Evaluating very nasty logarithmic integrals: Part III

This is part 3 of our series on very nasty logarithmic integrals. Please have a look at part 1 and part 2 before reading this post.

Integral #5

The first integral that we will evaluate in this post is the following: $$ I_1 = \int_0^1 \frac{\log^2(x) \arctan(x)}{1+x^2}dx $$ Of course, one can use brute force methods to find a closed form anti-derivative in terms of polylogarithms. Instead, a more elegant solution is possible by contour integration.

We’ll integrate the principal branch of $f(z) = \frac{\arctan(z)}{1+z^2}\left(\text{arctanh}^2(z) + \frac{\pi^2}{16} \right)$ around the following contour:

where

• $\gamma_{3,\epsilon}$ is an arc parameterized by $e^{it}$, where $\arctan\left(\frac{\epsilon \sqrt{4-\epsilon^2}}{2-\epsilon^2} \right)\leq t \leq \frac{\pi}{2} – \arctan\left(\frac{\epsilon \sqrt{4-\epsilon^2}}{2-\epsilon^2} \right)$.
• $\gamma_{2,\epsilon}$ and $\gamma_{4,\epsilon}$ are circular indents of radius $\epsilon$ around the branch points at $z=1$ and $z=i$, respectively.
• $\gamma_{1,\epsilon}$ is a straight line joining $0$ and $1-\epsilon$.
• $\gamma_{4,\epsilon}$ is a straight line joining $(1-\epsilon)i$ and $0$.

Note that $f$ is analytic on $|z| < 1$. It is easy to see that $$ \begin{aligned} \lim_{\epsilon\to 0^+} \int_{\gamma_{2,\epsilon}}f(z)\; dz &= 0 \\ \lim_{\epsilon\to 0^+} \int_{\gamma_{4,\epsilon}}f(z)\; dz &= 0 \end{aligned} $$ On $\gamma_{1,\epsilon}$, we have $$ \begin{aligned} \lim_{\epsilon\to 0^+} \int_{\gamma_{1,\epsilon}} f(z)\; dz &= \int_0^1\frac{\arctan(x)}{1+x^2}\left(\text{arctanh}^2(x) + \frac{\pi^2}{16} \right)dx \\ &= \int_0^1\frac{\arctan(x) \text{arctanh}^2(x) }{1+x^2}dx + \frac{\pi^2}{16} \frac{\arctan^2(x)}{2}\Big|_0^1 \\ &= \frac{1}{4}\int_0^1 \frac{\arctan\left(\frac{1-x}{1+x} \right)\log^2(x)}{1+x^2}dx + \frac{\pi^4}{512} \quad \left(x\mapsto \frac{1-x}{1+x} \right) \\ &= \frac{1}{4}\int_0^1 \frac{\left(\frac{\pi}{4}-\arctan(x) \right)\log^2(x)}{1+x^2}dx + \frac{\pi^4}{512} \\ &= -\frac{I_1}{4} +\frac{\pi}{16}\int_0^1 \frac{\log^2(x)}{1+x^2}dx +\frac{\pi^4}{512} \\ &= -\frac{I_1}{4} + \frac{3\pi^4}{512}\quad \color{blue}{\cdots (1)} \end{aligned} $$ Here, we used the fact that $\int_0^1 \frac{\log^2(x)}{1+x^2}dx = \frac{\pi^3}{16}$. Similarly, on $\gamma_{4,\epsilon}$ we have $$ \begin{aligned} \lim_{\epsilon\to 0^+} \int_{\gamma_{5,\epsilon}} f(z) \; dz &= \int_i^0 \frac{\arctan(x)}{1+x^2}\left(\text{arctanh}^2(x) + \frac{\pi^2}{16} \right)dx \\ &= \int_0^1 \frac{\text{arctanh}(x)}{1-x^2}\left(\frac{\pi^2}{16}-\arctan^2(x) \right)dx \quad \left(x\mapsto i x \right) \\ &= \int_0^1\frac{\arctan(x) \text{arctanh}^2(x) }{1+x^2}dx \quad (\text{IBP}) \\ &= -\frac{I_1}{4}+\frac{\pi^4}{256} \quad \color{blue}{\cdots (2)} \end{aligned} $$

