Correct or not? $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{x^2+\ln^2(2\sin(x))}dx\stackrel?=\frac{\pi}{8}\left[\frac{\zeta(2)}{2}+\ln(2\pi)\right]$

Question

I got the idea from here

(1) $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{x^2+\ln^2(2\sin(x))}dx=\frac{\pi}{8}\left[\frac{\zeta(2)}{2}+\ln(2\pi)\right]$$

I am not quite sure it corrects, because I check it with a calculator and it only gave 9 decimal places. So can anyone please help me to verify (1)?

Answer 1

Can anyone please help me to verify it?

The announced equality is not correct. We have

$$ \int_{0}^{\frac{\pi}{2}}\frac{x^2}{x^2+\ln^2(2\sin(x))}dx \color{red}{\neq}\frac{\pi}{8}\left[\frac{\zeta(2)}{2}+\ln(2\pi)\right], $$

as is seen with standard methods of numerical evaluation: $$ \begin{align} I=\int_{0}^{\frac{\pi}{2}}\frac{x^2}{x^2+\ln^2(2\sin(x))}dx&= \color{red}{1.04471}13415676906315535681507671\ldots \tag1 \\\\ J=\frac{\pi}{8}\left[\frac{\zeta(2)}{2}+\ln(2\pi)\right]&=\color{red}{1.04471}46850072181679012146131844\ldots \tag2 \end{align} $$

Evaluating the integral with a composite trapezoidal rule ensures that $I\color{red}{<}J$ ($f$ being the integrand, we have $\left|f^{(2)}(x)\right|<50$ over $[0,\pi/2]$, we take $n=10\,000$) .