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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 $P$?

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  • $\begingroup$ Just a clarification: do you mean $\arctan^2(x)$ or $\arctan\left(x^2\right)$ in the expression for $J$? $\endgroup$
    – an4s
    Commented Feb 7, 2020 at 19:44
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    $\begingroup$ 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. $\endgroup$
    – an4s
    Commented Feb 7, 2020 at 19:55
  • $\begingroup$ @an4s $\arctan(x)^2=\arctan(x)\cdot\arctan(x)$ $\endgroup$
    – clathratus
    Commented Feb 7, 2020 at 20:04
  • $\begingroup$ @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 $\endgroup$
    – clathratus
    Commented Feb 7, 2020 at 20:06
  • $\begingroup$ 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](math.stackexchange.com/questions/4586062/…). 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 $\zeta(3)$ $\endgroup$
    – user170231
    Commented Jan 7 at 0:00

1 Answer 1

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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): $$\frac{1}{2+\cos x}=\frac{1}{\sqrt 3}+\frac{2}{\sqrt 3}\sum_{n=1}^\infty (-1)^n(2-\sqrt 3)^n\cos(nx)$$ $$\Rightarrow J=\frac{\pi^3}{16\sqrt 3}+\frac{1}{\sqrt 3}\sum_{n=1}^\infty (-1)^n(2-\sqrt 3)^n\int_0^\frac{\pi}{2}x^2\cos(nx)dx$$ $$=\frac{\pi^3}{16\sqrt 3}+\frac{\pi^2}{4\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^n\sin\left(\frac{n\pi}{2}\right)}{n}$$ $$+\frac{\pi}{\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^n\cos\left(\frac{n\pi}{2}\right)}{n^2}-\frac{2}{\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^n\sin\left(\frac{n\pi}{2}\right)}{n^3}$$ $$\small =\frac{\pi^3}{16\sqrt 3}-\frac{\pi^2}{4\sqrt 3}\sum_{n=0}^\infty \frac{(-1)^n(2-\sqrt 3)^{2n+1}}{2n+1}+\frac{\pi}{\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^{2n}}{(2n)^2}+\frac{2}{\sqrt 3}\sum_{n=0}^\infty \frac{(-1)^n(2-\sqrt 3)^{2n+1}}{(2n+1)^3}$$ $$=\boxed{\frac{\pi^3}{24\sqrt 3}+\frac{\pi}{4\sqrt 3}\operatorname{Li}_2\left(-\left(2-\sqrt 3\right)^2\right)+\frac{2}{\sqrt 3}\operatorname{Ti}_3\left(2-\sqrt 3\right)}$$ $$\text{where }\operatorname{Li}_k(x)=\sum\limits_{n=1}^\infty \frac{x^n}{n^k},\ \operatorname{Ti}_k(x)=\Im\operatorname{Li}_k(ix)=\sum\limits_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)^k}.$$

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    $\begingroup$ dude you never cease to amaze me with your integral skills. Thank you very much (+1) $\endgroup$
    – clathratus
    Commented Feb 8, 2020 at 5:50
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    $\begingroup$ 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. $\endgroup$ Commented Feb 8, 2020 at 14:24

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