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I'm really stuck on this integral. I want to believe there is a closed form for it but I'm really unable to find it. WFA cannot find one either. I think it may be possible to use Feynman's Trick and reduce it to something easier but I can't find a good parameterisation that gives something workable. So far, I have factored the denominator to be $(x^2+x-1)(x^2+x+1)$ and have tried to parametrise like

$$I(a) = \displaystyle \int_{-\infty}^{\infty} \frac{\ln(a(x^2+1))}{(x^2+x-1)(x^2+x+1)}dx$$

Which looks promising but after differentiating and integrating (via symbolic math software) I get $I'(a)=\frac{\pi}{\sqrt{3}a}$ and I can't do anything to retrieve $I(1)$ since $I(0)$ is not defined. Performing partial fraction decomposition gives

$$\int_{-\infty}^{\infty} \frac{x\ln(x^2+1)+\ln(x^2+1)}{2(x^2+x+1)} - \int_{-\infty}^{\infty} \frac{x\ln(x^2+1)-\ln(x^2+1)}{2(x^2-x+1)}$$

I am unsure where to go from there. Can anyone give me a hint at the paramterisation if that even is the right approach or if a closed form for this integral even exists? WFA says the decimal expansion is 0.743763...

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  • $\begingroup$ Using contour integration and the Residue Theorem should be a good choice here. If you'd be content with such a solution, I can write one up. It looks like the exact form is $-\frac{\pi^2}{6} + \frac{\pi}{\sqrt{3}} \log(2 + \sqrt 3)$. $\endgroup$ Apr 30, 2023 at 0:17
  • $\begingroup$ I was really hoping I could avoid contour integration but it appears I am out of options so, yes, I’ll be happy with that $\endgroup$
    – Anik Patel
    Apr 30, 2023 at 0:20
  • $\begingroup$ A better way to apply Feynmans trick could be ${I(a) = \int_{-\infty}^{\infty}\frac{\log((ax)^2+1)}{x^4+x^2+1}dx}$. This will satisfy ${I(0) = 0}$ $\endgroup$ Apr 30, 2023 at 0:21
  • $\begingroup$ The integral that results from differentiating that is almost completely unworkable $\endgroup$
    – Anik Patel
    Apr 30, 2023 at 0:39
  • $\begingroup$ You don't have the correct factorization of the denominator. $\endgroup$
    – user317176
    Apr 30, 2023 at 2:44

8 Answers 8

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$$I=\int_{-\infty}^{\infty} \frac{\log(x^2+1)}{x^4+x^2+1}\,dx=2\int_{0}^{\infty} \frac{\log(x^2+1)}{x^4+x^2+1}\,dx$$ $$J(a)=\int_{0}^{\infty} \frac{\log(ax^2+1)}{x^4+x^2+1}\,dx$$ $$ J'(a)=\int_{0}^{\infty}\frac{x^2}{\left(x^4+x^2+1\right) \left(a x^2+1\right)}$$ Use the roots of unity and write (shorter) $$\frac{x^2}{\left(x^4+x^2+1\right) \left(a x^2+1\right)}=\frac{x^2}{(x^2-r)(x^2-s) \left(a x^2+1\right)}=$$ $$-\frac{a}{(a r+1) (a s+1) \left(a x^2+1\right)}+\frac{r}{(a r+1) (r-s) \left(x^2-r\right)}-\frac{s}{(a s+1) (r-s) \left(x^2-s\right)}$$ Simple integrals; using the bounds $$J'(a)=\frac {\pi \sqrt 3} 6\,\, \frac{a-\sqrt{3a} +1}{a^2-a+1}$$ $$\int_0^1 \frac{a-\sqrt{3a} +1}{a^2-a+1}\,da=2\int_0^1 \frac {db}{ b^2+b\,\sqrt{3}+1}=\log \left(2+\sqrt{3}\right)-\frac{\pi }{2 \sqrt{3}}$$ Recombining $$I=\frac{\pi }{\sqrt{3}}\log \left(2+\sqrt{3}\right)-\frac{\pi ^2}{6}$$

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  • $\begingroup$ This is the best solution IMO, using completely real methods. $\endgroup$ Sep 10, 2023 at 16:52
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Following Riemann'sPointyNose's suggestion, take $$I(a) := \int_{-\infty}^\infty \frac{\log((a x)^2 + 1)}{x^4 + x^2 + 1} dx,$$ so that the desired integral is $I(1)$. Then, $I(0) = 0$, and differentiating with respect to $a$ gives $$I'(a) = 2 a \int_{-\infty}^\infty \frac{x^2 \,dx}{(a^2 x^2 + 1) (x^4 + x^2 + 1)} ,$$ which is rational. One could decompose the integrand using partial fractions and integrate term-by-term. This computation is straightforward but tedious; it can at least be simplified some by exploiting the evenness of the integrand to reduce to a determination of $3$ coefficients instead of $6$.

