Expanding on an earlier question, I am trying to prove that
$$I = \int_0^\infty \frac{\log(x) \arctan(x)}{1 + x^2} \, dx = \frac78 \zeta(3)$$
using complex analysis but am running into some issues.
My attempt: Let $f(z) = \dfrac{\log^2(z) \arctan(z)}{a^2 + z^2}$ (where $\log^2(z) \equiv (\log(z))^2$) with $0<a<1$, and let
$$I(a) = \int_0^\infty \frac{\log(x) \arctan(x)}{a^2 + x^2} \, dx \quad .$$
Take $\mathcal C$ to be the same positively-oriented contour (redrawn here) with branch cuts along $[0,\infty)$ and $\pm[i,i\infty)$. The larger arc has radius $R>1$ and the smaller ones each have radii $\varepsilon$. The dotted circles indicate the poles at $z=\pm ia$.
I parameterized the banks to either side of the branch cuts (abusing notation) by
- $A$ : $z = x+i\varepsilon$, $x\in[\varepsilon, R]$
- $A'$ : $z = x - i\varepsilon$, $x\in[R,\varepsilon]$
- $B$ : $z=-\varepsilon+ix$, $x\in[1+\varepsilon, R]$
- $B'$ : $z=\varepsilon+ix$, $x\in[R,1+\varepsilon]$
- $C$ : $z=\varepsilon-ix$, $x\in[1+\varepsilon, R]$
- $C'$ : $z=-\varepsilon-ix$, $x\in[R,1+\varepsilon]$
As each bank approaches their respective branch cut, I believe we have
$$\begin{array}{c|ccc} A & f(x+i\varepsilon) & \to & \dfrac1{a^2+x^2} \log^2(x) \arctan(x) \\ A' & f(x - i\varepsilon) & \to & \dfrac1{a^2+x^2} \left(\log^2(x) + 4\pi i\log(x) - 4\pi^2\right) \arctan(x) \\ \hline B & f(-\varepsilon + ix) & \to & \dfrac1{a^2-x^2} \left(\log^2(x) + i\pi \log(x) - \dfrac{\pi^2}4\right) \left(-\dfrac i2 \log\left|\dfrac{1-x}{1+x}\right| - \dfrac\pi2\right) \\ B' & f(\varepsilon + ix) & \to & \dfrac1{a^2-x^2} \left(\log^2(x) + i\pi \log(x) - \dfrac{\pi^2}4\right) \left(-\dfrac i2 \log\left|\dfrac{1-x}{1+x}\right| + \dfrac\pi2\right) \\ \hline C & f(\varepsilon - ix) & \to & \dfrac1{a^2-x^2} \left(\log^2(x) - i\pi \log(x) - \dfrac{\pi^2}4\right) \left(-\dfrac i2 \log\left|\dfrac{1+x}{1-x}\right| + \dfrac\pi2\right) \\ C' & f(-\varepsilon - ix) & \to & \dfrac1{a^2-x^2} \left(\log^2(x) - i\pi \log(x) - \dfrac{\pi^2}4\right) \left(-\dfrac i2 \log\left|\dfrac{1+x}{1-x}\right| - \dfrac\pi2\right) \end{array}$$
I'm not sure that I fully grasp how to jump across branches so this might be where everything goes wrong.
Edit: The correction to make here (thanks to @Svyatoslav !) is along the banks $C$ and $C'$, for which we should have
$$C : f(\varepsilon - ix) \to \dfrac1{a^2-x^2} \left(\log^2(x)+\color{red}{3}i\pi\log(x)-\frac{\color{red}{9}\pi^2}4\right) \left(-\dfrac i2 \log\left|\dfrac{1+x}{1-x}\right| + \dfrac\pi2\right) \\ C' : f(-\varepsilon - ix) \to \dfrac1{a^2-x^2} \left(\log^2(x)+\color{red}{3}i\pi\log(x)-\frac{\color{red}{9}\pi^2}4\right) \left(-\dfrac i2 \log\left|\dfrac{1+x}{1-x}\right| - \dfrac\pi2\right)$$
The residues at the poles are
$$\begin{align*} \underset{z=ia}{\operatorname{Res}} f(z) &= \frac{\log^2(ia) \arctan(ia)}{2ia} \\[1ex] &= \frac1{2ia} \left(\log^2(a) + i\pi \log(a) - \frac{\pi^2}4\right) \left(-\frac i2 \log\left|\frac{1-a}{1+a}\right|+\frac12\arg\left(\frac{1-a}{1+a}\right)\right) \\[1ex] &= \frac1{2ia} \left(\log^2(a) + i\pi\log(a) - \frac{\pi^2}4\right) \arctan(ia) \\[3ex] \underset{z=-ia}{\operatorname{Res}} f(z) &= \frac{\log^2(-ia) \arctan(-ia)}{-2ia} \\[1ex] &= -\frac1{2ia} \left(\log^2(a) + 3i\pi \log(a) - \frac{9\pi^2}4\right) \left(-\frac i2 \log\left|\frac{1+a}{1-a}\right|+\frac12\arg\left(\frac{1+a}{1-a}\right)\right) \\[1ex] &= \frac1{2ia} \left(\log^2(a) + 3i\pi \log(a) - \frac{9\pi^2}4\right) \left(-\frac i2 \log\left|\frac{1-a}{1+a}\right| + \frac12 \arg\left(\frac{1-a}{1+a}\right)\right) \tag{*} \\[1ex] &= \frac1{2ia} \left(\log^2(a) + \color{red}{3}i\pi \log(a) - \frac{\color{red}{9}\pi^2}4\right) \arctan\left(ia\right) \end{align*}$$
