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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?

Booldy
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  • Have you tried the substitution $u = \pi^{-x^2}$? – Daniel Fischer Aug 22 '15 at 12:55
  • Edited.Daniel Fischer nothing happens,i just obtain some uggly expresions. – Booldy Aug 22 '15 at 13:01
  • Sure, it won't become beautiful all of a sudden. But maybe it's easier to bound the integral after the substitution (I haven't tried, maybe it's not easier). I don't think the estimate $\arcsin t \leqslant \frac{\pi}{2}t$ for $0 \leqslant t \leqslant 1$ will be enough to show the inequality. – Daniel Fischer Aug 22 '15 at 13:21
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    @DanielFischer: we need an extremely good approximation for the arcsine or some different idea, because the integral is very close to $1$. – Jack D'Aurizio Aug 22 '15 at 13:57
  • What is the true value of the integral? – marty cohen Aug 22 '15 at 14:14
  • @martycohen: it is about $0.99151$. – Jack D'Aurizio Aug 22 '15 at 14:26
  • With elementary techniques it is not difficult to prove that the integral is between $0.975$ and $1.054$, a pretty tight window, but a proof of point $b)$ still eludes me. – Jack D'Aurizio Aug 22 '15 at 14:46
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    It might be interesting to replace $\pi$ with a variable and see what the integral is as a function of that variable. Maybe differentiate wrt that variable. – marty cohen Aug 22 '15 at 16:28

1 Answers1

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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.

Jack D'Aurizio
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