 Solution

Assume a positive barrier $B\le K<S_0$. Under the zero-rate risk-neutral Black-Scholes model, $X_t=\log (S_t/B)$ starts at $x=\log (S_0/B)>0$ and is Brownian motion with variance rate ${\sigma }^2$ and drift $\mu =-{\sigma }^2/2$. The drifted transition density killed at zero isTo see the factor, multiply the zero-drift reflected Gaussian difference by $\exp [\mu (y-x)/{\sigma }^2-{\mu }^{2}t/(2{\sigma }^2)]$; its second term is the ordinary drifted Gaussian from starting point $-x$ times $e^{-2\mu x/{\sigma }^2}$.

Integrate the payoff $(B{e}^y-K{)}_{+}$ against this density. The first term is $C(S_0,K)$. Since $K\ge B$, the payoff is zero for $y\le 0$, so the second integral may be extended to all real $y$ and equals the ordinary call price from reflected spot $B{e}^{-x}=B^2/S_0$. The reflection weight is $e^x=S_0/B$. Thus the down-and-out European call value isThis assumes continuous monitoring and knock-out on first hitting, with no rebate.