I was reading "2001The Integral of Geometric Brownian Motion" am getting confused whether $A_t^{(\mu)}=\int_0^t e^{2 \mu \tau + 2 B_\tau} d\tau$ is a function of $t$ only or a function of $t$ and $B_t$, where $B_t$ is a standard Brownian motion. If the latter is the case, what is partial derivative of $A_t^{(\mu)}$ with respect to $B_t$?
I suppose the author consider this as a function of $t$ only based on Eq.(2.7) and Eq.(2.8).
My understanding is that stochastic differential equation only makes sense in the integral form. Since $A_0^{(\mu)}=0$, we have $$dA_t^{(\mu)}=e^{2 \mu t + 2 B_t} dt=e^{2 \mu t + 2 B_t} dt+0 dB_t.$$ Can I interpret this as $$\frac{\partial^n A_t^{(\mu)}}{\partial B_t^n}=0$$ for $n \in \mathbb{N}$?
However, from the second mean value theorem: if $f$ is integrable on $[a,b]$ and $g$ is monotone, then there exists $\eta \in [a,b]$ such that $$\int_a^bf(x)g(x)dx=g(a)\int_a^\eta f(x)dx+g(b)\int_\eta^bf(x)dx.$$ It reminds me that the integral could be represented as a function of integrand at $b$ for some $g$. Do we have other results for g not monotone?
Looking for your help.