The entrance law of a Brownian motion with absorbing boundary In the article  "Construction of Diffusion processes with Wentzell's Boundary conditions by means of poisson point processes of Browninan excursions"  one reads:

I tried to compute it for $n=1$
Then I guess we have:
\begin{align}
K(t,x) &= \frac{\sqrt{\pi t^3}}{\sqrt{2}} x \exp \bigg(-\frac{x^2}{2t}\bigg) \\
p^0_K(t,x,y) &= \frac{1}{\sqrt{2\pi t}} \Bigg(\exp \bigg(-\frac{x^2-y^2}{2t}\bigg)-\exp \bigg(-\frac{x^2+y^2}{2t}\bigg) \Bigg)
\end{align}
The question is how do we prove that 
$$\int_0^\infty K(t,x) p^0_K(s,x,y)\, dx  =  K(t+s, y)$$
 A: I doubt that the claimed identity holds true for $n=1$:

By the very definition of $K$ and $p_K^0$, we have
$$\begin{align*} \int_0^{\infty} K(t,x) p_K^0(s,x,y) \, dx = I_1+I_2 \end{align*}$$
where
$$\begin{align*} I_1&:= \frac{1}{2} \sqrt{\frac{t^3}{s}} \exp \left( \frac{y^2}{2s} \right) \int_{0}^{\infty}  x \exp \left(- \frac{x^2}{2} \left[ \frac{1}{t}+ \frac{1}{s} \right] \right) \, dx \\ &= - \frac{1}{2} \sqrt{\frac{t^3}{s}} \left[ \frac{1}{t}+\frac{1}{s} \right]^{-1} \exp \left( \frac{y^2}{2s} \right) \int_0^{\infty} \frac{d}{dx} \exp \left(- \frac{x^2}{2} \left[ \frac{1}{t}+ \frac{1}{s} \right] \right) \, dx \\ &= \sqrt{t^5 s} \frac{1}{t+s} \exp \left( \frac{y^2}{2t}  \right) \end{align*}$$
and
$$\begin{align*}I_2  &:= - \frac{1}{2} \sqrt{\frac{t^3}{s}} \exp \left( - \frac{y^2}{2t} \right) \int_0^{\infty} x \exp \left(- \frac{x^2}{2} \left[ \frac{1}{t}+ \frac{1}{s} \right] \right) \, dx \\ &= \dots = -\sqrt{t^5 s} \frac{1}{t+s} \exp \left(- \frac{y^2}{2t}  \right) \end{align*}$$
Hence,
$$ \int_0^{\infty} K(t,x) p_K^0(s,x,y) \, dx = 2\sqrt{t^5 s} \frac{1}{t+s} \sinh \left( \frac{y^2}{2t} \right) \neq K(t+s,y).$$

If I'm not mistaken, then the density of (one-dimensional) absorbed Brownian motion equals 
$$\frac{1}{\sqrt{2\pi t}} \left( \exp \left( - \frac{(x-y)^2}{2t} \right) - \exp \left( - \frac{(x+y)^2}{2t} \right) \right),$$
which does not agree with the given formula for $n=1$. 
