hard integral inequality with $\pi$ a) Prove that $\int_{0}^{\infty} \arcsin{\pi^{-x^2}} dx$ converges
b) Prove that $\int_{0}^{\infty} \arcsin{\pi^{-x^2}} dx<1$ 
I have tried to find suitable integral sum for b), unsuccessfully. Is there a special method?
 A: The point is that $\arcsin(\pi^{-x^2})$ is a function with a very fast decay, so it belongs to $L^1(\mathbb{R}^+)$.
For such a task, it is enough to exploit $\arcsin t\leq \frac{\pi}{2}\,t$ over $(0,1)$. However, that inequality does not prove the second point, neither it does the Shafer-Fink inequality
$$\forall t\in(0,1),\qquad \arcsin(t)\leq\frac{\pi t}{2+\sqrt{1-t^2}},$$
leading to an integral that equals $1.026\ldots$. 
If we consider that, through integration by parts and a substitution:
$$ I = \frac{2}{\sqrt{\log\pi}}\int_{0}^{+\infty}\frac{x^2\,dx}{\sqrt{e^{2x^2}-1}}$$
and that $\frac{x}{\sqrt{e^{2x^2}-1}}$ behaves like $\frac{1}{\sqrt{2}}e^{-x^2/2}$, the Cauchy-Schwarz inequality gives:
$$ I \leq \frac{2}{\sqrt{\log\pi}}\sqrt{\left(\int_{0}^{+\infty}x e^{-x^2/2}\,dx\right)\cdot\left(\int_{0}^{+\infty}\frac{x^3 e^{x^2/2}}{e^{2x^2}-1}\,dx\right)}$$
or:
$$ I \color{red}{\leq\sqrt{\frac{\pi^2-8K}{2\log\pi}}} = 1.053685\ldots $$
where $K$ is Catalan's constant, still no good. 
Maybe it is useful to switch to the inverse function and consider that:
$$ I = \frac{1}{\sqrt{\log\pi}}\int_{0}^{\pi/2}\sqrt{-\log\sin x}\,dx. $$
Since $\sqrt{x}$ is a concave function, Jensen's inequality gives the nice lower bound:
$$ I \color{red}{\geq} \frac{1}{\sqrt{\log\pi}}\sqrt{\int_{0}^{\pi/2}-\log\sin x\,dx} = \color{red}{\sqrt{\frac{\pi \log 2}{2\log \pi}}}=0.97526\ldots $$

Using 
$$\sqrt{-\log\cos x}+\sqrt{-\log\sin x}\leq \left(\sqrt{-\log x}+\frac{x}{\sqrt{2}}\right)\cdot\left(1-\frac{x^2}{12\log x}\right)$$
(over $(0,\pi/4)$) it is possible to prove that $\color{red}{I<1}$, but I hope a simpler solution will be found.
