Probability that a Wiener process has no zeroes on $[a,b]$ 
Let $(W_t)_{t\ge0}$ a Wiener process. We want to find $p:=\mathbb{P}(W_{t} \text{ has no zeroes on $[a,b]$})$.

I've considered
$$p = \mathbb{P}(W_{t} > 0, t \in [a,b]) + \mathbb{P}(W_{t} < 0, t \in [a,b]) = 2\mathbb{P}(W_{t} < 0) $$ 
How it's better to step then? It's better to use Bachelier theorem? It would be great to some hint , to start with.
 A: Approach I: By the continuity of the sample paths of Brownian motion, we have
$$\mathbb{P}(W_t \neq 0 \, \, \text{for all $t \in [a,b]$}) = \mathbb{P}(W_t < 0 \, \, \text{for all $t \in [a,b]$}) + \mathbb{P}(W_t > 0 \, \, \text{for all $t \in [a,b]$}).$$
By the symmetry of Brownian motion, this implies
$$\mathbb{P}(W_t \neq 0 \, \, \text{for all $t \in [a,b]$}) = 2\mathbb{P}(W_t < 0 \, \, \text{for all $t \in [a,b]$}). \tag{1}$$
Following the reasoning in this answer (with $[1,2]$ replaced by $[a,b]$) we obtain that
$$\mathbb{P}(W_t < 0 \, \, \text{for all $t \in [a,b]$}) = \mathbb{E} \left( \left[ 1-2\Phi\left(\frac{W_a}{\sqrt{b-a}} \right) \right] 1_{\{W_a<0\}} \right)$$ where $\Phi$ is the cdf of the centered Gaussian distribution with density $\varphi$. Hence,
$$\mathbb{P}(W_t < 0 \, \, \text{for all $t \in [a,b]$}) = \frac{1}{2} - 2 \int_{-\infty}^0 \Phi \left( \frac{\sqrt{a}}{\sqrt{b-a}} x \right) \varphi(x) \, dx.$$
Computing the latter integral (e.g. as in this answer) we conclude that
$$\mathbb{P}(W_t < 0 \, \, \text{for all $t \in [a,b]$})  =  \frac{1}{\pi} \arctan \left( \frac{\sqrt{a}}{\sqrt{b-a}} \right).$$
Finally, by $(1)$,
$$\mathbb{P}(W_t \neq 0 \, \, \text{for all $t \in [a,b]$}) = \frac{2}{\pi} \arctan \left( \frac{\sqrt{a}}{\sqrt{b-a}} \right).$$
Approach II: (Following the proof in the monograph Brownian motion - An introduction to stochastic processes by Schilling & Partzsch, Lemma 11.23, 2nd edition)
Since
$$\mathbb{P}(W_t \neq 0 \, \, \text{for all $t \in [a,b]$})= 1- \mathbb{P}(\exists t \in [a,b]: W_t = 0), \tag{2}$$
it clearly suffices to calculate tha latter probability. By the Markov property of Brownian motion, we have
$$ \mathbb{P}(\exists t \in [a,b]: W_t = 0) = \mathbb{E} \left( \mathbb{P}^{W_a}(\tau_0<b-a)\right)$$
where
$$\tau_0 := \inf\{t \geq 0; B_t = 0\}$$
is the hitting time of a Brownian motion $(B_t)_{t \geq 0}$ started at $B_0 = x$ under $\mathbb{P}^x$. Applying the reflection principle, we get
$$ \mathbb{P}(\exists t \in [a,b]: W_t = 0) = \mathbb{E} \left( \int_0^{b-a} \frac{|W_a|}{\sqrt{2\pi s^3}} \exp \left( - \frac{W_a^2}{2s} \right) \, ds \right).$$
Using $W_a \sim N(0,a)$ and applying Tonelli's theorem, we obtain that
\begin{align*} \mathbb{P}(\exists t \in [a,b]: W_t = 0) &= \frac{1}{2\pi} \int_0^{b-a} \int_{\mathbb{R}} \frac{|y|}{\sqrt{a s^3}} \exp \left( - \frac{1}{2} \left( \frac{1}{s} + \frac{1}{a} \right) y^2 \right) \, dy \, ds \\ &= \frac{1}{\pi} \int_0^{b-a} \frac{\sqrt{a}}{\sqrt{s} (s+a)} \, ds. \end{align*}
By a change of variables ($s=au^2$) this gives
$$\begin{align*} \mathbb{P}(\exists t \in [a,b]: W_t = 0) &= \frac{2}{\pi} \int_0^{\sqrt{(b-a)/a}} \frac{du}{1+u^2} \\ &= \frac{2}{\pi} \arctan \sqrt{\frac{b-a}{a}}. \end{align*}$$
As $$\arctan x = \frac{\pi}{2} - \arctan \frac{1}{x}, \qquad x>0,$$
we conclude from $(2)$ that
$$\mathbb{P}(W_t \neq 0 \, \, \text{for all $t \in [a,b]$}) = \frac{2}{\pi} \arctan \left( \frac{\sqrt{a}}{\sqrt{b-a}} \right).$$
