bookmark_borderEvaluating 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: I1=01log2(x)arctan(x)1+x2dx 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)=arctan(z)1+z2(arctanh2(z)+π216)f(z) = \frac{\arctan(z)}{1+z^2}\left(\text{arctanh}^2(z) + \frac{\pi^2}{16} \right) around the following contour:

where

  • γ3,ϵ\gamma_{3,\epsilon} is an arc parameterized by eite^{it}, where arctan(ϵ4ϵ22ϵ2)tπ2arctan(ϵ4ϵ22ϵ2)\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).
  • γ2,ϵ\gamma_{2,\epsilon} and γ4,ϵ\gamma_{4,\epsilon} are circular indents of radius ϵ\epsilon around the branch points at z=1z=1 and z=iz=i, respectively.
  • γ1,ϵ\gamma_{1,\epsilon} is a straight line joining 00 and 1ϵ1-\epsilon.
  • γ4,ϵ\gamma_{4,\epsilon} is a straight line joining (1ϵ)i(1-\epsilon)i and 00.
Note that ff is analytic on z<1|z| < 1. It is easy to see that limϵ0+γ2,ϵf(z)  dz=0limϵ0+γ4,ϵf(z)  dz=0 \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 γ1,ϵ\gamma_{1,\epsilon}, we have limϵ0+γ1,ϵf(z)  dz=01arctan(x)1+x2(arctanh2(x)+π216)dx=01arctan(x)arctanh2(x)1+x2dx+π216arctan2(x)201=1401arctan(1x1+x)log2(x)1+x2dx+π4512(x1x1+x)=1401(π4arctan(x))log2(x)1+x2dx+π4512=I14+π1601log2(x)1+x2dx+π4512=I14+3π4512(1) \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 01log2(x)1+x2dx=π316\int_0^1 \frac{\log^2(x)}{1+x^2}dx = \frac{\pi^3}{16}. Similarly, on γ4,ϵ\gamma_{4,\epsilon} we have limϵ0+γ5,ϵf(z)  dz=i0arctan(x)1+x2(arctanh2(x)+π216)dx=01arctanh(x)1x2(π216arctan2(x))dx(xix)=01arctan(x)arctanh2(x)1+x2dx(IBP)=I14+π4256(2) \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 γ3,ϵ\gamma_{3,\epsilon}, we will take advantage of the following identities: arctan(eiθ)=π4+i2log(1+tanθ21tanθ2),0θπ2arctanh(eiθ)=12log(tanθ2)+iπ4,0θπ2 \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 limϵ0+γ3,ϵf(z)  dz=i0π2f(eiθ)eiθ  dθ=i20π2(π4+i2log(1+tanθ21tanθ2))(iπ4log(tanθ2)+14log2(tanθ2))cosθdθ \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 Re[limϵ0+γ3,ϵf(z)  dz]=0π2116log2(tanθ2)log(1tanθ21+tanθ2)+π232log(tanθ2)cosθdθ=0118log2(u)log(1u1+u)+π216log(u)1u2du(u=tanθ2)=1801log2(u)log(1+u1u)1u2+π216(π28)=11601log(u)log2(1u1+u)uduπ4128(3) \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 01log(u)log2(1u1+u)udu=7401log(u)log2(1u)udu+201log(u)log2(1+u)udu=74(2ζ(4)2n=1Hnn3)+2(2Li4(1)+2n=1(1)n+1Hnn3)=7π4144+4n=1(1)n+1Hnn3 \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,   I12+5π4512+116(7π4144+4n=1(1)n+1Hnn3)π4128=0    I1=5π42304+12n=1(1)n+1Hnn3    I1=5π42304+12(11π4360+π212log2(2)log4(2)1274log(2)ζ(3)2Li4(12))    I1=151π411520+π224log2(2)log4(2)2478ζ(3)log(2)Li4(12) \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 n=1(1)n+1Hnn3\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^3}.

Integral #6

The next integral that we’ll evaluate is I2=01log3(x)arctan(x)1+x2dx(4) I_2 = \int_0^1 \frac{\log^3(x) \arctan(x)}{1+x^2}dx \quad \color{blue}{\cdots (4)} Using the transformation x1xx\mapsto \frac{1}{x}, we write the integral as: I2=1log3(x)(π2arctan(x))1+x2dx(5) \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: I2=π41log3(x)1+x2dx+120log3(x)arctan(x)1+x2dx=π401log3(x)1+x2dx+120log3(x)arctan(x)1+x2dx \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 01log3(x)1+x2dx=n=0(1)n01x2nlog3(x)  dx=6n=0(1)n(2n+1)4=6β(4) \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 I2=3π2β(4)+120log3(x)arctan(x)1+x2dx(6)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,)RF:[0,\infty) \to \mathbb{R} as F(s)=0log3(x)arctan(sx)1+x2dxF(s) = \int_0^\infty \frac{\log^3(x)\arctan(s x)}{1+x^2}dx. Now, we have F(s)=0xlog3(x)(1+x2)(1+s2x2)dx F'(s) = \int_0^\infty \frac{x \log^3(x)}{(1+x^2)(1+s^2 x^2)}dx Before evaluating F(s)F'(s), we will evaluate the simpler integral 0xlog(x)(1+x2)(1+s2x2)dx\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)=zlog2(z)(1+z2)(1+s2z2)g(z) = \frac{z\log^2(-z)}{(1+z^2)(1+s^2 z^2)} around the following “key hole” contour:

where:

