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?



Limit of an integral / Boundary behaviour of a Gaussian convolution / single layer potential
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., 
$$ 
\lim\_{x \nearrow a} q\_a(t,x) = k(t,a).
$$


