Let $A=(a,b)\times (c,d) \subset \mathbb{R}^2$ with $0 \in A$ and $(B_t)$ be standard two dimensional Brownian motion. Additionally, let $\tau_A := \inf \{t\geq 0: B_t \notin A\}$ and let $g:A \to \mathbb{R}$ be a smooth bounded function which can be written as $g(x,y)=u(x)v(y)$.
I am investigating the random variable $$\int_0^{\tau_A} g(B_s) ds$$ in particular I am interested in the expectation $$E[\int_0^{\tau_A} g(B_s) ds].$$ I know that there is a connection to the Dirichlet problem but I am interested in calculating or estimating (in both directions) this expression in a stochastical way. E.g., a bound, which contains the $L^1$ norm of $g$ would be very interesting. Since the domain $A$ is an "easy" one and $B_t$ consists of two one dimensional independent Brownian motions $B_t=(B_t^1, B_t^2)$, I have tried to reduce the problem into one dimension in the following way:
\begin{align*} E[\int_0^{\tau_A} g(B_s) ds] &= E^1 E^2 [\int_0^{\tau_{(a,b)}^1 \wedge \tau_{(c,d)}^2} g(B_s^1,B_s^2) ds] \\ &= \int_0^{\infty}E^1 \big[ 1_{[0, \tau^1_{(a,b)})}(s) u(B^1_s) \big] E^2 \big[1_{[0, \tau^2_{(c,d)})}(s) v(B^2_s)\big] d s \end{align*}
The superscripts $\{1,2\}$ refer to the distributions of the respective Brownian motion. Now I have no further ideas on how to proceed and am not familiar with tools that could help me here.
I would appreciate any help!
Sorry I don't know how to be properly stochastic about things and this might not be too helpful. Let me rewrite your last equation in a form more familiar to me. \begin{equation*} \mathbb{E}\left[g\right]=\int_{0}^{\infty} \left(\int_{a}^{b}\phi_{s}(x)u(x)dx\right) \left(\int_{c}^{d}\psi_{s}(y)v(y)dy\right)ds. \end{equation*} It's not necessary but for simplicity I'll assume $\left(a,b\right)=\left(c,d\right)=\left(-\pi/2,\pi/2\right)$. Given initial conditions $\phi_{0}(x)=\delta(x)$ and $\psi_{0}(y)=\delta(y)$, we solve with \begin{equation*} \phi_{s}(x)=\frac{2}{\pi}\sum_{n=0}^{\infty} e^{-\frac{1}{2}(2n+1)^{2}s} \cos\left(\left(2n+1\right)x\right) \end{equation*} \begin{equation*} \psi_{s}(y)=\frac{2}{\pi}\sum_{n=0}^{\infty} e^{-\frac{1}{2}(2n+1)^{2}s} \cos\left(\left(2n+1\right)y\right). \end{equation*} Then with $1\leq p\leq\infty$, \begin{equation*} \int_{-\pi/2}^{\pi/2}\phi_{s}(x)u(x)dx\leq k_{p} \left(\sum_{n=0}^{\infty}e^{-\frac{1}{2}(2n+1)^{2}s}\right) \left\lVert u\right\rVert_{L^{p}} \end{equation*} where, for example, $k_{1}=2/\pi$, $k_{2}=\sqrt{2/\pi}$ and $k_{\infty}=4/\pi$.
Edit2: As @Diger has pointed out in the comments, the better way to proceed (for $p=\infty$ and $p=2$ respectively) is by \begin{eqnarray*} \int_{-\pi/2}^{\pi/2}\phi_{s}(x)\,dx&=& \frac{2}{\pi}\int_{-\pi/2}^{\pi/2} \left(\sum_{n=0}^{\infty}e^{-\frac{1}{2}(2n+1)^{2}s}\cos\left((2n+1)x\right) \right)dx\\ &=&\frac{2}{\pi}\sum_{n=0}^{\infty}e^{-\frac{1}{2}(2n+1)^{2}s} \left(\frac{2(-1)^{n}}{2n+1}\right) \end{eqnarray*} implying \begin{equation*} \lVert \phi_{s} \rVert_{L^{1}} =\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n} e^{-\frac{1}{2}(2n+1)^{2}s}} {2n+1}, \end{equation*} and \begin{eqnarray*} \int_{-\pi/2}^{\pi/2} \phi_{s}(x)^{2}\,dx &=&\frac{4}{\pi^{2}}\int_{-\pi/2}^{\pi/2} \Bigg(\sum_{m,n=0}^{\infty}e^{-\frac{1}{2}(2m+1)^{2}s-\frac{1}{2}(2n+1)^{2}s}\\ &&\qquad\qquad\qquad\quad \cos\left((2m+1)x\right)\cos\left((2n+1)x\right)\Bigg)dx\\ &=&\frac{4}{\pi^{2}} \sum_{m,n=0}^{\infty}e^{-\frac{1}{2}(2m+1)^{2}s-\frac{1}{2}(2n+1)^{2}s} \left(\frac{\pi}{2}\delta_{m,n}\right)\\ &=&\frac{2}{\pi}\sum_{n=0}^{\infty}e^{-(2n+1)^{2}s} \end{eqnarray*} implying \begin{equation*} \left\lVert \phi_{s}\right\rVert_{L^{2}}= \left(\frac{2}{\pi}\sum_{n=0}^{\infty}e^{-(2n+1)^{2}s}\right)^{\frac{1}{2}}. \end{equation*} I am embarrassed by the mess I have made of this answer so I'll try and put the bounty back on the question.