Instead, we'll set up and evaluate an appropriate contour integral and apply the Residue Theorem. Denote $g(z) := \frac{2a z^2 \,dz}{(a^2 z^2 + 1) (z^4 + z^2 + 1)}$ and denote by $\Gamma_R$ the contour in the diagram, where $R > \operatorname{max}\left(1, \frac{1}{a}\right)$, and $\textrm{I}$ and $\textrm{II}$ the indicated arcs.

enter image description here

First, $$\lim_{R \to \infty} \int_\textrm{I} g(z) \,dz = 2 a \int_{-\infty}^\infty \frac{x^2 \,dx}{(a^2 x^2 + 1) (x^4 + x^2 + 1)} = I'(a) .$$ Comparing the degrees of the numerator and denominator $g(z)$ gives that $\int_\textrm{II} g(z) \,dz \in O(R^{-3})$; in particular $$\lim_{R \to \infty} \int_\textrm{II} g(z) \,dz = 0 ,$$ so $$I'(a) = \oint_{\Gamma_R} g(z) \,dz .$$ We evaluate the integral with the Residue Theorem. The poles of $g(z)$ inside the contour are at $\frac{i}{a}, e^{\pi i / 3}, e^{2 \pi i / 3}$, and all are simple; the residues of $g$ there are: \begin{align*} \operatorname{Res}\left(g(z); \frac{i}{a}\right) &= \frac{i a^2}{a^4 - a^2 + 1} \\ \operatorname{Res}\left(g(z); e^{\pi i / 3}\right) &= \frac{a}{2(a^4 - a^2 + 1)} \left(-(a^2 - 1) - \frac{1}{\sqrt{3}} (a^2 + 1) i \right)\\ \operatorname{Res}\left(g(z); e^{2 \pi i / 3}\right) &= \frac{a}{2(a^4 - a^2 + 1)} \left(\phantom{-}(a^2 - 1) - \frac{1}{\sqrt{3}} (a^2 + 1) i \right) \end{align*} So, taking the limit as $R \to \infty$ gives $$I'(a) = \oint_{\Gamma_R} g(z) \,dz = 2 \pi i \sum \operatorname{Res}(g(z), z_i) = \frac{2 \pi}{\sqrt{3}} \frac{a}{a^2 + \sqrt{3} a + 1} .$$

Thus, the original integral is \begin{align*} I(1) &= I(0) + \int_0^1 I'(a) \,da\\ &= \frac{2 \pi}{\sqrt{3}} \int_0^1 \frac{a \,da}{a^2 + \sqrt{3} a + 1} \\ &= \left.\frac{\pi}{\sqrt{3}} \log (a^2 + a \sqrt{3} + 1) - 2 \pi \arctan (2 a + \sqrt{3})\right\vert_0^1 \\ &= \boxed{\frac{\pi}{\sqrt{3}} \log(2 + \sqrt{3}) - \frac{\pi^2}{6}} , \end{align*} which agrees with the numerical computation in the question statement. In the last line we used that $\tan \frac{5 \pi}{12} = 2 + \sqrt{3}$, which can be derived, e.g., from the angle sum formula for $\tan$.

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    $\begingroup$ Could you explain how you got $I'(a)$ more? When I tried to do it using Maxima I got a massive expression. Only when asking Maxima to simplify it do I get the expression you obtained. Did you manage to skip this or is it just grueling computation? $\endgroup$
    – Anik Patel
    Apr 30, 2023 at 1:32
  • $\begingroup$ Computing the antiderivative is tedious though straightforward. We can avoid that computation altogether if we're willing to evaluate a contour integral, as it's a straightforward application of the Residue Theorem: The only poles in the upper half-plane are at $e^{\pi i / 3}, e^{2 \pi i / 3}, \frac{i}{a}$. Since the integrand is rational, you don't have to worry about branch cuts like you would if using the Residue Theorem to evaluate the original integral. $\endgroup$ Apr 30, 2023 at 1:43
  • $\begingroup$ If it would be useful to you, I can add to my answer the contour integration to evaluate the integral for $I'(a)$. $\endgroup$ Apr 30, 2023 at 1:47
  • $\begingroup$ Yes, I’ll probably end up accepting that answer since it’s most likely more straightforward and simple than Feynman’s Trick $\endgroup$
    – Anik Patel
    Apr 30, 2023 at 1:56
  • $\begingroup$ I meant, modifying this answer to include the contour integration to evaluate $I'(a) = 2 a \int_{-\infty}^\infty \frac{x^2 \,dx}{(a^2 x^2 + 1) (x^4 + x^2 + 1)}$, which is definitely easier than using a contour integration for the original integrand, but I could write up the latter, too. $\endgroup$ Apr 30, 2023 at 2:18
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Here's a somewhat overkill method.

