I came upon this problem while messing around with the function $\tan^{-1}\left(\frac{\ln\left(\sin x\right)}{x^{2}}\right)$ (Hence the recreational mathematics tag)
I noticed that the function's domain was $2n\pi<x<(2n+1)\pi$ for $n\in\mathbb{Z}$.
I also noticed the integral $\int_{2n\pi}^{\left(2n+1\right)\pi}\tan^{-1}\left(\frac{\ln\left(\sin x\right)}{x^{2}}\right)dx$ converges for all $n$ (as we have previously defined $n$ as any integer). I saw that as $n$ approached $-\infty$ and $\infty$, the integral approached $0$.
I then had an interesting idea: to sum the integrals for all $n$. That is how I got to the infinite sum $\sum_{n=-\infty}^{\infty}\int_{2n\pi}^{\left(2n+1\right)\pi}\tan^{-1}\left(\frac{\ln\left(\sin x\right)}{x^{2}}\right)dx$.
Using Desmos, I approximated it to $−1.46193837522$ with summation bounds $-155$ and $134$.
I do not even know how to begin solving this problem but my assumption would be to find an indefinite integral first.
I am looking for a closed form which may or may not exist.
Thank you for the help in advance!
Best Answer
"Abandon hope all ye who enter here" as wrote Dante Alighieri in "The Divine Comedy".
The summation converges very slowly. Consider $$S_p=\sum_{n=-100p}^{n=+100p} J_n\qquad \text{where} \qquad J_n=\int_{2n\pi}^{\left(2n+1\right)\pi}\tan^{-1}\left(\frac{\log\left(\sin (x)\right)}{x^{2}}\right)\,dx$$ $$\left( \begin{array}{cc} p & S_p \\ 1 & -1.461605315 \\ 2 & -1.462152793 \\ 3 & -1.462335893 \\ 4 & -1.462427557 \\ 5 & -1.462482592 \\ 6 & -1.462519297 \\ 7 & -1.462545523 \\ 8 & -1.462565196 \\ 9 & -1.462580500 \\ 10 & -1.462592744 \\ \end{array} \right)$$
which seems to converge to some $-1.4627030$ which is not recognized by inverse symbolic calculators.