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Evaluate the integral $$P=\int_0^{\pi/2}x\arctan\left(\tfrac{1}{\sqrt3}+\tfrac{2}{\sqrt3}\tan x\right)dx.$$

Context:

I started trying to evaluate the integral $$J=\int_0^\infty \frac{\arctan(x)^2}{x^2+x+1}dx,$$ and the integral $P$ is part of the process. At first, I tried $x\mapsto 1/x$, but it just ended up showing that $$J=\frac{\pi^2}{4}\int_0^\infty \frac{dx}{x^2+x+1}-\pi\int_0^\infty \frac{\arctan x}{x^2+x+1}dx+J,$$ which is of no use. Next, I tried integration by parts, using $$\int\frac{dx}{x^2+x+1}=\frac{2}{\sqrt3}\arctan\frac{2x+1}{\sqrt3},$$ so that $$J=\frac{\pi^3}{4\sqrt3}-\frac{4}{\sqrt3}\int_0^\infty\arctan(x)\arctan\left(\tfrac1{\sqrt3}+\tfrac{2}{\sqrt3}x\right)\frac{dx}{1+x^2}.$$ Then with $x\mapsto \tan x$ we have $$J=\frac{\pi^3}{4\sqrt3}-\frac{4}{\sqrt3}P.$$ Theoretically, integration by parts if possible from this point, as Wolfram provides an awful closed form for the anti-derivative of $\arctan\left(\tfrac{1}{\sqrt3}+\tfrac{2}{\sqrt3}\tan x\right)$, but I do not think this is really that realistic of an approach. Is there a better way to evaluate the integral $P$?

Zacky
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clathratus
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  • Just a clarification: do you mean $\arctan^2(x)$ or $\arctan\left(x^2\right)$ in the expression for $J$? – an4s Feb 07 '20 at 19:44
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    If the latter, then you could express $\arctan (z)$ in its logarithmic form which, after a few substitutions, should give you a result in the form of logarithms and dilogarithms. – an4s Feb 07 '20 at 19:55
  • @an4s $\arctan(x)^2=\arctan(x)\cdot\arctan(x)$ – clathratus Feb 07 '20 at 20:04
  • @an4s yes, I can see how it is evaluable in terms of $\mathrm{Li}_2$ and elementary functions, but it's extremely messy and I was hoping there was a simpler solution – clathratus Feb 07 '20 at 20:06
  • Contour integration might be feasible but cumbersome. IBP first to obtain$$\int_0^\infty\frac{\arctan^2x}{x+1+\frac1x},\frac{dx}x=\int_0^\infty\frac{(x^2-1)\arctan^2x\log x}{(x^2+x+1)^2},dx-2\int_0^\infty\frac{x\arctan x\log x}{(x^2+1)(x^2+x+1)},dx,$$then try to adapt this. I've lost myself in the details of my first attempt at the RHS's second integral, but there did appear some promising reductions of intermediate terms to closed forms containing polygammas and $\zeta(3)$ – user170231 Jan 07 '24 at 00:00

1 Answers1

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$$J=\int_0^\infty \frac{\arctan^2 x}{1+x+x^2}dx\overset{x=\tan t}=\int_0^\frac{\pi}{2}\frac{t^2}{1+\sin t\cos t}dt\overset{2t=\frac{\pi}{2}-x}=\frac14\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{\left(\frac{\pi}{2}-x\right)^2}{2+\cos x}dx$$ $$=\frac12\int_0^\frac{\pi}{2}\frac{\frac{\pi^2}{4}+x^2}{2+\cos x}dx=\frac{\pi^3}{24\sqrt 3}+\frac12\int_0^\frac{\pi}{2}\frac{x^2}{2+\cos x}dx$$ Now we will use the following Fourier series (refer to this thread): $$\frac{1}{2+\cos x}=\frac{1}{\sqrt 3}+\frac{2}{\sqrt 3}\sum_{n=1}^\infty (-1)^n(2-\sqrt 3)^n\cos(nx)$$ $$\Rightarrow J=\frac{\pi^3}{16\sqrt 3}+\frac{1}{\sqrt 3}\sum_{n=1}^\infty (-1)^n(2-\sqrt 3)^n\int_0^\frac{\pi}{2}x^2\cos(nx)dx$$ $$=\frac{\pi^3}{16\sqrt 3}+\frac{\pi^2}{4\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^n\sin\left(\frac{n\pi}{2}\right)}{n}$$ $$+\frac{\pi}{\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^n\cos\left(\frac{n\pi}{2}\right)}{n^2}-\frac{2}{\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^n\sin\left(\frac{n\pi}{2}\right)}{n^3}$$ $$\small =\frac{\pi^3}{16\sqrt 3}-\frac{\pi^2}{4\sqrt 3}\sum_{n=0}^\infty \frac{(-1)^n(2-\sqrt 3)^{2n+1}}{2n+1}+\frac{\pi}{\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^{2n}}{(2n)^2}+\frac{2}{\sqrt 3}\sum_{n=0}^\infty \frac{(-1)^n(2-\sqrt 3)^{2n+1}}{(2n+1)^3}$$ $$=\boxed{\frac{\pi^3}{24\sqrt 3}+\frac{\pi}{4\sqrt 3}\operatorname{Li}_2\left(-\left(2-\sqrt 3\right)^2\right)+\frac{2}{\sqrt 3}\operatorname{Ti}_3\left(2-\sqrt 3\right)}$$ $$\text{where }\operatorname{Li}_k(x)=\sum\limits_{n=1}^\infty \frac{x^n}{n^k},\ \operatorname{Ti}_k(x)=\Im\operatorname{Li}_k(ix)=\sum\limits_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)^k}.$$

Zacky
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    dude you never cease to amaze me with your integral skills. Thank you very much (+1) – clathratus Feb 08 '20 at 05:50
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    Very nice! I think that in the cosine sum the factor $\cos\left(\frac{2 n \pi}{2}\right) = (-1)^n$ got lost in the penultimate step. It would lead to $\operatorname{Li}_2 \left(\color{red}{-}(2-\sqrt{3})^2\right)$ in the final result, in agreement with numerical integration. – ComplexYetTrivial Feb 08 '20 at 14:24