  • γ1,ϵ,R\gamma_{1,\epsilon, R} is a line joining the points R2ϵ2iϵ\sqrt{R^2-\epsilon^2}-i\epsilon and iϵ-i\epsilon.
  • γ2,ϵ,R\gamma_{2,\epsilon, R} is a line joining the points iϵi\epsilon and R2ϵ2+iϵ\sqrt{R^2-\epsilon^2}+i\epsilon.
  • γ3,ϵ\gamma_{3,\epsilon} is parameterized by ϵeit\epsilon e^{-i t} where π2t3π2\frac{\pi}{2} \leq t \leq \frac{3\pi}{2}.
  • γ4,R\gamma_{4,R} is parameterized by ReitR e^{i t} where arctan(ϵR2ϵ2)t2πarctan(ϵR2ϵ2)\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 ϵ0+\epsilon\to 0^+ and RR\to \infty, the integrals along γ3,ϵ\gamma_{3,\epsilon} and γ4,R\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)g(z) at it’s poles is given by: Resz=i  g(z)=π28(1s2)Resz=i  g(z)=π28(1s2)Resz=i/s  g(z)=(log(s)+iπ2)22(1s2)Resz=i/s  g(z)=(log(s)iπ2)22(1s2) \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: 0x((log(x)iπ)2(log(x)+iπ)2)(1+x2)(1+s2x2)dx=2iπ(Resz=i  g(z)+Resz=i  g(z)+Resz=i/s  g(z)+Resz=i/s  g(z))    4iπ0xlog(x)(1+x2)(1+s2x2)dx=2iπlog2(s)1s2    0xlog(x)(1+x2)(1+s2x2)dx=log2(s)2(1s2)(7) \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)F'(s), we integrate the principal branch of h(z)=zlog4(z)(1+z2)(1+s2z2)h(z) = \frac{z \log^4(-z)}{(1+z^2)(1+s^2 z^2)} around the same contour. This time, the residues are: Resz=i  h(z)=π432(1s2)Resz=i  h(z)=π432(1s2)Resz=i/s  h(z)=(log(s)+iπ2)42(1s2)Resz=i/s  h(z)=(log(s)iπ2)42(1s2) \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: 0x((log(x)iπ)4(log(x)+iπ)4)(1+x2)(1+s2x2)dx=2iπ(Resz=i  h(z)+Resz=i  h(z)+Resz=i/s  h(z)+Resz=i/s  h(z))    8iπ0x(log3(x)π2log(x))(1+x2)(1+s2x2)dx=2iπ3π22log2(s)log4(s)1s2    0xlog3(x)(1+x2)(1+s2x2)dx=π2log2(s)+2log4(s)8(1s2)(8) \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: I2=3π2β(4)+1201F(s)ds=3π2β(4)+1201π2log2(s)+2log4(s)8(1s2)ds=3π2β(4)+π21601log2(s)1s2ds+1801log4(s)1s2ds=3π2β(4)+π216k=001s2klog2(s)ds+18k=001s2klog4(s)ds=3π2β(4)+π28k=01(2k+1)3+3k=01(2k+1)5=3π2β(4)+7π264ζ(3)+9332ζ(5) \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, I2I_2 can be reduced into an Euler sum which can be evaluated using contour integration.

bookmark_borderEvaluating very nasty logarithmic integrals: Part II

In this post, we’ll evaluate some more nasty logarithmic integrals. Please read part 1 of this series if you haven’t done so already.

Integral #3

We’ll start by finding a closed form for the integral: I1=01log2(1+x2)1+x2dx I_1 = \int_0^1 \frac{\log^2(1+x^2)}{1+x^2}dx This integral can be reduced to Euler sums like our previous problem. But this time, the resulting Euler sums cannot be evaluated using the method of residues. Therefore, we’ll have to use a different approach.