Let $I$ be the original integral. Note that $\displaystyle I=2\int_{0}^{\infty}\frac{\ln\left(1+x^{2}\right)}{1+x^{2}+x^{4}}dx$. Define $f(z) = \displaystyle \frac{\log\left(1+z^{2}\right)}{1+z^{2}+z^{4}}$ where the principal log's argument is defined as $\Im\log{\left(1+z^2\right)} \in (-\pi,\pi]$. Let $C = [0,R] \cup \Gamma \cup [iR,i+ir] \cup \gamma \cup [i-ir,0]$, for sufficiently large $R$ and sufficiently small $r > 0$, and traverse it counterclockwise. A visual is provided below.

Contour C

From the picture above, let each simple pole be $\displaystyle z_{1}=\frac{1}{2}+\frac{i\sqrt{3}}{2}$, $\displaystyle z_{2}=-\frac{1}{2}+\frac{i\sqrt{3}}{2}$, $\displaystyle z_{3}=-\frac{1}{2}-\frac{i\sqrt{3}}{2}$, and $\displaystyle z_{4}=\frac{1}{2}-\frac{i\sqrt{3}}{2}$. By Cauchy's Residue Theorem, we can write $\displaystyle \oint_C f(z)dz$ as

$$ \begin{align} 2\pi i \operatorname{Res}(f(z), z = z_1) =& \text{ } \left(\int_{0}^{R}+\int_{\Gamma}^{ }+\int_{iR}^{i+ir}+\int_{\gamma}^{ }+\int_{i-ir}^{0}\right)f(z)dz \\ \implies \int_0^R f(z)dz =& \text{ } \Re\operatorname{PV}\int_0^{iR}f(z)dz - \Re\int_{\Gamma}f(z)dz - \Re\int_{\gamma}f(z)dz + \Re 2\pi i \operatorname{Res}(f(z), z = z_1) \\ \implies I =& \text{ } 2\lim_{R \to \infty} \Re \operatorname{PV} \int_0^{iR}f(z)dz - 2\lim_{R \to \infty} \Re \int_{\Gamma} f(z)dz - 2\Re\lim_{r \to 0^+}\int_{\gamma} f(z)dz \\ &+ \Re 4\pi i \operatorname{Res}(f(z), z = z_1). \end{align} $$ From left to right, let each expression be $I_1$, $I_2$, $I_3$, and $I_4$, respectively.


To prove $I_2 = 0$, recall if $z \in \Gamma$, then $|z| = R$. The ML-inequality yields

$$ 0 \leq \left|\int_{\Gamma} f(z)dz\right| \leq M\frac{\pi R}{2}. $$ We can get an upper bound $M$ by this:

$$ \left|\frac{\log\left(1+z^{2}\right)}{1+z^{2}+z^{4}}\right|\le\frac{1+\left|z\right|^{2}+\left|\operatorname{arg}\left(1+z^{2}\right)\right|}{\left|z\right|^{4}+\left|z\right|^{2}-1} \leq \frac{1+R^{2}+\pi}{R^{4}+R^{2}-1}:=M. $$

From there, we just use the Squeeze Theorem and eventually conclude that $I_2 = 0$.


To prove $I_3 = 0$, parameterize $z = i + re^{it}$ where $t \in \displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Then

$$ \begin{align} I_3 &= -2\Re\lim_{r \to 0^+}\int_{-\pi/2}^{\pi/2}f\left(i+re^{it}\right)d\left(i+re^{it}\right) \\ &= -2\Re i\lim_{r \to 0^+} \int_{-\pi/2}^{\pi/2}\frac{r\log\left(1+\left(i+re^{it}\right)^{2}\right)e^{it}}{\left(i+re^{it}\right)^{4}+\left(i+re^{it}\right)^{2}+1}dt \\ &= -2\Re i \int_{-\pi/2}^{\pi/2} e^{it}\lim_{r \to 0^+}\frac{1}{\left(i+re^{it}\right)^{4}+\left(i+re^{it}\right)^{2}+1}\lim_{r \to 0^+}\left(r\log\left(2ire^{it}\right)-\frac{ir^{2}e^{it}}{2}+ O\left(r^{3}\right)\right) \\ &= -2\Re i \int_{-\pi/2}^{\pi/2}\frac{e^{it}}{i^{4}+i^{2}+1}\left(0-0+O\left(0\right)\right)dt \\ &= 0. \\ \end{align} $$


