Let $k(t,x)$ be the transition density of Brownian motion $$ k(t,x) := \frac{1}{\sqrt{2 \pi t}} \exp \left\{ \frac{-x^2}{2t} \right\} , \quad t \geq 0, x \in {\mathbb R.}$$
Question Let $0 < x < a$. Show that $$ \lim_{x \nearrow a}\int_0^t \frac{a-x}{s} k(s,x-a)k(t-s,a)ds = k(t,a).$$ Can someone offer some intuition as to why this is true?
First note that $$ \partial_x k(s, x) = -\frac{x}{s}k(s,x), \quad s > 0, x \in {\mathbb R},$$ so your integral becomes $$ q_a(t,x) := \int^t_0 \partial_x k(s,x-a)k(t-s,a)ds. $$ Now suddenly your integral $q_a(t,x)$ becomes a representation of the unique classical solution to the boundary value problem for the heat equation (see e.g Cannon's book One-Dimensional Heat Equation, Ch 4): \begin{align} \partial_x q_a(t,x) &= \frac{1}{2} q_a(t,x), \\ q_a(0,x) &= 0, \quad x < a, \\ q_a(t,a) &= k(t,a) \quad t > 0. \end{align} The constant boundary $x = a$ is regular for this problem, and the solution achieves the boundary values continuously, i.e., $$ \lim_{x \nearrow a} q_a(t,x) = k(t,a). $$