Let us first consider the following integral: I2=01log2(1+ix)1+x2dxI_2 = \int_0^1 \frac{\log^2(1+ix)}{1+x^2}dx Throughout this post, log\log denotes the principal branch of the logarithmic function defined by logz=logz+iarg(z)\log z = \log|z| + i\text{arg}(z), with π<arg(z)π-\pi < |\text{arg}(z)| \leq \pi. We have I2=1201log2(1+ix)(11+ix+11ix)dx=log3(1+ix)6i01+1201log2(1+ix)1ixdx=log3(1+i)6i+i2121i2log2(2(1x))xdx=log3(1+i)6i+i2121i2log2(1x)+log2(2)+2log(2)log(1x)xdx=log3(1+i)6i+ilog2(2)log(1i)2+ilog(2)[Li2(12)Li2(1i2)]+i2121i2log2(1x)xdx(1) \begin{aligned} I_2 &= \frac{1}{2}\int_0^1\log^2(1+ix)\left(\frac{1}{1+ix}+\frac{1}{1-ix} \right)dx \\ &= \frac{\log^3(1+ix)}{6i}\Big|_0^1 + \frac{1}{2}\int_0^1 \frac{\log^2(1+ix)}{1-ix}dx \\ &= \frac{\log^3(1+i)}{6i} + \frac{i}{2}\int_{\frac{1}{2}}^{\frac{1-i}{2}}\frac{\log^2(2(1-x))}{x}dx \\ &= \frac{\log^3(1+i)}{6i} + \frac{i}{2}\int_{\frac{1}{2}}^{\frac{1-i}{2}}\frac{\log^2(1-x)+\log^2(2)+2\log(2)\log(1-x)}{x}dx \\ &= \frac{\log^3(1+i)}{6i} + \frac{i\log^2(2)\log\left(1-i\right)}{2} + i\log(2) \left[\text{Li}_2\left(\frac{1}{2}\right)-\text{Li}_2\left(\frac{1-i}{2}\right) \right] \\ &\quad + \frac{i}{2}\int_{\frac{1}{2}}^{\frac{1-i}{2}}\frac{\log^2(1-x)}{x}dx \quad \color{blue}{\cdots (1)} \end{aligned} We can use equation (2) from (B) to evaluate 121+i2log2(1x)xdx\int_{\frac{1}{2}}^{\frac{1+i}{2}}\frac{\log^2(1-x)}{x}dx. 121i2log2(1x)xdx=log2(1+i2)log(1i2)+2log(1+i2)Li2(1+i2)2Li3(1+i2)+log3(2)+2log(2)Li2(12)+2Li3(12)(2) \begin{aligned} \int_{\frac{1}{2}}^{\frac{1-i}{2}}\frac{\log^2(1-x)}{x}dx &= \log^2\left( \frac{1+i}{2}\right)\log\left(\frac{1-i}{2}\right) + 2\log\left(\frac{1+i}{2} \right)\text{Li}_2\left(\frac{1+i}{2}\right) \\ &\quad -2\text{Li}_3\left(\frac{1+i}{2}\right) + \log^3(2) + 2\log(2)\text{Li}_2\left(\frac{1}{2}\right) + 2\text{Li}_3\left(\frac{1}{2} \right) \quad \color{blue}{\cdots (2)} \end{aligned}

To simplify Li2(1+i2)\text{Li}_2\left(\frac{1+i}{2} \right) and Li2(1i2)\text{Li}_2\left(\frac{1-i}{2} \right), we can use the following Dilogarithm identity: Li2(1z)+Li2(1z1)=12log2(z) \text{Li}_2(1-z) + \text{Li}_2\left(1-z^{-1} \right) = -\frac{1}{2}\log^2(z) This is easy to verify by differentiating both sides of the above equation with respect to zz. Plugging in z=1+i2z=\frac{1+i}{2} gives Li2(1i2)=Li2(i)12log2(1+i2)=5π296log2(2)8+i(G+π8log(2))Li2(1+i2)=Li2(1i2)=5π296log2(2)8i(G+π8log(2)) \begin{aligned} \text{Li}_2\left(\frac{1-i}{2}\right) &= -\text{Li}_2(i) - \frac{1}{2}\log^2\left(\frac{1+i}{2}\right) = \frac{5\pi^2}{96}-\frac{\log^2(2)}{8}+i\left(-G + \frac{\pi}{8}\log(2)\right) \\ \text{Li}_2\left(\frac{1+i}{2}\right) &= \overline{\text{Li}_2\left(\frac{1-i}{2}\right)} = \frac{5\pi^2}{96}-\frac{\log^2(2)}{8}-i\left(-G + \frac{\pi}{8}\log(2)\right) \end{aligned}

