Let $\gamma_s(x)=\frac{1}{\sqrt{2\pi s}}\exp(\frac{-x^2}{2s})$ be a gaussian density with variance $s$ and zero mean. Is the integral $$\int_{\{x\geq y\geq 0\}\subset\mathbb{R}^2}\gamma_s(x)\gamma_t(y)dxdy$$ something that can be computed? i am particlarly interested in the case $t=1-s$.
I was asked for context, so the context is trying to do part 3 of this exercise brings up this integral ( $B_t$ is a brownian motion, $S_1$ is its supremum over $0\leq t\leq 1$)
Namely $$\{T\leq s\}=\{sup_{t\leq s}B_t-B_s\leq sup_{1\geq t\geq s}B_t-B_s\}$$ And the two sides of the latter inequality are distributed as (the absolute value of) independent Gaussian variables with variances $s$ and $1-s$. So this gives rise to the integral that interests me. I continue to get close votes, so i'm not sure how much more context you require.