Here is the calculation of $I_4$:

$$ \Re 4\pi i \operatorname{Res}\left(\frac{\log\left(1+z^{2}\right)}{1+z^{2}+z^{4}}, z = z_1\right) = 4\pi \Re i \lim_{z \to z_1}\frac{\left(z-z_{1}\right)\log\left(1+z^{2}\right)}{\left(z-z_{1}\right)\left(z-z_{2}\right)\left(z-z_{3}\right)\left(z-z_{4}\right)} = \frac{\pi^2}{3}. $$


Here is the calculation of $I_1$:

$$ \begin{align} I_1 &= 2\lim_{R \to \infty} \Re \operatorname{PV} \int_0^{iR} \frac{\log\left(1+z^{2}\right)}{1+z^{2}+z^{4}}dz \\ &= 2\Re\lim_{R \to \infty}\lim_{r \to 0^+}\int_{0}^{i-ir}\frac{\log\left(1+z^{2}\right)}{1+z^{2}+z^{4}}dz + 2\Re\lim_{R \to \infty}\lim_{r \to 0^+}\int_{i+ir}^{iR}\frac{\log\left(1+z^{2}\right)}{1+z^{2}+z^{4}}dz \\ &= 2\Re\lim_{r \to 0^+}\int_{0}^{1-r}\frac{\log\left(1+\left(ix\right)^{2}\right)}{1+\left(ix\right)^{2}+\left(ix\right)^{4}}d(ix) + 2\Re\lim_{R \to \infty}\lim_{r \to 0^+}\int_{1+r}^{R}\frac{\log\left(1+\left(ix\right)^{2}\right)}{1+\left(ix\right)^{2}+\left(ix\right)^{4}}d(ix) \\ &= 2\Re i\lim_{r \to 0^+}\int_{0}^{1-r}\frac{\log\left(1-x^{2}\right)}{1-x^{2}+x^{4}}dx + 2\Re i\lim_{R \to \infty}\lim_{r \to 0^+}\int_{1+r}^{R}\frac{\log\left(1-x^{2}\right)}{1-x^{2}+x^{4}}dx \\ &= 2 \cdot 0 + 2\Re i\lim_{R \to \infty}\lim_{r \to 0^+}\int_{1+r}^{R}\frac{\log\left|1-x^{2}\right|}{1-x^{2}+x^{4}}dx - 2\Re \lim_{R \to \infty}\lim_{r \to 0^+}\int_{1+r}^{R}\frac{\operatorname{arg}\left(1-x^2\right)}{1-x^{2}+x^{4}}dx \\ &= 2\cdot 0 + 2 \cdot 0 -2\pi\int_{1}^{\infty}\frac{dx}{1-x^{2}+x^{4}}. \end{align} $$

To avoid divergent integrals and messy bounds, consider the following calculation:

$$ \begin{align} J :=& \int_{ }^{ }\frac{dx}{x^{4}-x^{2}+1} \\ =& \int_{ }^{ }\frac{dx}{\left(x^{2}-\sqrt{3}x+1\right)\left(x^{2}+\sqrt{3}x+1\right)} \\ =& \text{ } \frac{1}{2\sqrt{3}}\int_{ }^{ }\frac{x+\sqrt{3}}{x^{2}+\sqrt{3}x+1}dx-\frac{1}{2\sqrt{3}}\int_{ }^{ }\frac{x-\sqrt{3}}{x^{2}-\sqrt{3}x+1}dx \\ =& \text{ } \frac{1}{2\sqrt{3}}\left(\frac{1}{2}\int_{ }^{ }\frac{2x+\sqrt{3}}{x^{2}+\sqrt{3}x+1}dx+\frac{\sqrt{3}}{2}\int_{ }^{ }\frac{dx}{x^{2}+\sqrt{3}x+1}\right) \\ &-\frac{1}{2\sqrt{3}}\left(\frac{1}{2}\int_{ }^{ }\frac{2x-\sqrt{3}}{x^{2}-\sqrt{3}x+1}dx-\frac{\sqrt{3}}{2}\int_{ }^{ }\frac{dx}{x^{2}-\sqrt{3}x+1}\right) \\ =& \text{ } \frac{1}{2\sqrt{3}}\left(\frac{1}{2}\int_{ }^{ }\frac{d\left(x^{2}+\sqrt{3}x+1\right)}{x^{2}+\sqrt{3}x+1}+\frac{\sqrt{3}}{2}\int_{ }^{ }\frac{dx}{\left(x+\frac{3}{2}\right)^{2}+\frac{1}{4}}\right) \\ &-\frac{1}{2\sqrt{3}}\left(\frac{1}{2}\int_{ }^{ }\frac{d\left(x^{2}-\sqrt{3}x+1\right)}{x^{2}-\sqrt{3}x+1}-\frac{\sqrt{3}}{2}\int_{ }^{ }\frac{dx}{\left(x-\frac{3}{2}\right)^{2}+\frac{1}{4}}\right) \\ =& \text{ } \frac{1}{2\sqrt{3}}\left(\frac{1}{2}\ln\left(x^{2}+\sqrt{3}x+1\right)+\sqrt{3}\arctan\left(2x+\sqrt{3}\right)\right) \\ &-\frac{1}{2\sqrt{3}}\left(\frac{1}{2}\ln\left(x^{2}-\sqrt{3}x+1\right)-\sqrt{3}\arctan\left(2x-\sqrt{3}\right)\right)+C. \\ \end{align} $$