Now, we have everything needed to simplify equation (1). This is a tedious task so I used Mathematica to do it. The final result is I2=3π3128Glog(2)2+7πlog2(2)32+i(Gπ4+7π2log(2)192+log3(2)48+78ζ(3)Li3(1+i2))(3) \begin{aligned} I_2 &= -\frac{3\pi^3}{128}- \frac{G \log(2) }{2} + \frac{7\pi \log^2(2)}{32} + i\Bigg( -\frac{ G \pi }{4} + \frac{7\pi^2 \log(2)}{192} + \frac{\log^3(2)}{48}+\frac{7}{8}\zeta(3) \\ &\quad - \text{Li}_3\left(\frac{1+i}{2} \right)\Bigg) \quad \color{blue}{\cdots (3)} \end{aligned} As of now, I am not aware of a closed form expression for Li3(1+i2)\text{Li}_3\left(\frac{1+i}{2} \right). So, we’ll leave it as it is. We can now extract I1I_1 from the real part of I2I_2. Re I2=0114log2(1+x2)arctan2(x)1+x2dx=1401log2(1+x2)1+x2dxarctan3(x)301=14I1π3192 \begin{aligned} \text{Re }I_2 &= \int_0^1 \frac{\frac{1}{4}\log^2(1+x^2) - \arctan^2(x)}{1+x^2}dx \\ &= \frac{1}{4}\int_0^1 \frac{\log^2(1+x^2)}{1+x^2}dx - \frac{\arctan^3(x)}{3}\Big|_0^1 \\ &= \frac{1}{4}I_1 - \frac{\pi^3}{192} \end{aligned} Therefore, I1=2Glog(2)7π396+7πlog2(2)8+4 Im Li3(1+i2)(4) \boxed{I_1 = -2G\log(2) - \frac{7\pi^3}{96} + \frac{7\pi \log^2(2)}{8} + 4\text{ Im }\text{Li}_3\left(\frac{1+i}{2} \right)}\quad \color{blue}{\cdots (4)} Now, let’s turn our attention to another integral: I3=01log(x)log(1+x2)1+x2dxI_3 = \int_0^1 \frac{\log(x)\log(1+x^2)}{1+x^2}dx Notice that with the help of some algebra, we can write: I3=1201log2(x1+x2)1+x2dx+1201log2(1+x2)1+x2dx+1201log2(x)1+x2dx I_3 = -\frac{1}{2}\int_0^1 \frac{\log^2\left(\frac{x}{1+x^2} \right)}{1+x^2}dx + \frac{1}{2}\int_0^1 \frac{\log^2(1+x^2)}{1+x^2}dx + \frac{1}{2}\int_0^1 \frac{\log^2(x)}{1+x^2}dx The leftmost integral can be dealt with the trigonometric substitution x=tanθx=\tan \theta: 01log2(x1+x2)1+x2dx=0π4log2(sinθcosθ)  dθ=0π4log2(sin(2θ)2)  dθ=120π2log2(sinθ2)  dθ=12lims1d2ds20π2(sinθ2)s1dθ=12lims1d2ds2[2sπΓ(s2)Γ(1+s2)]=π348+πlog2(2) \begin{aligned} \int_0^1 \frac{\log^2\left(\frac{x}{1+x^2} \right)}{1+x^2}dx &= \int_0^{\frac{\pi}{4}} \log^2\left(\sin \theta \cos\theta \right) \; d\theta \\ &= \int_0^{\frac{\pi}{4}} \log^2\left(\frac{\sin(2\theta)}{2} \right)\; d\theta \\ &= \frac{1}{2}\int_0^{\frac{\pi}{2}}\log^2\left(\frac{\sin \theta}{2} \right)\; d\theta \\ &= \frac{1}{2}\lim_{s\to 1}\frac{d^2}{ds^2}\int_0^{\frac{\pi}{2}}\left(\frac{\sin\theta}{2} \right)^{s-1}d\theta \\ &= \frac{1}{2}\lim_{s\to 1}\frac{d^2}{ds^2}\left[\frac{2^{-s}\sqrt{\pi}\Gamma\left(\frac{s}{2}\right)}{\Gamma\left(\frac{1+s}{2} \right)} \right] \\ &= \frac{\pi^3}{48}+ \pi \log^2(2) \end{aligned} The middle integral has already been evaluated. As for the rightmost integral, we have: 01log2(x)1+x2dx=n=0(1)n01x2nlog2(x)  dx=2n=0(1)n(2n+1)3=π316 \begin{aligned} \int_0^1 \frac{\log^2(x)}{1+x^2}dx &= \sum_{n=0}^\infty (-1)^{n}\int_0^1 x^{2n}\log^2(x)\; dx \\ &= 2\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3} \\ &= \frac{\pi^3}{16} \end{aligned} This gives us I3=π364Glog(2)πlog2(2)16+2 Im Li3(1+i2)(5) \boxed{I_3 =-\frac{\pi^3}{64} -G \log(2) - \frac{\pi \log^2(2)}{16} +2 \text{ Im }\text{Li}_3\left(\frac{1+i}{2}\right)} \quad \color{blue}{\cdots (5)}

