Finding $\mathbb{P}(\max_{t\leq 1} (W_t+t)\geq 1)$ I was trying to find $\mathbb{P}(\max_{t\leq 1}  (W_t+t)\geq 1),$ where $W_t$ is a one-dimensional Brownian motion for $t\in [0,1].$ I first applied the Girsanov's theorem to get $$\mathbb{P}\Big(\max_{t\leq 1}  (W_t+t)\geq 1\Big)=\mathbb{E}\Big[\mathbb{1}_{\max_{t\leq 1}  (W_t)\geq 1} \exp\Big(W_1-\frac{1}{2}\Big)\Big],$$ where $\mathbb{1}_A$ is the indicator function of set $A$.
A given hint here is to use $\mathbb{P}(\max_{t\leq 1} W_t\geq 1, W_1\leq b)$ and use the fact that if $\mathbb{P}(\xi\leq a ,\eta\leq b)=\int_{-\infty}^b f(x) \ dx$ for every $b$, then $\mathbb{E}[g(\eta)\mathbb{1}_{\xi\leq a}]=\int_\mathbb{R}g(x)f(x) \ dx.$
Hence, I've found
$$
\mathbb{P}(\max_{t\leq 1} W_t\geq 1, W_1\leq b) = \left\{
        \begin{array}{ll}
            \mathbb{P}(W_1\geq 2-b) & \quad b \leq 1 \\
            2\mathbb{P}(W_1\geq 1)-\mathbb{P}(W_1\geq b) & \quad b \geq 1
        \end{array}
    \right.
$$
and we have $g(x)=\exp(x-\frac{1}{2})$ (using the same notation as the hint), since we want to compute $\mathbb{E}\Big[\mathbb{1}_{\max_{t\leq 1}  (W_t)\geq 1} \exp\Big(W_1-\frac{1}{2}\Big)\Big].$
Thus, by using the above result, we get for $b\leq 1$ $$\mathbb{P}(\max_{t\leq 1} W_t\geq 1, W_1\leq b)=\int_{-\infty}^b \frac{1}{\sqrt{2\pi}} \ e^{-(2-x)^2/2} \ dx.$$
However, I could not find an integral of this form for the case $b\geq 1$. Can someone offer me guidance on how to proceed from here?
The answer, by the way, is $$ \mathbb{P}(\max_{t\leq 1}  (W_t+t)\geq 1)=\int_1^\infty e^{x-\frac{1}{2}}\frac{1}{\sqrt{2\pi}} \ e^{-x^2/2} \ dx + \int_{-\infty}^1 e^{x-\frac{1}{2}} \frac{1}{\sqrt{2\pi}} \ e^{-(2-x)^2/2} \ dx.$$
 A: First, as you noticed, $\tilde{W}_t:=W_t+t$, $t\in[0,1]$, is a Brownian motion under $\mathsf{Q}$ s.t. $d\mathsf{P}=\exp(\tilde{W}_1-1/2)d\mathsf{Q}$.
Then, letting $A:=\{\max_{0\le t\le 1}\tilde{W}_t\ge 1\}$,
$$
\mathsf{P}(A)=\mathsf{E}_{\mathsf{Q}}[1_{A}\exp(\tilde{W}_1-1/2)]=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{x-\frac{1}{2}}e^{-\frac{(|x-1|+1)^2}{2}}\,dx
$$
because for a $\mathsf{Q}$-Brownian motion $\{B_t\}$ and $y\ge 0$,
$$
\mathsf{Q}\!\left(\max_{0\le s\le t}B_s\ge y, B_t\in dx\right)=\frac{1}{\sqrt{2\pi t}}e^{-\frac{(|x-y|+y)^2}{2t}}\, dx.
$$
(See, for example, Eq. 1.1.8 on page 154 here)

The last formula can be derived as follows. For $\epsilon>0$ and $y\le x-\epsilon$,
$$
\mathsf{Q}\!\left(\max_{0\le s\le t}B_s\ge y, x-\epsilon< B_t< x+\epsilon\right)=\mathsf{Q}\!\left(x-\epsilon< B_t< x+\epsilon\right),
$$
and for $x+\epsilon<y$, using the strong Markov property,
$$
\mathsf{Q}\!\left(\max_{0\le s\le t}B_s\ge y, x-\epsilon< B_t< x+\epsilon\right)=\mathsf{Q}\!\left(2y-x-\epsilon< B_t< 2y-x+\epsilon\right).
$$
Sending $\epsilon\to 0$ yields the result.