Since $\displaystyle \frac{1}{x^{4}-x^{2}+1}$ is improperly integrable on $[1,\infty)$, we use FTC and get

$$ \begin{align} I_1 &= -2\pi \Big[\frac{1}{4\sqrt{3}}\ln\left(\frac{x^{2}+\sqrt{3}x+1}{x^{2}-\sqrt{3}x+1}\right)+\frac{1}{2}\arctan\left(2x+\sqrt{3}\right)+\frac{1}{2}\arctan\left(2x-\sqrt{3}\right)\Big]_1^{\infty} \\ &= \frac{\pi}{2\sqrt{3}}\ln\left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right)-\frac{\pi^{2}}{2}. \\ \end{align} $$


We conclude with

$$ I= \frac{\pi}{2\sqrt{3}}\ln\left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right)-\frac{\pi^{2}}{2} - 0 - 0 + \frac{\pi^{2}}{3} = \frac{\pi}{\sqrt{3}}\ln\left(2+\sqrt{3}\right)-\frac{\pi^{2}}{6}. $$

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Utilize $\int_0^\infty \frac{\ln(x^2+1)}{x^2+a^2}dx=\frac\pi a\ln(1+a)$ to integrate \begin{align} &\int_{-\infty}^{\infty} \frac{\ln(x^2+1)}{x^4+x^2+1}\ dx\\ = &\int_{0}^{\infty} \frac{2\ln(x^2+1)}{(x^2+e^{-\frac{i\pi}3})(x^2+e^{\frac{i\pi}3})}\ dx =\frac4{\sqrt3}\ \Im \int _0^\infty \frac{\ln(x^2+1)}{x^2+e^{-\frac{i\pi}3}}\ dx\\ =& \ \frac4{\sqrt3}\ \Im \frac{\pi}{e^{-\frac{i\pi}6}}\ln (1+ e^{-\frac{i\pi}6}) = \frac{\pi }{\sqrt{3}}\ln \left(2+\sqrt{3}\right)-\frac{\pi ^2}{6} \end{align}

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Here's an approach just using contour integration and the Residue Theorem.

Denote the integrand by $$g(z) := \frac{\log(1 + z^2)}{z^4 + z^2 + 1}$$ and the contour indicated in the diagram by $\Gamma_{\epsilon, R}$, $0 < \epsilon < \frac{\sqrt 3 - 1}{\sqrt 2}$, $1 < R$; we use the standard choice of branch cut for logarithm, so that $\arg z = \Im z \in (-\pi, \pi)$. The wavy red curve indicates the part of the consequent branch cut for $\log(1 + z^2)$ in the upper half-plane.

The contour of integration

We evaluate $$\oint_{\Gamma_{\epsilon, R}} g(z) \,dz$$ in two ways: using direct integration and via the Residue Theorem. We denote by $\textrm{I}, \ldots, \textrm{VI}$ the arcs so labeled in the diagram.

By design, $$\lim_{R \to \infty} \int_\textrm{I} g(z) \,dz = \int_{-\infty}^\infty \frac{\log(x^2 + 1) \,dx}{x^4 + x^2 + 1}$$ is the integral whose value we want to compute.

Now, $$\left\vert\int_\textrm{II} g(z) \,dz\right\vert \leq \int_\textrm{II} |g(z)| \,|dz| \leq \frac{\pi}{2} \cdot \frac{\log(R^2 + 1)}{R^4 - R^2 - 1},$$ which (by, e.g., l'Hôpital's Rule) approaches to $0$ as $R \to \infty$. So, $\lim_{R \to \infty} \int_\textrm{II} g(z) \,dz = 0$. For the same reason, $\lim_{R \to \infty} \int_\textrm{VI} g(z) \,dz = 0$.