For the integral over $\gamma_{3,\epsilon}$, we will take advantage of the following identities: $$ \begin{aligned} \arctan(e^{i\theta}) &= \frac{\pi}{4}+\frac{i}{2} \log\left(\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}} \right), \quad 0\leq \theta \leq \frac{\pi}{2} \\ \text{arctanh}(e^{i\theta}) &= -\frac{1}{2}\log \left(\tan\frac{\theta}{2} \right)+\frac{i\pi}{4}, \quad 0\leq \theta \leq \frac{\pi}{2}\\ \end{aligned} $$ We have $$ \begin{aligned} \lim_{\epsilon\to 0^+} \int_{\gamma_{3,\epsilon}}f(z) \; dz &= i \int_0^{\frac{\pi}{2}}f(e^{i\theta}) e^{i\theta} \; d\theta \\ &= \frac{i}{2} \int_0^{\frac{\pi}{2}}\frac{\left( \frac{\pi}{4}+\frac{i}{2} \log\left(\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}} \right)\right)\left( -\frac{i\pi}{4}\log \left(\tan\frac{\theta}{2} \right)+\frac{1}{4}\log^2\left(\tan \frac{\theta}{2} \right) \right)}{\cos \theta}d\theta \end{aligned} $$ The real part of the above integral is $$ \begin{aligned} \text{Re}\left[\lim_{\epsilon\to 0^+} \int_{\gamma_{3,\epsilon}}f(z) \; dz \right] &= \int_0^\frac{\pi}{2} \frac{\frac{1}{16}\log^2\left(\tan\frac{\theta}{2} \right)\log\left(\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}} \right)+\frac{\pi^2}{32}\log\left(\tan\frac{\theta}{2} \right)}{\cos \theta} d\theta \\ &= \int_0^1 \frac{\frac{1}{8}\log^2\left(u \right)\log\left(\frac{1-u}{1+u} \right)+\frac{\pi^2}{16}\log\left(u \right)}{1-u^2} du \quad \left(u= \tan\frac{\theta}{2} \right) \\ &= -\frac{1}{8} \int_0^1 \frac{\log^2(u)\log\left(\frac{1+u}{1-u}\right)}{1-u^2} + \frac{\pi^2}{16}\left(-\frac{\pi^2}{8} \right) \\ &= \frac{1}{16}\int_0^1 \frac{\log(u) \log^2\left(\frac{1-u}{1+u} \right)}{u}du – \frac{\pi^4}{128} \quad \color{blue}{\cdots (3)} \end{aligned} $$ Using the results from part 1, we get $$ \begin{aligned} \int_0^1 \frac{\log(u)\log^2\left(\frac{1-u}{1+u} \right)}{u}du &= \frac{7}{4}\int_0^1 \frac{\log(u)\log^2(1-u)}{u}du + 2\int_0^1 \frac{\log(u)\log^2(1+u)}{u}du \\ &= \frac{7}{4}\left(2\zeta(4) – 2\sum_{n=1}^\infty \frac{H_n}{n^3} \right)+2\left(2\text{Li}_4(-1)+2\sum_{n=1}^\infty\frac{(-1)^{n+1} H_n}{n^3} \right) \\ &= -\frac{7\pi^4}{144}+4\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n^3} \end{aligned} $$ By Cauchy’s integral theorem, the sum of (1), (2) and (3) is equal to zero. Therefore, $$ \begin{aligned} &\; -\frac{I_1}{2} + \frac{5\pi^4}{512}+\frac{1}{16}\left(-\frac{7\pi^4}{144}+4 \sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n^3}\right)-\frac{\pi^4}{128} = 0 \\ &\implies I_1 = -\frac{5\pi^4}{2304}+\frac{1}{2}\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^3} \\ &\implies I_1 = -\frac{5\pi^4}{2304}+\frac{1}{2}\left(\frac{11\pi^4}{360}+\frac{\pi^2}{12}\log^2(2)-\frac{\log^4(2)}{12} -\frac{7}{4}\log(2)\zeta(3)-2\text{Li}_4\left(\frac{1}{2} \right)\right) \\ &\implies \boxed{I_1 = \frac{151 \pi ^4}{11520}+\frac{\pi ^2}{24} \log ^2(2)-\frac{\log ^4(2)}{24} -\frac{7}{8} \zeta (3) \log (2)-\text{Li}_4\left(\frac{1}{2}\right)} \end{aligned} $$ Refer to this post for the evaluation of $\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^3}$.