\begin{multline*} \int_{h}^{\infty} \left(\int_{-\pi/2}^{\pi/2}\phi_{s}(x)u(x)dx\right) \left(\int_{-\pi/2}^{\pi/2}\psi_{s}(y)v(y)dy\right) ds\\ \leq k_{p}k_{q} \left\lVert u\right\rVert_{L^{p}} \left\lVert v\right\rVert_{L^{q}} \int_{h}^{\infty} \left(\sum_{n=0}^{\infty}e^{-\frac{1}{2}(2n+1)^{2}s}\right)^{2}ds. \end{multline*} My guess is that the integral converges for $h>0$. But is it the sort of thing you are after? For the integral up to $h$ we might have to rely on the fact that $\left\lVert \phi_{s}\right\rVert_{L^{1}}\leq 1$ and therefore be more constrained in the choice of norm. \begin{equation*} \int_{-\pi/2}^{\pi/2}\phi_{s}(x)u(x)dx\leq \left\lVert \phi_{s}\right\rVert_{L^{1}} \left\lVert u\right\rVert_{L^{\infty}} \leq \left\lVert u\right\rVert_{L^{\infty}} \end{equation*} \begin{multline*} \int_{0}^{h} \left(\int_{-\pi/2}^{\pi/2}\phi_{s}(x)u(x)dx\right) \left(\int_{-\pi/2}^{\pi/2}\psi_{s}(y)v(y)dy\right) ds\\ \leq h \left\lVert u\right\rVert_{L^{\infty}} \left\lVert v\right\rVert_{L^{\infty}}= h \left\lVert g\right\rVert_{L^{\infty}}. \end{multline*} I was surprised that you wanted estimates in terms of $\left\lVert g\right\rVert_{L^{1}}$ because to me the $L^{\infty}$ norm seems more natural. We are basically integrating against distributions.
Edit1: I have changed $\epsilon$ to $h$ in the above. It was misleading notation because I did not intend $\epsilon\to 0$. In fact according to Mathematica its optimal value in this setup is about $h=0.636$. Let's assume henceforth that we are only interested in $p=q=\infty$. By the above, \begin{multline*} \mathbb{E}[g]\leq h\left\lVert u\right\rVert_{L^{\infty}}\left\lVert v\right\rVert_{L^{\infty}} +k_{\infty}k_{\infty}\lVert u\rVert_{L^{\infty}}\lVert v\rVert_{L^{\infty}} \int_{h}^{\infty} \left(\sum_{n=0}^{\infty}e^{-\frac{1}{2}(2n+1)^{2}s}\right)^{2}ds\\ =\left( h+\left(\frac{4}{\pi}\right)^{2} \int_{h}^{\infty} \left(\sum_{n=0}^{\infty}e^{-\frac{1}{2}(2n+1)^{2}s}\right)^{2}ds \right)\left\lVert g\right\rVert_{L^{\infty}}. \end{multline*} Mathematica is happy to evaluate that prefactor (with $h$ as above), giving \begin{equation*} \mathbb{E}[g]\leq 1.522 \left\lVert g\right\rVert_{L^{\infty}}. \end{equation*}
Is this likely to be a good estimate? Well let's perform the same procedure in the one-dimensional case for comparative purposes. \begin{multline*} \mathbb{E}[u]= \int_{0}^{\infty}\left( \int_{-\pi/2}^{\pi/2}\phi_{s}(x)u(x)dx\right)ds\\ \leq \left(h+ \frac{4}{\pi} \int_{h}^{\infty} \left(\sum_{n=0}^{\infty}e^{-\frac{1}{2}(2n+1)^{2}s}\right)ds \right)\lVert u \rVert_{L^{\infty}}= 2.505 \left\lVert u \right\rVert_{L^{\infty}}. \end{multline*} In the one-dimensional case, we know the expected survival time is $\pi^{2}/4$ (see @Diger's comment below). Consider $u(x)=1$. Then $\mathbb{E}[u]=\pi^{2}/4=2.467$, only slightly less than $2.505$. So these estimates aren't looking terrible!
BUT $\pi^{2}/4$ is clearly the better estimate, if only slightly. Why is that? In the above I calculated \begin{equation*} \int_{-\pi/2}^{\pi/2}\left\lvert \cos \left((2n+1) x\right)\right\rvert dx=2. \end{equation*} However in the comments @Diger calculated \begin{equation*} \int_{-\pi/2}^{\pi/2} \cos \left((2n+1) x\right) dx= 2(-1)^{n}/(2n+1). \end{equation*} In fact I think @Diger's approach is valid, and of course it gives a slightly better estimates. Its validity stems from the nonnegativity of the probability distribution throughout its domain.