We parameterize the integral over $\textrm{V}$ by $z = \epsilon - i t$, $dz = i \,dt$, where $t$ varies from $1$ to $R$: $$\int_\textrm{V} g(z) \,dz = \int_1^R \frac{\log[(-\epsilon + it)^2 + 1]}{(-\epsilon + i t)^4 + (\epsilon + i t)^2 + 1},$$ and in the limit as $\epsilon \searrow 0$, this is $$\lim_{\epsilon \searrow 0} \int_\textrm{V} g(z) \,dz = i \int_1^R \frac{\log(t^2 - 1) - \pi i}{t^4 - t^2 + 1}\,dt .$$ Parameterizing the integral over $\textrm{III}$ by $z = \epsilon + i t$, $dz = i \,dt$ similarly gives $$\lim_{\epsilon \searrow 0} \int_\textrm{III} g(z) \,dz = -i \int_1^R \frac{\log(t^2 - 1) + \pi i}{t^4 - t^2 + 1}\,dt,$$ so $$\lim_{\epsilon \searrow 0} \left[\int_\textrm{III} g(z) \,dz + \int_\textrm{V} g(z) \,dz\right] = 2 \pi \int_1^R \frac{dt}{t^4 - t^2 + 1} ,$$ and in the limit as $R \to \infty$, \begin{multline*} \lim_{\epsilon \searrow 0, R \to \infty} \left[\int_\textrm{III} g(z) \,dz + \int_\textrm{V} g(z) \,dz\right] \\ = 2 \pi \int_1^\infty \frac{dt}{t^4 - t^2 + 1} = 2 \pi \left(\frac{\pi}{4} - \frac{\log(2 + \sqrt{3})}{2 \sqrt{3}}\right) = \frac{\pi^2}{2} - \frac{\pi}{\sqrt 3} \log (2 + \sqrt 3) .\end{multline*}

Finally, parameterizing the half-circle $\textrm{IV}$ by $z = i + \epsilon e^{i \theta}$, $dz = i \epsilon e^{i \theta} \,d\theta$ gives $$\int_\textrm{IV} g(z) \,dz = i \int_0^{-\pi} \frac{\epsilon \log [1 + (i + \epsilon e^{i \theta})^2] e^{i \theta} \,d\theta}{(i + \epsilon e^{i \theta})^4 + (i + \epsilon e^{i \theta})^2 + 1}.$$ We have the bound $$\left\vert\int_\textrm{IV} g(z) \,dz\right\vert \leq \int_\textrm{IV} |g(z)| \,|dz| \leq \frac{\epsilon (2 + 2 \pi + |\log \epsilon| + \epsilon)}{1 - 12 \epsilon} \cdot \pi\epsilon \in O(\epsilon^2 \log \epsilon),$$ so by l'Hôpital's Rule $$\lim_{\epsilon \searrow 0} \int_\textrm{IV} g(z) \, dz = 0 .$$

In summary we have $$\phantom{(\ast)} \qquad \oint_{\Gamma_{\epsilon, R}} g(z) \,dz = \int_{-\infty}^\infty \frac{\log(x^2 + 1) \,dx}{x^4 + x^2 + 1} + \left[\frac{\pi^2}{2} - \frac{\pi}{\sqrt 3} \log (2 + \sqrt 3)\right]. \qquad (\ast)$$

On the other hand, the only poles of $g$ inside the contour are the simple poles at $e^{\pi i / 3}, e^{2 \pi / 3}$, and computing directly gives $$ \operatorname{Res}\left(g; e^{\pi i / 3}\right) = \frac{\pi}{12 \sqrt{3}} - \frac{\pi}{12} i \qquad \textrm{and} \qquad \operatorname{Res}\left(g; e^{2 \pi i / 3}\right) = -\frac{\pi}{12 \sqrt{3}} - \frac{\pi}{12} i ,$$ so the Residue Theorem yields $$ \phantom{(\ast\ast)} \qquad \oint_{\Gamma_{\epsilon, R}} g(z) \,dz = 2 \pi i \left[\operatorname{Res}\left(g; e^{\pi i / 3}\right) + \operatorname{Res}\left(g; e^{2\pi i / 3}\right)\right] = 2 \pi i \left(-\frac{\pi i}{6}\right) = \frac{\pi^2}{3}. \qquad (\ast\ast) $$ Combining $(\ast)$ and $(\ast\ast)$ gives the desired value: $$\int_{-\infty}^\infty \frac{\log(x^2 + 1) \,dx}{x^4 + x^2 + 1} = \boxed{\frac{\pi}{\sqrt 3} \log (2 + \sqrt 3) - \frac{\pi^2}{6}} .$$

