I'm trying to do an exercise, for a brownian motion $B_t$ show the PDF of $T=\inf \{t|B_t=\sup_{0\leq s\leq 1}B_s \} $ is $\frac{1}{\pi \sqrt{t(1-t)}} \textbf{1}_{(0,1)}(t)$. The CDF of this is $\mathbb{P}(T\leq t)=\frac{2}{\pi}\sin^{-1}(\sqrt{t})$, for $0\leq t\leq 1$.
In my attempt, I arrived at this integral $\int_{0}^{\infty} \frac{2}{\sqrt{t\pi}}\exp \left(-\frac{x^2}{2t} \right) erf(\frac{x}{\sqrt{2-2t}}) dx$, this is supposed to equal to $\mathbb{P}(T\leq t)$. I thought this couldn't possibly be equal to $\sin^{-1}$ because it only contains $\exp$ and $erf$, no trig functions or their derivatives, and wasted a lot of time checking my reasoning. But then I calculated the integral numerically for various $t$, and it always agrees with $\frac{2}{\pi}\sin^{-1}(\sqrt{t})$.
How do I show $\int_{0}^{\infty} \frac{2}{\sqrt{t\pi}}\exp \left(-\frac{x^2}{2t} \right) erf(\frac{x}{\sqrt{2-2t}}) dx = \frac{2}{\pi}\sin^{-1}(\sqrt{t})$
By substituting $x\mapsto x/\sqrt{2t}$ and $t=s^2$, the identity can be rewritten as $$I=\int_0^\infty\int_0^{x\tan s}e^{-x^2-u^2}\,du\,dx=\frac s2$$ for all $0\le s<\pi/2$ using the integral definition of $\operatorname{erf}$.
The substitution $v=\pi u/(2x\tan s)$ yields \begin{align}I&=\int_0^\infty\int_0^{\pi/2}\frac{2x\tan s}\pi e^{-x^2[1+(2\tan s)^2v^2/\pi^2]}\,dv\,dx\\&=\int_0^{\pi/2}\frac{\tan s}{\pi(1+(2v\tan s/\pi)^2)}\,dv\end{align} upon swapping the order of integration and the integral is now easily seen to equal $s/2$.