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I am trying to evaluate this integral:$$K = \displaystyle \int\limits_0^1 {\frac{{{{\arctan }^2}\left( {\frac{x}{{1 - {x^2}}}} \right)}}{x}dx}$$

And here is my attempt so far:

$$\begin{align} K&=\int\limits_0^1 {\frac{{{{\arctan }^2}\left( {\frac{x}{{1 - {x^2}}}} \right)}}{x}dx} \xrightarrow{{{\text{IBP}}}} - 2\int\limits_0^1 {\frac{{\left( {{x^2} + 1} \right)\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx}\\&= - 2\int\limits_0^1 {\frac{{{x^2}\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} - 2\int\limits_0^1 {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} \hfill \\ \\&=- 2\int\limits_0^1 {\frac{{{x^2}\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} - 2\int\limits_0^\infty {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} \\&\quad\quad+ 2\underbrace {\int\limits_1^\infty {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} }_{x \to \frac{1}{x}} \\&= - 2\int\limits_0^\infty {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} \hfill \\ \end{align}$$

The last integral seems can be evaluated by using complex integration technique. But I don't know how to use it. And can I ask for help or another approach? Thank you so much.

The closed form for $K$ is: $$K=\frac{4\pi}{3}\cdot G-\frac{35}{36}\cdot\zeta(3)-\frac{5\pi}{36\sqrt{3}}(\psi'(1/3)-\psi'(2/3))$$

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  • $\begingroup$ Similar question math.stackexchange.com/questions/4405138/… $\endgroup$ Commented Jun 12, 2023 at 10:49
  • $\begingroup$ @Zima In fact, there is closed form for this integral. $\endgroup$ Commented Jun 12, 2023 at 13:09
  • $\begingroup$ @ErikSatie Thank you for your suggestion. But these 2 problem are different too much. $\endgroup$ Commented Jun 12, 2023 at 13:10
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    $\begingroup$ @Zima $$\frac{4 \pi G}{3}-\frac{35 \zeta (3)}{36}-\frac{5 \pi \left(\psi ^{(1)}\left(\frac{1}{3}\right)-\psi ^{(1)}\left(\frac{2}{3}\right)\right)}{36 \sqrt{3}}$$. Thank you. $\endgroup$ Commented Jun 12, 2023 at 13:19
  • $\begingroup$ The integral is equivalent to $$\frac{\pi^2}{4}\log2+\frac74 \zeta(3)-2\int_0^{\frac{\pi}{2}}t \log(\sqrt{4\tan^2 t+1}-1) dt$$ $\endgroup$ Commented Jun 12, 2023 at 14:07

2 Answers 2

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A key observation is that

$$ \arctan\left(\frac{x}{1-x^2}\right) = \arctan(x)+\arctan(x^3). \tag{1} $$

Since $\int_{0}^{1}\frac{\arctan^2(x)}{x}\,\mathrm{d}x = \frac{\pi}{2}G-\frac{7}{8}\zeta(3)$ is fairly well-known (it is enough to enforce the substitution $x=\tan\theta$ and exploit suitable Fourier series) and $\int_{0}^{1}\frac{\arctan^2(x^3)}{x}\,\mathrm{d}x$ is just a multiple of the previous integral (by the substitution $x\mapsto z^{1/3}$) the whole problem boils down to the evaluation of

$$ \int_{0}^{1}\arctan(x)\arctan(x^3)\frac{\mathrm{d}x}{x}=\int_{0}^{1}\sum_{m\geq 0}\frac{(-1)^m x^{2m+1}}{2m+1}\sum_{n\geq 0}\frac{(-1)^n x^{6n+3}}{2n+1}\frac{\mathrm{d}x}{x}\\=\sum_{m\geq 0}\sum_{n\geq 0}\frac{(-1)^{m+n}}{(2m+1)(2n+1)(2m+6n+4)}. \tag{2} $$

This Euler series can be evaluated in terms of $\psi'(s)=\frac{d^2}{ds^2}\left(\log\Gamma(s)\right)$ and we are done.

In order to address the comments: $$2\int_{0}^{1}\frac{x^3\,dx}{(1+a^2 x^2)(1+b^2 x^6)} =\int_{0}^{1}\frac{x\,dx}{(1+a^2 x)(1+b^2 x^3)}\\ \stackrel{\text{PFD}}{=}\frac{a^2\log(1+a^2)}{b^2-a^6}+\frac{1}{a^6-b^2}\int_{0}^{1}\frac{a^4-b^2 x+a^2 b^2 x^2}{1+b^2 x^3}\,dx $$ is a linear combination, with coefficients given by algebraic functions of $\alpha$ and $\beta^{1/3}$, of logarithms and arctangents evaluated at polynomials of $\alpha$ or polynomials of $\beta^{1/3}$. It is almost an unbearable task by hand, but a CAS is able to integrate such thing over $\alpha\in(0,1)$ and $\beta\in(0,1)$, making $\int_{0}^{1}\arctan(x)\arctan(x^3)\frac{dx}{x}$ solvable through Feynman's trick.

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    $\begingroup$ Could you give a hint how to evaluate the series? $\endgroup$ Commented Jul 15 at 20:44
  • $\begingroup$ @AmrutAyan here $\beta(s)$ doesn't refer to the dirichlet beta function. $\beta(s)=\frac{1}{2}\left[\psi\left(\frac{s+1}{2}\right)-\psi\left(\frac{s}{2}\right)\right]$ $\endgroup$ Commented Jul 16 at 19:43
  • $\begingroup$ Thank you Jack, but the series is still very hard to evaluate. It's better if you give some hints to crack it (+1). $\endgroup$ Commented Jul 18 at 4:44
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The integral J is what you are looking for You may want to consider this (remember that it is not my solution):

https://www.facebook.com/groups/170225716325580/permalink/6514069418607813/?app=fbl

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