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By using the upper-semi-circle contour and Residue theorem, I tried to do a simplified computation, fitting the given answer: $$\begin{align} \frac{I}{2}=\int_0^{\infty}\frac{\ln(x^2+1)}{x^4+x^2+1}dx&=\Re\int_C\frac{\ln(z+i)}{z^4+z^2+1}dz\\ &=\Re\,\,2\pi i(Res_{z=\large{e^{\pi i/3}}}+Res_{z=e^{2\pi i/3}})\\ &=\Re\,\,2\pi i\left(\frac{\ln(e^{\pi i/3}+i)}{2\sqrt3\,i\,e^{\pi i/3}}+\frac{\ln(e^{2\pi i/3}+i)}{-2\sqrt3\,i\,e^{2\pi i/3}}\right)\\ &=\frac\pi{\sqrt3}\Re\,\,e^{-\pi i/3}\ln(\sqrt{2+\sqrt3}\,e^{5\pi i/12})-e^{-2\pi i/3}\ln(\sqrt{2+\sqrt3}\,e^{7\pi i/12})\\ &=\frac\pi{\sqrt3}(\frac14\ln(2+\sqrt3)+\frac{5\sqrt3\pi}{24}+\frac14\ln(2+\sqrt3)-\frac{7\sqrt3\pi}{24})\\ &=\frac{\pi}{2\sqrt3}\ln(2+\sqrt3)-\frac{\pi^2}{12} \end{align}$$

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First, we will manipulate our integral into a form we can easily evaluate using double integrals, a replacement for Feynman's trick. $$I=2\int_0^\infty\frac{\ln(x^2+1)}{x^4+x^2+1}dx=\int_0^\infty\frac{\ln(x^4+2x^2+1)}{x^4+x^2+1}dx$$ $$=\int_0^\infty\left(\frac{\ln(x^2+y(x^4+x^2+1))}{x^4+x^2+1}\Bigg|_{y=0}^{y=1} +\frac{\ln(x^2)}{x^4+x^2+1}\right)dx$$ $$=\int_0^\infty\int_0^1\frac{1}{yx^4+(y+1)x^2+y}dy\space dx+2\int_0^\infty\frac{\ln x}{x^4+x^2+1}dx=J+2K$$ Our integral has split into 2 more integrals. We will name and evaluate the first integral J using a formula to cut out the tedium. $$J=\int_0^\infty\int_0^1\frac{1}{yx^4+(y+1)x^2+y}dy\space dx$$ $$\int_0^\infty\frac{dx}{ax^4+bx^2+c}=\frac{\pi}{2\sqrt c\sqrt{b+2\sqrt{ac}}}$$ $$J=\pi\int_0^1\frac{ dy}{2\sqrt{3y+1}\sqrt y}=\frac{\pi}{\sqrt3}\log(2+\sqrt3)$$ Our second integral $K$ seems tricky, but the denominator allows for tricks. We'll start by splitting the integral at $x=1$ and inverting to form an integral in $[0,1]$. This allows us to use geometric series. We can use the Basel sum to evaluate onwards. $$K=\int_0^\infty\frac{\ln x}{x^4+x^2+1}dx=\int_0^1\frac{(1-x^2)\ln x}{x^4+x^2+1}dx=\int_0^1\frac{(1-x^2)^2\ln x}{1-x^6}dx$$$$=\int_0^1(1-2x^2+x^4)\ln x \space \sum_{n=0}^\infty x^{6n}dx$$$$=\sum_{n=0}^\infty\int_0^1x^{6n}\ln x\space dx-2\sum_{n=0}^\infty\int_0^1x^{6n+2}\ln x\space dx+\sum_{n=0}^\infty\int_0^1x^{6n+4}\ln x\space dx$$$$=-\sum_{n=0}^\infty\frac{1}{(6n+1)^2}+2\sum_{n=0}^\infty\frac{1}{(6n+3)^2}-\sum_{n=0}^\infty\frac{1}{(6n+5)^2}$$$$=-\sum_{n=0}^\infty\frac{1}{(6n+1)^2}-\sum_{n=0}^\infty\frac{1}{(6n+3)^2}-\sum_{n=0}^\infty\frac{1}{(6n+5)^2}+3\sum_{n=0}^\infty\frac{1}{(6n+3)^2}$$$$=-\sum_{n=1}^\infty\frac{1}{(2n+1)^2}+\frac{1}{3}\sum_{n=0}^\infty\frac{1}{(2n+1)^2}$$ $$K=-\frac{2}{3}\frac{\pi^2}{8}=-\frac{\pi^2}{12}$$ Finally $$I=\left(\frac{\pi}{\sqrt3}\log(2+\sqrt3)\right)+2\left(-\frac{\pi^2}{12}\right)=\frac{\pi}{\sqrt3}\log(2+\sqrt3)-\frac{\pi^2}{6}$$