One can also evaluate I4=01log(1+x2)arctan(x)xdxI_4 = \int_0^1\frac{\log(1+x^2)\arctan(x)}{x}dx by noting that I4=Im01log2(1+ix)xdx=Im0ilog2(1x)xdxI_4 = \text{Im}\int_0^1 \frac{\log^2(1+ix)}{x}dx = \text{Im}\int_0^{-i}\frac{\log^2(1-x)}{x}dx and using equation (3) from (B).The end result is: I4=3π364+Glog(2)πlog2(2)16+2 Im Li3(1+i2)(6) \boxed{I_4 =-\frac{3\pi^3}{64}+G\log(2)-\frac{\pi \log^2(2)}{16}+2\text{ Im }\text{Li}_3\left(\frac{1+i}{2}\right) } \quad \color{blue}{\cdots (6)} Using I3I_3 and I4I_4, we can evaluate I5=01log(x)arctan(x)x(1+x2)dxI_5 = \int_0^1 \frac{\log(x)\arctan(x)}{x(1+x^2)}dx as follows: I5=01log(x)arctan(x)(1xx1+x2)dx=01log(x)arctan(x)xdx01xlog(x)arctan(x)1+x2dx=1201log2(x)1+x2dx+1201log(x)log(1+x2)1+x2dx+1201log(1+x2)arctan(x)xdx(IBP)=π332+I3+I42=π316πlog2(2)16+2 Im Li3(1+i2) \begin{aligned} I_5 &= \int_0^1 \log(x)\arctan(x)\left(\frac{1}{x}-\frac{x}{1+x^2} \right) dx \\ &= \int_0^1 \frac{\log(x)\arctan(x)}{x}dx - \int_0^1 \frac{x\log(x)\arctan(x)}{1+x^2}dx \\ &= -\frac{1}{2}\int_0^1 \frac{\log^2(x)}{1+x^2}dx + \frac{1}{2} \int_0^1 \frac{\log(x)\log(1+x^2)}{1+x^2} dx + \frac{1}{2}\int_0^1 \frac{\log(1+x^2)\arctan(x)}{x}dx \quad (\text{IBP}) \\ &= -\frac{\pi^3}{32} + \frac{I_3 + I_4}{2} \\ &= -\frac{\pi^3}{16} - \frac{\pi \log^2(2)}{16} + 2\text{ Im }\text{Li}_3\left(\frac{1+i}{2}\right) \end{aligned} On the other hand, we have: I5=01log(x)(n=0(1)nH~nx2n)dx=n=0(1)nH~n01x2nlog(x)  dx=n=0(1)nH~n(2n+1)2 \begin{aligned} I_5 &= \int_0^1 \log(x) \left(\sum_{n=0}^\infty (-1)^n \tilde{H}_n x^{2n} \right) dx\\ &= \sum_{n=0}^\infty (-1)^n \tilde{H}_n \int_0^1 x^{2n}\log(x) \; dx \\ &= -\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+1)^2} \end{aligned} where H~n=j=0n12j+1\tilde{H}_n = \sum_{j=0}^n \frac{1}{2j+1}. This gives us an interesting Euler sum: n=0(1)nH~n(2n+1)2=π316+πlog2(2)162 Im Li3(1+i2)(7) \boxed{\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+1)^2} = \frac{\pi^3}{16} + \frac{\pi \log^2(2)}{16} - 2\text{ Im }\text{Li}_3\left(\frac{1+i}{2}\right)}\quad \color{blue}{\cdots (7)} Of course, one can proceed in a similar manner to create more crazy integrals. The following problem is left as an exercise for the reader.

Exercise 1: Using the method of residues, show that n=0(1)nH~n2n+1=G2+πlog(2)8(8)\sum_{n=0}^\infty \frac{(-1)^n\tilde{H}_n}{2n+1} = \frac{G}{2}+\frac{\pi \log(2)}{8} \quad \color{blue}{\cdots (8)}

Integral #4

Many years ago, I encountered the following integral: I6=01xarctan(x)log(1x2)1+x2dxI_6 = \int_0^1 \frac{x \arctan(x)\log(1-x^2)}{1+x^2}dx At that time, I couldn’t find a solution to this problem. Hence, I ended up asking it on math.stackexchange.com. The answers that I received there involved evaluating complex logarithmic integrals by brute force. Recently, I discovered a much simpler way to solve it using the method of residues.

Let’s start by breaking down I6I_6 into Euler sums. I6=01xlog(1x2)(n=0(1)nH~nx2n+1)dx=n=0(1)nH~n01x2n+2log(1x2)dx=n=0(1)n+1H~n(ψ0(n+52)+γ2n+3)=n=0(1)n+1(H~n+112n+3)(2log(2)+2H~n+12n+3)=n=0(1)n(H~n12n+1)(2log(2)+2H~n2n+1)=2log(2)n=0(1)nH~n2n+1+2Glog(2)+2n=0(1)n(H~n)22n+12n=0(1)nH~n(2n+1)2=Glog(2)πlog2(2)4+2n=0(1)n(H~n)22n+12n=0(1)nH~n(2n+1)2(9) \begin{aligned} I_6 &= \int_0^1 x\log(1-x^2) \left(\sum_{n=0}^\infty (-1)^n \tilde{H}_n x^{2n+1} \right)dx \\ &= \sum_{n=0}^\infty (-1)^n \tilde{H}_n \int_0^1 x^{2n+2}\log(1-x^2) dx \\ &= \sum_{n=0}^\infty (-1)^{n+1} \tilde{H}_n \left(\frac{\psi_0\left(n+\frac{5}{2} \right)+\gamma}{2n+3} \right) \\ &= \sum_{n=0}^\infty (-1)^{n+1}\left(\tilde{H}_{n+1}-\frac{1}{2n+3} \right)\left(\frac{-2\log(2)+2\tilde{H}_{n+1}}{2n+3} \right) \\ &= \sum_{n=0}^\infty (-1)^n \left(\tilde{H}_{n}-\frac{1}{2n+1} \right)\left(\frac{-2\log(2)+2\tilde{H}_{n}}{2n+1} \right) \\ &= -2\log(2)\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{2n+1}+2G\log(2) + 2\sum_{n=0}^\infty \frac{(-1)^n (\tilde{H}_n)^2}{2n+1} -2\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+1)^2} \\ &= G\log(2) - \frac{\pi \log^2(2)}{4} + 2\sum_{n=0}^\infty \frac{(-1)^n (\tilde{H}_n)^2}{2n+1} -2\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+1)^2} \quad \color{blue}{\cdots (9)} \end{aligned} The result of exercise 1 was used in the last step.

