I need to evaluate this integral: $$I=\int_0^{\pi/2}\arctan^2\!\left(\frac{\sin x}{\sqrt3+\cos x}\right)dx$$ Maple and Mathematica cannot evaluate it in this form. Its numeric value is $$I\approx0.156371391375711701230837603266631522020409597791339398428...$$ that is not recognized by WolframAlpha and Inverse Symbolic Calculator+.

Is it possible to evaluate this integral in a closed form?

I found similar questions here, here and here, but approaches shown in the answers do not seem to be directly applicable here.

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    $\begingroup$ Please see this post for the similar problem: math.stackexchange.com/questions/1593334/…. $\endgroup$ – xpaul Dec 31 '15 at 19:21
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    $\begingroup$ Since $$\arctan^2\left(\frac{r\sin x}{1+r\cos x}\right)=-\frac14\ln^2\left(\frac{1+re^{ix}}{1+re^{-ix}}\right)$$ one may simply find the antiderivative of $$\frac i{4z}\ln^2\left(\frac{1+rz}{1+rz^{-1}}\right)$$ either by hand or with Mathematica and plug in the integration limits at $z=1$ and $z=i$ (the integrand is analytic in the first quadrant) . Cleo's answer should follow from some simplifications through the standard polylogarithm identities. $\endgroup$ – M.N.C.E. Jan 1 '16 at 7:44
  • $\begingroup$ @M.N.C.E. That should work, but the antiderivative looks scary. I hope there is a different approach to this problem. $\endgroup$ – Vladimir Reshetnikov Jan 2 '16 at 4:08
  • $\begingroup$ By the way, if you want, write \mathrm dx to generate $\mathrm dx$ as opposed to $dx$. $\endgroup$ – Mr Pie Apr 25 '18 at 4:58

With the same approach I took in this answer, we have:

$$\begin{eqnarray*}\arctan\left(\frac{\sin x}{\sqrt{3}+\cos x}\right) &=& \text{Im}\log\left(\sqrt{3}+e^{ix}\right)\\&=&\text{Im}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n\sqrt{3}^n}\,e^{inx}\\&=&\sum_{n\geq 1}\frac{(-1)^{n+1}}{n\sqrt{3}^n}\,\sin(nx)\tag{1}\end{eqnarray*} $$ and by computing $\int_{0}^{\pi/2}\sin(nx)\sin(mx)\,dx$ it follows that: $$\begin{eqnarray*}I&=&\int_{0}^{\pi/2}\arctan^2\left(\frac{\sin x}{\sqrt{3}+\cos x}\right)\,dx\\&=&\frac{\pi}{4}\,\text{Li}_2\left(\frac{1}{3}\right)+\sum_{m\neq n}\frac{(-1)^{n+m}}{nm\sqrt{3}^{n+m}}\int_{0}^{\pi/2}\sin(nx)\sin(mx)\,dx\tag{2}\end{eqnarray*}$$ but the last integral is zero if $n$ and $m$ have the same parity.

It follows that the last series in $(2)$ can be written as:

$$\begin{eqnarray*}-\sqrt{3}\sum_{a\geq 1}\sum_{b\geq 1}\frac{1}{(2b-1)3^{a+b}}\cdot\frac{(-1)^{a+b}}{(2a)^2-(2b-1)^2}\tag{3}\end{eqnarray*}$$ and by reindexing the last double series on $a+b=s$ we get:

$$\begin{eqnarray*}-\sqrt{3}\sum_{s\geq 2}\frac{(-1)^s}{3^s}\sum_{b=1}^{s-1}\frac{1}{2b-1}\cdot\frac{1}{(2s-2b)^2-(2b-1)^2}\tag{3}\end{eqnarray*}$$ than can be dealt through partial fraction decomposition and dilogarithm identities.


$$I=\frac\pi{20}\ln^23+\frac\pi4\operatorname{Li_2}\left(\tfrac13\right)-\frac15\operatorname{Ti}_3\left(\sqrt3\right),$$ where $$\operatorname{Ti}_3\left(\sqrt3\right)=\Im\Big[\operatorname{Li}_3\left(i\sqrt3\right)\Big]=\frac{\sqrt3}8\Phi\left(-3,3,\tfrac12\right)=\frac{5\sqrt3}4\,{_4F_3}\!\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\tfrac34\right)-\frac{5\pi^3}{432}-\frac\pi{16}\ln^23.$$

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    $\begingroup$ This is not an answer. $\endgroup$ – Carl Mummert Dec 31 '15 at 20:44
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    $\begingroup$ @CarlMummert: And this is not a pipe. :-$)$ $\endgroup$ – Lucian Dec 31 '15 at 20:57
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    $\begingroup$ @BenLongo I doubt very much it could be made any simpler than the forms already shown here. See this answer and comments below it. $\endgroup$ – Vladimir Reshetnikov Jan 1 '16 at 4:15
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    $\begingroup$ Is there a table somewhere of known hypergeometric forms? $\endgroup$ – Ben Longo Jan 1 '16 at 19:42
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    $\begingroup$ @BenLongo The most extensive list I know (but still far from being complete): for $_2F_1$, for $_pF_q$. You also can download corresponding Mathematica notebooks from there. $\endgroup$ – Vladimir Reshetnikov Jan 1 '16 at 23:00

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