Need help with $\int_0^\pi\arctan^2\left(\frac{\sin x}{2+\cos x}\right)dx$

Question

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...

Answer 1

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:

$$ \int_{0}^{\pi}\arctan^2\left(\frac{\sin x}{2+\cos x}\right)\,dx=\frac{\pi}{2}\sum_{n\geq 1}\frac{1}{n^2 4^n}=\color{red}{\frac{\pi}{2}\cdot\text{Li}_2\left(\frac{1}{4}\right)}.$$

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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 $\mathbb{Z}^+$.

Answer 2

Generalizing Jack's answer. Take $a\in\Bbb R\setminus\{0\}$, and write $$J(a)=\int_0^\pi\arctan^2\left(\frac{\sin x}{a+\cos x}\right)dx.$$ From symmetry, we can write $$J(a)=\frac12\int_{-\pi}^\pi\arctan^2\left(\frac{\sin x}{a+\cos x}\right)dx$$ Then $$\arctan\left(\frac{\sin x}{a+\cos x}\right)=\Im\log(a+e^{ix})=\sum_{n\ge1}\frac{(-1)^{n+1}}{na^n}\sin(nx).$$ Then from Parseval's theorem, $$\frac1\pi\int_{-\pi}^\pi\arctan^2\left(\frac{\sin x}{a+\cos x}\right)dx=\sum_{n\ge1}\left(\frac{(-1)^{n+1}}{na^n}\right)^2,$$ so that $$J(a)=\frac\pi2\mathrm{Li}_2\left(\tfrac1{a^2}\right).$$ Unfortunately, your integral, given by $J(2)=\frac\pi2\mathrm{Li}_2(1/4)$, does not appear to have a closed form.

Answer 3

$$\begin{align*} I &= 2\int_0^\infty\arctan^2\left(\frac{2t}{3+t^2}\right)\,\frac{dt}{1+t^2} \\[1ex] &= 2\sqrt3\int_0^\infty\arctan^2\left(\frac{2\sqrt3 t}{3+3t^2}\right)\,\frac{dt}{1+3t^2} \tag{1} \\[1ex] &= 2\sqrt3\int_0^1\left(\frac1{1+3t^2}+\frac1{3+t^2}\right)\arctan^2\left(\frac1{\sqrt3}\frac{2t}{1+t^2}\right)\,dt \tag{2} \\[1ex] &= 4\sqrt3\int_0^1\arctan^2\left(\frac t{\sqrt3}\right)\,\frac{dt}{(t^2+3)\sqrt{1-t^2}} \tag{3} \\[1ex] &= 4\int_0^{\frac1{\sqrt3}=\cot\left(\frac\pi3\right)} \arctan^2(t)\,\frac{dt}{(t^2+1)\sqrt{1-3t^2}} \tag{4} \\[1ex] &= 4\int_{\frac\pi3}^{\frac\pi2} \left(\frac\pi2-t\right)^2 \,\frac{dt}{\sqrt{1-3\cot^2(t)}} \tag{5} \\[1ex] &=2\int_{\frac\pi3}^{\frac{2\pi}3} \left(\frac\pi2-t\right)^2 \,\frac{dt}{\sqrt{1-3\cot^2(t)}} \tag{6} \\[1ex] &= 2\int_{-\frac\pi6}^{\frac\pi6} \frac{t^2}{\sqrt{1-3\tan^2(t)}} \, dt \tag{7} \\[1ex] &= \frac{\pi^3}{36} - 2 \int_{-\frac\pi6}^{\frac\pi6} t \arcsin(2\sin(t)) \, dt \tag{8} \\[1ex] &= \frac{\pi^3}{36} - 2 \int_{-1}^1 \arcsin\left(\frac t2\right) \arcsin(t) \, \frac{dt}{\sqrt{4-t^2}} \tag{9} \\[1ex] &= \frac{\pi^3}{36} - 2\int_{-\frac12}^{\frac12} \arcsin(t) \arcsin(2t) \, \frac{dt}{\sqrt{1-t^2}} \tag{10} \\[1ex] &= 2 \int_{-\frac12}^{\frac12} \arcsin^2(t) \frac{dt}{\sqrt{1-4t^2}} \tag{11} \\[1ex] &= 2 \sum_{n=1}^\infty \frac{2^{2n-1}}{n^2 \binom{2n}n} \int_{-\frac12}^{\frac12} \frac{t^{2n}}{\sqrt{1-4t^2}} \, dt \tag{12} \\[1ex] &= \sum_{n=1}^\infty \frac1{n^2 \binom{2n}n} \int_0^1 \frac{t^{2n}}{\sqrt{1-t^2}} \, dt \tag{13} \\[1ex] &= \frac12 \sum_{n=1}^\infty \frac1{n^2 \binom{2n}n} \int_0^1 t^{n-\frac12} (1-t)^{-\frac12} \, dt \tag{14} \\[1ex] &= \frac\pi2 \sum_{n=1}^\infty \frac1{n^24^n} \tag{15} \\[1ex] &= \boxed{\frac\pi2 \operatorname{Li}_2\left(\frac14\right)} \tag{16} \end{align*}$$

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Steps

• $(1)$ : substitute $t\mapsto\sqrt3\,t$
• $(2)$ : substitute $t\mapsto\frac1t$ in the integral over $[1,\infty)$ and join them
• $(3)$ : substitute $t\mapsto\frac{1+\sqrt{1-t^2}}t$
• $(4)$ : substitute $t\mapsto\sqrt 3 t$
• $(5)$ : substitute $t\mapsto\operatorname{arccot}(t)$
• $(6)$ : exploit symmetry about $t=\frac\pi2$
• $(7)$ : substitute $t\mapsto\frac\pi2-t$
• $(8)$ : integrate by parts using the elementary antiderivative of $\frac1{\sqrt{1-3\tan^2(t)}}$, which simplifies nicely on $\left[-\frac\pi6,\frac\pi6\right]$
• $(9)$ : substitute $t\mapsto\arcsin\left(\frac t2\right)$
• $(10)$ : substitute $t\mapsto2t$
• $(11)$ : integrate by parts
• $(12)$ : power series of $\arcsin^2(t)$
• $(13)$ : substitute $t\mapsto\frac t2$ and exploit symmetry about $t=0$
• $(14)$ : substitute $t\mapsto\sqrt t$
• $(15)$ : beta function, and simplify beta/gammas/factorials
• $(16)$ : dilogarithm