Now, integrate the function f(z)=πcsc(πz)(γ+ψ0(z+32))22z+1f(z) = \pi\csc(\pi z) \frac{\left( \gamma + \psi_0 \left(-z+\frac{3}{2} \right)\right)^2}{-2z+1} over the positively oriented square, CNC_N, with vertices ±(N+14)±i(N+14)\pm \left(N+\frac{1}{4}\right)\pm i\left(N+\frac{1}{4} \right). It takes a bit of effort to see that limNCNf(z)  dz=0\lim_{N\to \infty}\int_{C_N} f(z)\; dz = 0 This implies that the sum of residues of f(z)f(z) at it’s poles is equal to 00. We have

Resz=nf(z)=(1)n(γ+ψ0(n+32))22n+1=(1)n(2log(2)+2H~n)22n+1,n{0,1,2,}Resz=nf(z)=(1)n1(γ+ψ0(n+32))22n1=(1)n1(2log(2)+2H~n122n1)22n1,n{1,2,3,}Resz=2n+12f(z)=(1)n1πHnn(1)n13π2n2,n{1,2,3,} \begin{aligned} \mathop{\text{Res}}\limits_{z=-n} f(z) &= (-1)^n \frac{\left(\gamma +\psi_0\left(n+\frac{3}{2}\right) \right)^2}{2n+1} = (-1)^n \frac{\left(-2\log(2)+2\tilde{H}_n \right)^2}{2n+1}, \quad n\in\{0,1,2,\cdots\} \\ \mathop{\text{Res}}\limits_{z=n} f(z) &= (-1)^{n-1} \frac{\left(\gamma+\psi_0\left(-n+\frac{3}{2} \right) \right)^2}{2n-1} = (-1)^{n-1} \frac{\left(-2\log(2)+2\tilde{H}_{n-1} -\frac{2}{2n-1}\right)^2}{2n-1}, \quad n\in\{1,2,3,\cdots\} \\ \mathop{\text{Res}}\limits_{z=\frac{2n+1}{2}} f(z) &= (-1)^{n-1} \frac{\pi H_n}{n} - (-1)^{n-1} \frac{3\pi}{2n^2}, \quad n\in\{1,2,3,\cdots\} \end{aligned}

Summing up the residues and performing some algebraic simplifications gives:   8n=0(1)n(H~n)22n+18n=0(1)nH~n(2n+1)216log(2)n=0(1)nH~n2n+1+4n=0(1)n(2n+1)3+8log(2)n=0(1)n(2n+1)2+πn=1(1)n1Hnn3π2n=1(1)n1n2=0    8(n=0(1)n(H~n)22n+1n=0(1)nH~n(2n+1)2)+π38+π(π212log2(2)2)3π2(π212)=0    n=0(1)n(H~n)22n+1n=0(1)nH~n(2n+1)2=π396+πlog2(2)16(10) \begin{aligned} &\; 8\sum_{n=0}^\infty \frac{(-1)^n (\tilde{H}_n)^2}{2n+1}-8\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+1)^2}-16\log(2)\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{2n+1}+ 4\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3} \\ &\quad + 8\log(2)\sum_{n=0}^\infty \frac{(-1)^n }{(2n+1)^2} + \pi\sum_{n=1}^\infty \frac{(-1)^{n-1}H_n}{n} - \frac{3\pi}{2}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2} = 0 \\ &\implies 8\left(\sum_{n=0}^\infty \frac{(-1)^n (\tilde{H}_n)^2}{2n+1}-\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+1)^2} \right) + \frac{\pi^3}{8} + \pi \left(\frac{\pi^2}{12}-\frac{\log^2(2)}{2}\right)-\frac{3\pi}{2}\left(\frac{\pi^2}{12} \right) = 0 \\ &\implies \sum_{n=0}^\infty \frac{(-1)^n (\tilde{H}_n)^2}{2n+1}-\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+1)^2} = -\frac{\pi^3}{96}+\frac{\pi \log^2(2)}{16} \quad \color{blue}{\cdots (10)} \end{aligned} In the above calculation, we used the result of exercise 1 and that n=1(1)n1Hnn=π212log2(2)2\sum_{n=1}^\infty \frac{(-1)^{n-1}H_n}{n} = \frac{\pi^2}{12}-\frac{\log^2(2)}{2} This follows from (A). Finally, plugging equation (10) into (9), gives I6=π348π8log2(2)+Glog(2)(11) \boxed{I_6 = -\frac{\pi^3}{48}-\frac{\pi}{8}\log^2 (2) +G\log (2)} \quad \color{blue}{\cdots (11)}

bookmark_borderEvaluating very nasty logarithmic integrals: Part I

In this post, we’ll continue our exploration of logarithmic integrals and Euler sums. We’ll also discuss the contour integration method for evaluating Euler sums. It is recommended that the reader goes through the previous posts, (A) and (B), before reading this post.