Integral #6

The next integral that we’ll evaluate is $$ I_2 = \int_0^1 \frac{\log^3(x) \arctan(x)}{1+x^2}dx \quad \color{blue}{\cdots (4)} $$ Using the transformation $x\mapsto \frac{1}{x}$, we write the integral as: $$ \begin{aligned} I_2 &= -\int_1^\infty\frac{\log^3(x)\left(\frac{\pi}{2}-\arctan(x) \right)}{1+x^2}dx \quad \color{blue}{\cdots (5)} \end{aligned} $$ Adding up equations (4) and (5) and dividing both sides by 2 gives us: $$ \begin{aligned} I_2 &= -\frac{\pi}{4}\int_1^\infty \frac{\log^3(x)}{1+x^2}dx + \frac{1}{2}\int_0^\infty \frac{\log^3(x) \arctan(x)}{1+x^2}dx\\ &= \frac{\pi}{4}\int_0^1 \frac{\log^3(x)}{1+x^2}dx + \frac{1}{2}\int_0^\infty \frac{\log^3(x) \arctan(x)}{1+x^2}dx \end{aligned} $$ Note that $$ \begin{aligned} \int_0^1 \frac{\log^3(x)}{1+x^2}dx &= \sum_{n=0}^\infty (-1)^n \int_0^1 x^{2n}\log^3(x) \; dx \\ &= -6\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^4} \\ &= -6\beta(4) \end{aligned} $$ So, we have $$ I_2 = – \frac{3\pi}{2}\beta(4) + \frac{1}{2}\int_0^\infty \frac{\log^3(x) \arctan(x)}{1+x^2}dx \quad \color{blue}{\cdots (6)} $$ To evaluate the integral in the above equation, we will use the Feynman technique. Define the function $F:[0,\infty) \to \mathbb{R}$ as $F(s) = \int_0^\infty \frac{\log^3(x)\arctan(s x)}{1+x^2}dx$. Now, we have $$ F'(s) = \int_0^\infty \frac{x \log^3(x)}{(1+x^2)(1+s^2 x^2)}dx $$ Before evaluating $F'(s)$, we will evaluate the simpler integral $\int_0^\infty \frac{x \log(x)}{(1+x^2)(1+s^2 x^2)}dx$. To do this, integrate the principal branch of $g(z) = \frac{z\log^2(-z)}{(1+z^2)(1+s^2 z^2)}$ around the following “key hole” contour:

where:

• $\gamma_{1,\epsilon, R}$ is a line joining the points $\sqrt{R^2-\epsilon^2}-i\epsilon$ and $-i\epsilon$.
• $\gamma_{2,\epsilon, R}$ is a line joining the points $i\epsilon$ and $\sqrt{R^2-\epsilon^2}+i\epsilon$.
• $\gamma_{3,\epsilon}$ is parameterized by $\epsilon e^{-i t}$ where $\frac{\pi}{2} \leq t \leq \frac{3\pi}{2}$.
• $\gamma_{4,R}$ is parameterized by $R e^{i t}$ where $\arctan\left(\frac{\epsilon}{\sqrt{R^2-\epsilon^2}} \right) \leq t \leq 2\pi – \arctan\left(\frac{\epsilon}{\sqrt{R^2-\epsilon^2}} \right)$.

