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I have been trying to find the closed form for integral below $$\int_1^{\infty}\frac{x^2\tan^{-1}(ax)}{x^4+x^2+1}dx ,\; \; a>0 $$ My progress to this integral $$\cong\frac{\pi^2}{8\sqrt 3}+\frac{\pi}{8}\log(3)-\frac{\pi}{6a\sqrt 2}+\frac{1}{2a^3}\left(\frac{\log(3)}{4} -\frac{\pi}{12\sqrt 3}\right)-\frac{1}{5a^5}\left(\frac{1}{2}-\frac{\pi}{12\sqrt 2}-\frac{\log(3)}{4}\right)+\frac{1}{7a^7}\left(\frac{\pi}{6\sqrt 3}-\frac{1}{2}\right)+\frac{1}{108a^9} +\frac{1}{9a^9}\left(\frac{\log(3)}{9}-\frac{\pi}{12\sqrt 3}\right)-\frac{1}{24a^{11}}+\cdots $$ Using the series of $\tan(ax)$ the above form is obtained. However, I dont find closed form for it. If $a\to\infty+$,then it is equal to $\frac{\pi^2}{8\sqrt 3}+\frac{\pi}{8}\log(3)$.

Naren
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    Is there a particular reason you are interested in this integral? – Yuriy S Apr 27 '21 at 10:51
  • If we see the integral a bit carefully we can see the occurrence of $\log(3)/4$, $\pi/(12\sqrt 2)$ . I don't understand why this is happening. What makes me interested about it, will the terms like mentioned here occur even if we go with higher terms? Or does it can have an explicit closed form? There was a proposed issue somewhere in RMM to find the closed form of $$\int_0^{\infty} \frac{x^2\tan^{-1}(ax)}{x^4+x^2+1}dx$$ – Naren Apr 27 '21 at 11:05

2 Answers2

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We have \begin{align}\int_1^{\infty}\frac{x^2\arctan ax}{x^4+x^2+1}\,dx&=\int_0^1\frac{\arctan a/u}{u^2(1/u^4+1/u^2+1)}\frac{du}{u^2}\\&=\int_0^1\frac{\pi/2-\arctan u/a}{u^4+u^2+1}\,dx\\&=\frac\pi2\left(\frac14\log3+\frac{\pi\sqrt3}{12}\right)-\int_0^1\frac{\arctan bu}{u^4+u^2+1}\,du\end{align} using $u^4+u^2+1=(u^2+u+1)(u^2-u+1)$ and $b=1/a$. The remaining integral can be represented in terms of dilogarithms, such as in this answer, but I doubt there is a clean result with an arbitrary $b>0$.

In short, this integral does have a closed form, but is unlikely to be "simple"; e.g. case $a=7$.

(As $a\to+\infty$ the term $\int_0^1\frac{\arctan bu}{u^4+u^2+1}\,du$ has contribution $\to0$ so your observation is true.)

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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You could simplify the problem using $$\frac{x^2}{x^4+x^2+1}=\frac {x^2}{(x^2-r)(x^2-s)}=\frac 1{r-s} \left(\frac{r}{x^2-r}-\frac{s}{x^2-s}\right)$$ where $$r=-\frac{1+i \sqrt{3}}{2} \qquad \text{and} \qquad s=-\frac{1-i \sqrt{3}}{2}$$ So, the problem boils down to the computation of $$I(t)=\int_1^\infty \frac{\tan ^{-1}(a x)}{x^2-t} \,dx$$ where $t$ is a complex number.

Surprising or not, there is an antiderivative (a monster found by a CAS).

So, there is a closed form. I saw it !

Claude Leibovici
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