Integral #1

Our first integral is the following: I=01log(t)log(1t)log(1+t)tdt \begin{aligned} I &= \int_0^1 \frac{\log(t)\log(1-t)\log(1+t)}{t}dt \end{aligned} We will use the integral J=01log(t)log2(1t)tdtJ = \int_0^1 \frac{\log(t)\log^2(1-t)}{t}dt as a starting point. Make the substitution tt2t\mapsto t^2 in JJ to obtain: J=201log(t2)log2(1t2)tdt=401log(t)log2(1t)tdt+401log(t)log2(1+t)tdt+801log(t)log(1t)log(1+t)tdt=4J+401log(t)log2(1+t)tdt+8I \begin{aligned} J &= 2\int_0^1 \frac{\log(t^2) \log^2(1-t^2)}{t}dt \\ &= 4\int_0^1 \frac{\log(t)\log^2(1-t)}{t}dt + 4\int_0^1 \frac{\log(t)\log^2(1+t)}{t}dt + 8\int_0^1 \frac{\log(t)\log(1-t) \log(1+t)}{t}dt \\ &= 4J + 4\int_0^1 \frac{\log(t)\log^2(1+t)}{t}dt + 8I \end{aligned}

Solving the above equation for II gives us: I=38J1201log(t)log2(1+t)tdt(1) I = -\frac{3}{8}J - \frac{1}{2} \int_0^1 \frac{\log(t)\log^2(1+t)}{t}dt \quad \color{blue}{\cdots (1)} Applying integration by parts and using the generating function of the Harmonic number yields: 01log(t)log2(1+t)tdt=01log2(t)log(1+t)1+tdt=n=1(1)n+1Hn01tnlog2(t)  dt=2n=1(1)n+1Hn(n+1)3=2n=0(1)n+1Hn+11n+1(n+1)3=2n=1(1)n+1Hnn32n=1(1)n+1n4=2Li4(1)+2n=1(1)n+1Hnn3(2) \begin{aligned} \int_0^1 \frac{\log(t)\log^2(1+t)}{t}dt &= -\int_0^1 \frac{\log^2(t)\log(1+t)}{1+t}dt \\ &= -\sum_{n=1}^\infty (-1)^{n+1} H_n \int_0^1 t^{n} \log^2(t) \; dt \\ &= -2\sum_{n=1}^\infty (-1)^{n+1} \frac{H_n}{(n+1)^3} \\ &= -2\sum_{n=0}^\infty (-1)^{n+1} \frac{H_{n+1}-\frac{1}{n+1}}{(n+1)^3} \\ &= 2\sum_{n=1}^\infty (-1)^{n+1} \frac{H_n}{n^3} - 2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^4} \\ &= 2\text{Li}_4(-1) + 2\sum_{n=1}^\infty (-1)^{n+1} \frac{H_n}{n^3} \quad \color{blue}{\cdots (2)} \end{aligned} A similar calculation shows that: J=2ζ(4)2n=1Hnn3(3) J = 2\zeta(4) - 2\sum_{n=1}^\infty \frac{H_n}{n^3} \quad \color{blue}{\cdots (3)} From (A), we know how to evaluate n=1(1)n+1Hnn3\sum_{n=1}^\infty (-1)^{n+1} \frac{H_n}{n^3} and n=1Hnn3\sum_{n=1}^\infty \frac{H_n}{n^3}. n=1Hnn3=π472n=1(1)n+1Hnn3=11π4360+π212log2(2)log4(2)1274log(2)ζ(3)2Li4(12) \begin{aligned} \sum_{n=1}^\infty \frac{H_n}{n^3} &= \frac{\pi^4}{72} \\ \sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^3} &= \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) \end{aligned} Finally, putting everything together gives us: I=3π4160π212log2(2)+log4(2)12+74log(2)ζ(3)+2Li4(12) \boxed{I = -\frac{3\pi^4}{160}-\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)}

Integral #2

The next integral on our list is K=01log(x)log(1+x2)arctan(x)xdxK = \int_0^1 \frac{\log(x)\log(1+x^2)\arctan(x)}{x}dx This integral was originally posted by the user FDP on math.stackexchange.com. My solution is posted there as well.