As $\epsilon\to 0^+$ and $R\to \infty$, the integrals along $\gamma_{3,\epsilon}$ and $\gamma_{4,R}$ tend to 0. Therefore, we are only left with the integrals above and below the branch cut. The residues of $g(z)$ at it’s poles is given by: $$ \begin{aligned} \mathop{\text{Res}}\limits_{z=i} \; g(z) &= -\frac{\pi^2}{8(1-s^2)} \\ \mathop{\text{Res}}\limits_{z=-i} \; g(z) &= -\frac{\pi^2}{8(1-s^2)} \\ \mathop{\text{Res}}\limits_{z=i/s} \; g(z) &= -\frac{\left(\log(s)+\frac{i\pi}{2} \right)^2}{2(1-s^2)} \\ \mathop{\text{Res}}\limits_{z=-i/s} \; g(z) &= -\frac{\left(\log(s)-\frac{i\pi}{2} \right)^2}{2(1-s^2)} \end{aligned} $$ Now, the residue theorem gives us: $$ \begin{aligned} \int_0^\infty \frac{x\left((\log (x)-i\pi)^2 – (\log (x)+i\pi)^2 \right)}{(1+x^2)(1+s^2 x^2)}dx &= 2i\pi \Big(\mathop{\text{Res}}\limits_{z=i} \; g(z)+ \mathop{\text{Res}}\limits_{z=-i}\; g(z) + \mathop{\text{Res}}\limits_{z=i/s}\; g(z) \\ &\quad + \mathop{\text{Res}}\limits_{z=-i/s}\; g(z) \Big) \\ \implies -4i\pi \int_0^\infty \frac{x\log(x)}{(1+x^2)(1+s^2 x^2)}dx &= -2i\pi \frac{\log^2(s)}{1-s^2} \\ \implies \int_0^\infty \frac{x\log(x)}{(1+x^2)(1+s^2 x^2)}dx &= \frac{\log^2(s)}{2(1-s^2)} \quad \color{blue}{\cdots (7)} \end{aligned} $$ To evaluate $F'(s)$, we integrate the principal branch of $h(z) = \frac{z \log^4(-z)}{(1+z^2)(1+s^2 z^2)}$ around the same contour. This time, the residues are: $$ \begin{aligned} \mathop{\text{Res}}\limits_{z=i} \; h(z) &= \frac{\pi^4}{32(1-s^2)} \\ \mathop{\text{Res}}\limits_{z=-i} \; h(z) &= \frac{\pi^4}{32(1-s^2)} \\ \mathop{\text{Res}}\limits_{z=i/s} \; h(z) &= -\frac{\left(\log(s)+\frac{i\pi}{2} \right)^4}{2(1-s^2)} \\ \mathop{\text{Res}}\limits_{z=-i/s} \; h(z) &= -\frac{\left(\log(s)-\frac{i\pi}{2} \right)^4}{2(1-s^2)} \end{aligned} $$ The residue theorem gives us: $$ \begin{aligned} \int_0^\infty \frac{x\left((\log (x)-i\pi)^4 – (\log (x)+i\pi)^4 \right)}{(1+x^2)(1+s^2 x^2)}dx &= 2i\pi \Big(\mathop{\text{Res}}\limits_{z=i} \; h(z)+ \mathop{\text{Res}}\limits_{z=-i}\; h(z) + \mathop{\text{Res}}\limits_{z=i/s}\; h(z) \\ &\quad + \mathop{\text{Res}}\limits_{z=-i/s}\; h(z) \Big) \\ \implies -8i\pi \int_0^\infty \frac{x\left(\log^3(x) -\pi^2 \log(x) \right)}{(1+x^2)(1+s^2 x^2)}dx &= 2i\pi \frac{\frac{3\pi^2}{2}\log^2(s)-\log^4(s)}{1-s^2} \\ \implies \int_0^\infty \frac{x\log^3(x)}{(1+x^2)(1+s^2 x^2)}dx &= \frac{\pi^2\log^2(s)+2\log^4(s)}{8(1-s^2)} \quad \color{blue}{\cdots (8)} \end{aligned} $$ Note that we used equation (7) to the get the above result. We can now calculate our original integral as follows: $$ \begin{aligned} I_2 &= -\frac{3\pi}{2}\beta(4) + \frac{1}{2}\int_0^1 F'(s) ds \\ &= -\frac{3\pi}{2}\beta(4) + \frac{1}{2}\int_0^1 \frac{\pi^2\log^2(s)+2\log^4(s)}{8(1-s^2)} ds \\ &= -\frac{3\pi}{2}\beta(4) + \frac{\pi^2}{16}\int_0^1\frac{\log^2(s)}{1-s^2}ds+\frac{1}{8}\int_0^1 \frac{\log^4(s)}{1-s^2} ds \\ &= -\frac{3\pi}{2}\beta(4) + \frac{\pi^2}{16}\sum_{k=0}^\infty\int_0^1 s^{2k}\log^2(s)ds+\frac{1}{8}\sum_{k=0}^\infty\int_0^1 s^{2k}\log^4(s) ds \\ &= -\frac{3\pi}{2}\beta(4) + \frac{\pi^2}{8}\sum_{k=0}^\infty\frac{1}{(2k+1)^3}+3\sum_{k=0}^\infty\frac{1}{(2k+1)^5} \\ &= \boxed{-\frac{3\pi}{2}\beta(4) + \frac{7\pi^2}{64}\zeta(3) +\frac{93}{32}\zeta(5)} \end{aligned} $$ Interestingly, $I_2$ can be reduced into an Euler sum which can be evaluated using contour integration.