
is a pathwise integral, i.e. it is defined as
$$\int_0^t e^{W_s(\omega)} \, ds$$
for each fixed $\omega \in \Omega$. Since the Brownian motion has (almost surely) continuous sample paths, we know that $s \mapsto W_s(\omega)$ is continuous, and therefore $s \mapsto e^{W_s(\omega)}$ is continuous for each fixed $\omega \in \Omega$. Applying the fundamental theorem of calculus, we find
$$\frac{d}{dt} \int_0^t e^{W_s(\omega)} \, ds = e^{W_t(\omega)}$$
for all $\omega \in \Omega$ and $t \geq 0$.