# Using complex analysis, prove $\int\limits_0^\infty \frac{\log(x) \arctan(x)}{1 + x^2} \, dx = \frac78 \zeta(3)$

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, 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.

• Is it $\log$ or $\log^2$?
– dfnu
Nov 27, 2022 at 9:48
• @dfnu It's $\log$ in the integral $I$, but I use $\log^2$ in $f(z)$ in the complex plane. It's a fairly well-known technique. See e.g. here Nov 27, 2022 at 10:00
• Just a quick observation which could help: $\log x$ is define by a cut along the positive part of axis $X$, and you go counterclockwise, making finally a full turn. At B, B' you get $\ln (xe^{\pi i/2})=\ln x+\pi i/2$; at C,C' $\ln (xe^{3\pi i/2})=\ln x+3\pi i/2$ Nov 27, 2022 at 10:03
• Thanks @Svyatoslav , that seems to be exact what I was missing. My solution is still incorrect but I'm getting much closer :) Nov 27, 2022 at 21:15

After a bit of investigation, I found the last mistake in the evaluation of $$K(1)$$:
\begin{align*} K(a) &= \int_1^\infty \frac{\log^2(x)}{a^2-x^2} \, dx \\[1ex] &= -\frac1a \int_0^a \frac{\log^2(\frac xa)}{1 - x^2} \, dx \tag{1} \\[3ex] \implies K(1) &= -\int_0^1 \frac{\log^2(x)}{1-x^2} \, dx \\[1ex] &= - \sum_{n=0}^\infty \int_0^1 x^{2n} \log^2(x) \, dx \tag{2} \\[1ex] &= \sum_{n=0}^\infty \frac1{2n+1} \int_0^1 x^{2n} \log(x) \, dx \tag{3} \\[1ex] &= - \sum_{n=0}^\infty \frac{\color{red}2}{(2n+1)^2} \int_0^1 x^{2n} \, dx \tag{3} \\[1ex] &= - \sum_{n=0}^\infty \frac2{(2n+1)^3} \\ &= -\frac74 \zeta(3) \end{align*}
• $$(1)$$ : $$x\mapsto\frac ax$$
• $$(2)$$ : exploit Maclaurin series for $$\frac1{1-x}$$
• $$(3)$$ : integrate by parts