I'm Trying to prove that given $$ X_t=\mu t+W_t $$
with $W_t$ a Brownian Motion and $$\tau_m=\min \{ t \geq 0 ; X_t = m\}$$
is the $m$ - first hitting time.
My goal is to prove that $\tau_m$ is finite almost surely : $P[\tau_m<\infty]=1$
My Attempt :
We know that the joint density of $W_t$ and $\tau = \min \{ t \geq 0 \ ; W_t = p \} $ is given by (using the reflection principle) :
$$ P[\tau < t , W_t \leq w ] = P[W_t \geq 2p-w] $$
Writing $\tau_m$ as :
\begin{align*}\tau_m &= \min \{ t \geq 0 ; X_t = m \} \\ &= \min \{ t \geq 0 ; W_t = m - \mu t \} \end{align*}
and putting $p=m - \mu t$ we get :
\begin{align*} P[\tau_m < t , X_t \leq x ] &= P[\tau_m < t , W_t \leq x - \mu t ] \\ &= P[W_t \geq 2(m-\mu t) - x + \mu t] \end{align*}
This gives : \begin{align*} P[\tau_m<t] &= 2 P[\tau_m < t , X_t \leq x ] \\ &= 2 P[W_t \geq 2(m-\mu t) - x + \mu t] \\ &= 2 \int _{2m - x - \mu t}^{\infty} \frac{1}{\sqrt{2\pi t} } e^{-\frac{1}{2t}y^2} dy \end{align*}
Now when I take the limit $t \rightarrow \infty$ this gives zero. Where did I go wrong? Any help is appreciated.