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Too long for comment:

$$\begin{align*} I &= \int_{-\infty}^\infty \frac{\ln\left(x^2+1\right)}{x^4+x^2+1} \, dx \\ &= 2 \int_0^\infty \frac{\ln\left(x^2+1\right)}{x^4+x^2+1} \, dx \tag1 \\ &= 2 \left\{\int_0^1 + \int_1^\infty\right\} \frac{\ln\left(x^2+1\right)}{x^4+x^2+1} \, dx \\ &= 2 \int_0^1 \frac{\ln\left(x^2+1\right)}{x^4+x^2+1} \, dx + 2 \int_0^1 \frac{\ln\left(\frac1{x^2}+1\right)}{\frac1{x^4}+\frac1{x^2}+1} \, \frac{dx}{x^2} \tag2 \\ &= 2 \int_0^1 \frac{\ln\left(x^2+1\right)}{x^4+x^2+1} \, dx + 2 \int_0^1 \frac{x^2 \left(\ln\left(x^2+1\right) - \ln\left(x^2\right)\right)}{x^4+x^2+1} \, dx \\ &= 2 \underbrace{\int_0^1 \frac{x^2+1}{x^4+x^2+1} \ln\left(x^2+1\right) \, dx}_J - 4 \underbrace{\int_0^1 \frac{x^2}{x^4+x^2+1} \ln(x) \, dx}_K \\[2ex] \hline J &= \int_0^1 \frac{x^2+1}{x^4+x^2+1} \left[\int_0^1 \frac{x^2}{1+x^2y} \, dy\right] \, dx \tag3 \\ &= \frac12 \int_0^1 \frac1{y^2-y+1} \left[\int_0^1 \left(\frac{x+y-1}{x^2-x+1}-\frac{x-y+1}{x^2+x+1} - \frac{2(y-1)}{1+yx^2}\right) \, dx \right] \, dy \\ &= \frac\pi{4\sqrt3} \int_0^1 \frac{2y-1}{y^2-y+1} \, dy - \frac{\ln(3)}4 \int_0^1 \frac{dy}{y^2-y+1} - \int_0^1 \frac{y-1}{y^2-y+1} \cdot \frac{\arctan\left(\sqrt y\right)}{\sqrt y} \, dy \\ &= -\frac{\pi\ln(3)}{6\sqrt3} - 2 \underbrace{\int_0^1 \frac{y^2-1}{y^4-y^2+1} \arctan(y) \, dy}_{L} \tag4 \\[2ex] \hline K &= \int_0^1 \frac{x^4-x^2}{x^6-1} \ln(x) \, dx \tag5 \\ &= \sum_{n=0}^\infty \int_0^1 \left(x^{6n+2} - x^{6n+4}\right) \ln(x) \, dx \tag6 \\ &= \sum_{n=0}^\infty \left(\frac1{(6n+5)^2}-\frac1{(6n+3)^2}\right) \\ &= \frac{\psi^{(1)}\left(\frac56\right) - \psi^{(1)}\left(\frac12\right)}{36} \tag7 \end{align*}$$


  • $(1)$ : symmetry
  • $(2)$ : substitute $x\mapsto\dfrac1x$ in the latter integral
  • $(3)$ : integral definition of $\ln\left(x^2+1\right)$
  • $(4)$ : substitute $y\mapsto y^2$
  • $(5)$ : introduce a factor of $x^2-1$
  • $(6)$ : exploit the series expansion of $\dfrac1{1-x}$
  • $(7)$ : trigamma function, or first derivative of digamma

I'm currently stuck on $L$. Partial evaluation follows with substituting $y=\tan(z)$, then replacing $z\mapsto\dfrac z2$ and integrating by parts.

$$\begin{align*} L &= \int_0^1 \frac{y^2-1}{y^4-y^2+1} \arctan(y) \, dy \\ &= \int_0^{\tfrac\pi4} \frac{\tan^4z-1}{\tan^4z-\tan^2z+1} z \sec^2(z) \, dz \\ &= - \int_0^{\tfrac\pi2} \frac{2z\cos(z)}{3\cos(2z)+5} \, dz \\ &= \frac\pi{4\sqrt3} \ln(2+\sqrt3) + \frac1{4\sqrt3} \int_0^{\tfrac\pi2} \ln\left(\frac{2+\sqrt3\sin(w)}{2-\sqrt3\sin(w)}\right) \, dw \end{align*}$$

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