Question:
Define $F(r)=\int_1^r\dfrac{dx}{\sqrt{(1+\frac{r^4-1}{\log r}\log x)^\frac{1}{2}-x^2}}$, $r>1$, show $F(r)>\dfrac{\pi}{2}$ by numerical methods.
I've tried using some $\arcsin$ function to compare but it seems not working well.
Any help will be appreciated.
I think that it would be difficult to find better than @copper.hat's result.
Consider $$f(x)=1+\frac{(r^4-1)}{\log (r)}\log(x)$$ For $x\in (1,r)$, it is an increasing function $$1 \leq f(x) \leq r^4 \quad \implies \quad 1 \leq\sqrt{f(x)}\leq r^2$$
$$ 1-x^2 \leq\sqrt{f(x)}-x^2\leq r^2-x^2\quad \implies \quad \sqrt{1-x^2} \leq\sqrt{\sqrt{f(x)}-x^2}\leq \sqrt{r^2-x^2}$$ $$\frac 1 {\sqrt{r^2-x^2}} \leq \frac 1 {\sqrt{\sqrt{f(x)}-x^2}} \leq \frac 1 {\sqrt{1-x^2}}$$ $$\int_1^r \frac {dx} {\sqrt{r^2-x^2}} \leq \int_1^r \frac {dx} {\sqrt{\sqrt{f(x)}-x^2}}\leq \int_1^r \frac {dx} {\sqrt{1-x^2}} $$
$$F(r) \geq \sec ^{-1}(r)=\frac{\pi }{2}-\sin ^{-1}\left(\frac{1}{r}\right)=\frac \pi 2-\frac{1}{\sqrt{\pi }}\sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{1}{2}\right)}{(2 n+1) \Gamma (n+1)} r^{-(2n+1)}$$