- $(*)$ : $\arg(x)=\arg\left(\frac1x\right)$ for all real $x\neq0$
As $\varepsilon\to0$ and $R\to\infty$,
$$\begin{align*} \left\{\int_A + \int_{A'}\right\} f(z) \, dz &= \int_\varepsilon^R (f(x+i\varepsilon)-f(x-i\varepsilon)) \, dx \\[1ex] &\to -4\pi i \, I(a) + 4\pi^2 \underbrace{\int_0^\infty \frac{\arctan(x)}{a^2+x^2} \, dx}_{J(a)} \\[2ex] \left\{\int_B + \int_{B'}\right\} f(z) \, dz &= i \int_{1+\varepsilon}^R (f(-\varepsilon+ix)-f(\varepsilon+ix)) \, dx \\[1ex] &\to -i\pi \underbrace{\int_1^\infty \frac{\log^2(x)}{a^2-x^2} \, dx}_{K(a)} + \pi^2 \underbrace{\int_1^\infty \frac{\log(x)}{a^2-x^2} \, dx}_{L(a)} + \frac{i\pi^3}4 \underbrace{\int_1^\infty \frac{dx}{a^2-x^2} \, dx}_{M(a)} \\[2ex] \left\{\int_C + \int_{C'}\right\} f(z) \, dz &= -i \int_{1+\varepsilon}^R (f(-\varepsilon-ix) - f(\varepsilon-ix)) \, dx \\[1ex] &\to -i\pi \, K(a) \color{red}{+ 3}\pi^2 \, L(a) + \frac{\color{red}{9}i\pi^3}4 \, M(a) \end{align*}$$
and the integrals along the circular arcs vanish.
Let $a\to1^-$. We already know $J(1)=\dfrac{\pi^2}8$, and it's easy to show that $L(1)=-\dfrac{\pi^2}8$ (correct) and $K(1)=-\dfrac78\zeta(3)$ (not correct!). The last integral is
$$M(a) = \int_1^\infty \frac{dx}{a^2-x^2} = \frac1{2a} \log\left(\frac{1-a}{1+a}\right) \quad .$$
Now by the residue theorem,
$$\begin{align*} \oint_{\mathcal C} f(z) \, dz &= 2\pi i \cdot -\frac{2\log^2(a) + 4i\pi \log(a) - \frac{5\pi^2}2}{4a} \log\left(\frac{1-a}{1+a}\right) \\[1ex] &= -4\pi i \, I(a) + 4\pi^2 \, J(a) - 2i\pi \, K(a) + 4\pi^2 \, L(a) + \frac{5i\pi^3}2 M(a) \\[3ex] \implies I(a) &= \frac{4\log^2(a) + 8i\pi \log(a) - 5\pi^2}{16a} \log\left(\frac{1-a}{1+a}\right) \\[1ex] &\qquad - \pi i \, J(a) - \frac12 K(a) - \pi i \, L(a) + \frac{5\pi^2}8 M(a) \\[3ex] \implies I &= \frac{5\pi^2}{16} \lim_{a\to1^-} \underbrace{\left(2\,M(a) - \frac1a \log\left(\frac{1-a}{1+a}\right)\right)}_{=0} - \frac{i\pi^3}8 + \frac7{16}\zeta(3) + \frac{i\pi^3}8 \\[1ex] &= \frac7{16}\zeta(3) \neq \frac78 \zeta(3) \end{align*}$$
All that's left is to figure out where that extra factor of $\frac12$ came from ...
Where is the mistake?
There was this very similar integral with $1+x$ in place of $1+x^2$, but with no obvious use of contour integration among its answers.
I also found $I$'s value appearing in another question and slightly less conspicuously in this integral, which leads to a nice identity that helps verify the result.
Substituting $e^{-x}\mapsto x$ and integrating by parts yields
$$\int_0^\infty \arctan^2(e^{-x}) \, dx = \int_0^1 \frac{\arctan^2(x)}x \, dx = -2 \int_0^1 \frac{\log(x) \arctan(x)}{1+x^2} \, dx$$
and by subsequently replacing $x\mapsto\frac1x$ one gets
$$\frac\pi2 G = \left\{\int_1^\infty -\int_0^1\right\} \frac{\log(x) \arctan(x)}{1+x^2} \, dx$$
Now
$$\begin{align*} \int_0^\infty \arctan^2(e^{-x}) \, dx &= \frac\pi2 G - \frac78 \zeta(3) \\[1ex] \implies \int_0^1 \frac{\log(x) \arctan(x)}{1+x^2} \, dx &= -\frac\pi4 G + \frac7{16} \zeta(3) \\[1ex] \implies \int_1^\infty \frac{\log(x) \arctan(x)}{1+x^2} \, dx &= \frac\pi4 G + \frac7{16} \zeta(3) \end{align*}$$
and the result follows.
