Problem:
$dX_t = \sigma X_tdB_t$, $X_0=x$.
$dY_t=X_tdt-Z_tdt$ find $Y_t$, where $Z_t$ is a control and $B_t$ is standard Brownian motion.
My attempt:
From Ito's lemma,
$\partial_BX_t=\sigma X_t$, therefore $X_t=c(t)\exp(\sigma B_t)$.
$\partial_tX_t+\frac{1}{2}\partial^2_BX_t=0$, substitue in above expression for $X_t$ and
$c^\prime(t)+\sigma^2\frac{1}{2}c(t)=0$ which implies $c(t)=a\exp(-\sigma^2\frac{t}{2}).$ Putting all together $$X_t=a\exp(-\sigma^2\frac{t}{2}+B_t\sigma)$$ which can be solved for $X_t=x$ to fine $a$.
Then $Y_s-Y_r=\int_r^s a\exp(-\sigma^2\frac{t}{2}+B_t\sigma)dt+\int^s_rP(t)dt$. How can I solve the time integral of exponential of Brownian motion? A hint is sufficient. I would like to do it myself. Thank you all.
Edit: I need help to find a closed form expression for $\int_r^s a\exp(B_t\sigma)dt.$
As mentioned in the comments the expression $\int e^{B_s}ds$ is as closed form as it can get.
But if you interested to know more about its law, there is a large literature on "integrated Brownian motion" eg. "The Integral of Geometric Brownian Motion" by Daniel Dufresne.