 Solution

For zero drift, the reflection principle gives the known Brownian first-passage time densityTo obtain a general drift, on a fixed finite horizon use the exponential martingale $M_t=e^{c{B}_t-c^{2}t/2}$ as change-of-measure density. The Novikov condition holds for constant $c$, and the Girsanov theorem says that the coordinate process $B$ has drift $c$ under the new measure. Thus the hitting law of $B$ under that measure is the law of $B+ct$ under the original measure. Conditional expectation of $M_T$ at a hitting time bounded by $T$ equals its stopped value, by the martingale optional sampling theorem. At $H_a=t$ this value is $e^{ca-c^{2}t/2}$. The drifted Brownian first-passage density is obtained by multiplying $h_0$, giving $\boxed{h_c(t)=a{e}^{-(a-ct{)}^2/(2t)}/\sqrt{2\pi t^3}}$.

To verify the integrated formula, put $u=(a-cT)/\sqrt{T}$ and $v=(-a-cT)/\sqrt{T}$. The identity $e^{2ac}\varphi (v)=\varphi (u)$ for the standard normal density givesBoth terms vanish as $T\downarrow 0$, so $\boxed{\mathbb{P}(H_a\le T)=\overline{\Phi }((a-cT)/\sqrt{T})+e^{2ac}\Phi ((-a-cT)/\sqrt{T})}$. The distribution has total mass one for $c\ge 0$ and mass $e^{2ac}$ for $c<0$; in the latter case the remaining probability is an atom at infinity.