Find $P(B(s)>0, B(t)>0)$, where $B(s)$ is Brownian motion, $sExercise 7.1.1 in Textbook by Durret. I try to find $P(B(s)>0, B(t)>0)$, where $B(s)$ is Brownian motion, $s<t$. But how to write the probability in terms of $B(s)$ and $B(t)-B(s)$? If it works, I have got
$$P(B(s)>0, B(t)-B(s)>0)=P(B(s)>0)P(B(t)-B(s)>0)$$
$$=\int_{0}^{\infty}\frac{1}{\sqrt{2\pi s}}\exp\Big(-\frac{x^2}{2s}\Big)\int_{0}^{\infty}\frac{1}{\sqrt{2\pi (t-s)}}\exp\Big(-\frac{x^2}{2(t-s)}\Big).$$
 A: The formulation is correct, it thus remains to compute the integral. 
Let $P$ be the desired probability, write $\Phi(\cdot)$ and $\phi(\cdot)$ for the standard Gaussian c.d.f. and p.d.f.,
note that
$$
P =\int_0^{\infty}  \Phi\left(\frac{a}{\sqrt{t-s}}\right) \phi\left(\frac{a}{\sqrt{s}}\right)\mathrm{d} \frac{a}{\sqrt{s}}
        =\int_0^{\infty}\Phi(ba)\phi(a)\mathrm{d} a
$$
with
$$
b := \sqrt{\frac{s}{t-s}}.
$$
Denote by $I(b)$ the above integral, we have
$$
I(0) = \frac{1}{4}, \quad \partial_b I(b)= \int_0^{\infty} a\phi(ba)\phi(a)\mathrm{d} a
=\frac{1}{2\pi\sqrt{1+b^2}} 
            \int_0^{\infty} a\mathrm{e}^{-\frac{1+b^2}{2} a^2} \mathrm{d} a
        =\frac{1}{2\pi} \frac{1}{1+b^2},
$$
where the differentiation inside the integral side is ensured by the Leibniz rule.
Integrating back, we have
$$
I(b) = I(0) + \frac{1}{2\pi} \int_0^b 
            \frac{\mathrm{d} c}{ 1+c^2}
        = \frac{1}{4} + \frac{1}{2\pi} \tan^{-1}(b).
$$
Finally, since $\tan^{-1}(\sqrt{s/(t-s)}) = \sin^{-1}(\sqrt{s/t})$, we have
$$
P = I(b) = \frac{1}{4} + \frac{1}{2\pi} \sin^{-1}\left(\sqrt{\frac{s}{t}}\right).
$$
Note that always do we have
$$
\frac{1}{4} \le P \le \frac{1}{2},
$$
and as $s \uparrow t$, $P \to 1/2$.
The idea of taking derivative wrt the parameter was inspired by @Davide Giraudo's 
answer to integrating products of Gaussian c.d.f. and p.d.f.
EDIT (Apr 27, 2020) Fixed an error in $\partial_b I(b)$ and updated the answer.
A: Let $X=B(s),Y=B(t)-B(s)$. You have to find $P\{X>0,X+Y>0\}$. Integrate the joint density over $\{(x,y):x>0,x+y>0\}$. Split it onto two parts: one with $y>0$ and the other with $y<0$. When $y>0$ the only condition is $x>0$ so the integral is over $(0,\infty) \times(0,\infty )$. When $y<0$ we have the condition $x>-y$ so integrate with respect to $x$ from $-y$ to $\infty$ and then with respect to $y$ from  $-\infty $ to 0. 
