I want to find the transition density of a $1$-d Brownian motion killed at two points. Specifically, let $B_t$ be a Brownian motion that starts from $x \in D = [0,\pi]$ and stops when reaching $\partial D = \{0, \pi \}$. I want to find the function $p(t, x,y)$ such that for $a, b \in (0, \pi)$, $\tau$ the stopping condition,
$$\mathbf{P}\{B_t \in (a,b); t < \tau; B_0 = x \} = \int_a^bp(t,x,y)dy.$$
Now by examining the properties of brownian motion, I am able to determine the following constraints for $p(t,x,y)$:
$$\partial_t p(t,x,y) = \frac{1}{2} \partial^2_{yy} p(t,x,y) \;\;\;\; p(t,x,0) = p(t,x,\pi)=0 \;\;\;\; \forall t>0, x\in intD $$
$$p(0, x,y) = \delta(x-y),$$
where $\delta$ is the dirac delta function. I want to view it as a PDE with respect to $t$ and $y$ and use Fourier series to solve this question following the procedure here. Using the strategy there, I have determined the fourier coefficients as $$D_n = \frac{2}{\pi}\int_0^\pi \delta(x-y)sin(ny)dy = \frac{2}{\pi}sin(nx).$$
Thus $$p(t,x,y) = \frac{2}{\pi}\sum_{n=1}^\infty sin(nx)sin(ny)e^{-\frac{n^2t}{2}}.$$
However, I'm not sure if my result is correct. Asymptotically, we should have for $t \to \infty,$ $$p(t,x,y) \sim c_1(x)c_2(y)e^{-\lambda t},$$ for some functions with respect to $x$ and $y$ and an exponential function over $t$. However, my solution is of the form of an infinite sum and seems not converge to this clean form. So I wonder if I have done anything wrong. Any comments will be much appreciated.