I was reading the paper NOISY 1-BIT COMPRESSED SENSING EMBEDDINGS ENJOY A RESTRICTED ISOMETRY PROPERTY
In the paper, the author does the following to obtain the geodesic distance on the unit sphere:
Let $g=(g_1,\dots, g_n)$ be a random vector with i.i.d standard normal entries. For $x,y\in\mathbb{S}^{n-1}$, define the following events $$H_x=\{\langle x,g\rangle>0\}$$ $$W_{x,y}=H_x\triangle H_y$$ where $\langle \cdot,\cdot\rangle$ denotes the usual dot product and $\triangle$ denotes the symmetric difference between two sets.
Then set $Z=(\langle x,g\rangle,\langle y,g\rangle)$, which is a 2 dimensional normal distribution with covariance matrix $$\Sigma=\begin{bmatrix}1, &\langle x,y\rangle\\ \langle x,y\rangle,&1\end{bmatrix}$$
Thus it follows that $$\mathbb{P}(W_{x,y})=\frac{1}{\pi\sqrt{1-\langle x,y\rangle^2}}\int_0^\infty\int_0^\infty\exp\left(\frac{2uv\langle x,y\rangle-u^2-v^2}{2-2\langle x,y\rangle^2}\right)dudv=\frac{1}{\pi}\cos^{-1}(\langle x,y\rangle)\quad(*).$$
What I was trying to do is deriving equation $(*)$. For convenience set $$u=\langle x,g\rangle,\quad v=\langle y,g\rangle.$$ Then it is easy to observe that $g\in W_{x,y}$ if and only if $uv\leq 0$. So by plug in the density function of a 2 dimensional normal distribution, we get $$\begin{aligned}\mathbb{P}(W_{x,y})&=\frac{1}{2\pi\sqrt{1-\langle x,y\rangle^2}}\int_{\{uv\leq 0\}}\exp\left(\frac{2uv\langle x,y\rangle-u^2-v^2}{2-2\langle x,y\rangle^2}\right)dudv\\ &=\frac{1}{\pi\sqrt{1-\langle x,y\rangle^2}}\int_0^\infty\int_{-\infty}^0 \exp\left(\frac{2uv\langle x,y\rangle-u^2-v^2}{2-2\langle x,y\rangle^2}\right)dudv \end{aligned}$$ which has a different integration domain than the one in equation (*). My first question is: where did I made the mistake?
Secondly, even with the integral in $(*)$, I still cannot derive the last equality in $(*)$. I first tried the polar coordinates, then I got $$\begin{aligned}&\int_0^\infty\int_0^\infty\exp\left(\frac{2uv\langle x,y\rangle-u^2-v^2}{2-2\langle x,y\rangle^2}\right)dudv\\ =&\int_0^{\frac{\pi}{2}}\int_0^\infty r\exp\left(\frac{r^2\sin(2\theta)\langle x,y\rangle-r^2}{2-2\langle x,y\rangle^2}\right)drd\theta\\ =&\frac{1}{2}\int_0^{\pi}\frac{1-\langle x,y\rangle^2}{\langle x,y\rangle\sin(\theta)-1}d\theta\\ =&\frac{1}{2}\int_0^\infty \frac{2-2\langle x,y\rangle^2}{w^2-\langle x,y\rangle w+1}dw\quad (\text{using the substitution }\tan\frac{\theta}{2}=w)\\ =&\frac{1}{2}\int_0^\infty \frac{2-2\langle x,y\rangle^2}{(w-\frac{1}{2}\langle x,y\rangle)^2+1-\frac{1}{4}\langle x,y\rangle^2}dw \end{aligned}$$ So the final answer is something involving $\tan^{-1}$, but no $\cos^{-1}$ involved. My second question is: how to deduce the last equality in $(*)$?.
Thanks for your attentions and help.
Answer to your second question; it is not contradictory.
Notations: I use normalized notations acos, asin instead of $\cos^{-1}, \sin^{-1}$.
Setting $a=\langle x,y\rangle$, assuming $0<a<1$, your integral (given by Mathematica) becomes:
$$\ \tag{1} \int_0^{\infty} \dfrac{1-a^2}{a \sin(t)-1}dt=-2\sqrt{1-a^2}(\dfrac{\pi}{2}+asin(a))$$
But : asin$(a)$+acos$(a)=\dfrac{\pi}{2}$;
Thus, you can write (1) as a function of acos$(a)$.