Please help me to evaluate this integral:
$$\int_0^\pi\arctan^2\left(\frac{\sin x}{2+\cos x}\right)dx$$
Using substitution $x=2\arctan t$ it can be transformed to:
$$\int_0^\infty\frac{2}{1+t^2}\arctan^2\left(\frac{2t}{3+t^2}\right)dt$$
Then I tried integration by parts, but without any success…
Calculus – Solving Integral of Arctan Squared Function
calculusclosed-formdefinite integralsintegrationtrigonometry
Best Answer
A Fourier analytic approach. If $x\in(0,\pi)$, $$\begin{eqnarray*}\arctan\left(\frac{\sin x}{2+\cos x}\right) &=& \text{Im}\log(2+e^{ix})\\&=&\text{Im}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n 2^n}\,e^{inx}\\&=&\sum_{n\geq 1}\frac{(-1)^{n+1}}{n 2^n}\,\sin(nx),\end{eqnarray*}$$ hence by Parseval's theorem:
As a side note, we may notice that $\text{Li}_2\left(\frac{1}{4}\right)$ is quite close to $\frac{1}{4}$.
By applying summation by parts twice we get:
$$ \sum_{n\geq 1}\frac{1}{n^2 4^n} = \color{red}{\frac{1}{3}-\frac{1}{12}}+\sum_{n\geq 1}\frac{1}{9\cdot 4^n}\left(\frac{1}{n^2}-\frac{2}{(n+1)^2}+\frac{1}{(n+2)^2}\right)$$ and the last sum is positive but less than $\frac{11}{486}$, since $f:n\mapsto \frac{1}{n^2}-\frac{2}{(n+1)^2}+\frac{1}{(n+2)^2}$ is a positive decreasing function on $\mathbb{Z}^+$.