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^4}}{2-\epsilon^2} \right)\leq t \leq \frac{\pi}{2} - \arctan\left(\frac{\epsilon \sqrt{4-\epsilon^4}}{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:\mathbb{R}_{> 0}\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 \left(\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) + \mathop{\text{Res}}\limits_{z=-i/s}\; g(z) \right) \\ &\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 \boxed{\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 can 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 \left(\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) + \mathop{\text{Res}}\limits_{z=-i/s}\; h(z) \right) \\ &\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 \boxed{\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.

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