Using integration by parts gives us: K=1201log2(x)log(1+x2)1+x2dx01xlog2(x)arctan(x)1+x2dx K = - \frac{1}{2}\int_0^1 \frac{\log^2(x)\log(1+x^2)}{1+x^2}dx - \int_0^1 \frac{x \log^2(x) \arctan(x)}{1+x^2}dx Now, we can use the following series expansions to reduce KK into Euler sums: arctan(x)1+x2=n=0(1)nH~nx2n+1,x<1log(1+x2)1+x2=n=1(1)n+1Hnx2n,x<1 \begin{aligned} \frac{\arctan (x)}{1+x^2} &= \sum_{n=0}^\infty (-1)^n \tilde{H}_n x^{2n+1} , \quad |x| < 1\\ \frac{\log(1+x^2)}{1+x^2} &= \sum_{n=1}^\infty (-1)^{n+1} H_n x^{2n} , \quad |x| < 1 \end{aligned} where H~n=i=0n12i+1\tilde{H}_n = \sum_{i=0}^n \frac{1}{2i+1}. This gives us: K=n=0(1)nH~n01x2n+2log2(x)  dx12n=1(1)n+1Hn01x2nlog2(x)  dx=2n=0(1)nH~n(2n+3)3n=1(1)n+1Hn(2n+1)3=2n=1(1)n+1H2n(2n+1)3(4) \begin{aligned} K &= -\sum_{n=0}^\infty (-1)^n \tilde{H}_n\int_0^1 x^{2n+2} \log^2(x)\; dx-\frac{1}{2}\sum_{n=1}^\infty (-1)^{n+1} H_n\int_0^1 x^{2n}\log^2(x)\; dx \\ &= -2\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+3)^3} - \sum_{n=1}^\infty \frac{(-1)^{n+1} H_n}{(2n+1)^3} \\ &= -2\sum_{n=1}^\infty \frac{(-1)^{n+1} H_{2n}}{(2n+1)^3} \quad \color{blue}{\cdots (4)} \end{aligned} We’ll employ the contour integration method to evaluate the above Euler sum. It is a very powerful tool that can handle a large class of Euler sums (see, for e.g. Euler sums and contour integral representations by Philippe Flajolet and Bruno Salvy).

We’ll integrate the function f(z)=πcsc(πz)γ+ψ0(2z+1)(2z+1)3f(z) = \pi \csc(\pi z) \frac{\gamma+\psi_0(-2z+1)}{(-2z+1)^3} around the positively oriented square, CNC_N, with vertices ±(N+14)±(N+14)i\pm \left(N+\frac{1}{4} \right)\pm \left(N+\frac{1}{4} \right)i. It is easy to see that limNCNf(z)  dz=0 \lim_{N\to \infty}\int_{C_N}f(z)\; dz = 0 Hence, the sum of all residues of f(z)f(z) at its poles is equal to 00. A straightforward computation shows that the residues are: Resz=nf(z)=(1)nH2n(2n+1)3,n{0,1,2,}Resz=2n+12f(z)=(1)n+1π16n3,n{1,2,3,}Resz=nf(z)=(1)n+1H2n1(2n1)33(1)n+1(2n1)4,n{1,2,3,}Resz=12f(z)=πζ(3)2 \begin{aligned} \mathop{\text{Res}}\limits_{z=-n} f(z) &= (-1)^n \frac{H_{2n}}{(2n+1)^3} , \quad n\in \{0,1,2,\cdots\} \\ \mathop{\text{Res}}\limits_{z=\frac{2n+1}{2}} f(z) &= \frac{(-1)^{n+1} \pi}{16 n^3} , \quad n\in \{1,2,3,\cdots\} \\ \mathop{\text{Res}}\limits_{z=n} f(z) &= \frac{(-1)^{n+1}H_{2n-1}}{(2n-1)^3}- 3\frac{(-1)^{n+1}}{(2n-1)^4}, \quad n\in \{1,2,3,\cdots \}\\ \mathop{\text{Res}}\limits_{z=\frac{1}{2}} f(z) &= \frac{\pi \zeta(3)}{2} \end{aligned} The list of local expansions of basic kernels given on page 6 of the above mentioned paper are quite useful for carrying out these computations. Now, adding up all the residues gives us: πζ(3)2+n=1(1)nH2n(2n+1)3+π16n=1(1)n+1n3+n=1(1)n+1H2n1(2n1)33n=1(1)n+1(2n1)4=0    πζ(3)2+n=1(1)nH2n(2n+1)3+π16(3ζ(3)4)+n=1(1)nH2n(2n+1)32n=1(1)n+1(2n1)4=0    2n=1(1)n+1H2n(2n+1)3+35πζ(3)642β(4)=0    n=1(1)n+1H2n(2n+1)3=β(4)+35πζ(3)128(5) \begin{aligned} \frac{\pi \zeta(3)}{2}+\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{(2n+1)^3} + \frac{\pi}{16}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3} + \sum_{n=1} ^\infty \frac{(-1)^{n+1}H_{2n-1}}{(2n-1)^3} -3\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-1)^4}&= 0\\ \implies \frac{\pi \zeta(3)}{2}+\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{(2n+1)^3} + \frac{\pi}{16}\left(\frac{3\zeta(3)}{4} \right) + \sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{(2n+1)^3} -2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-1)^4} &= 0 \\ \implies -2\sum_{n=1}^\infty \frac{(-1)^{n+1}H_{2n}}{(2n+1)^3} + \frac{35\pi \zeta(3)}{64} -2\beta(4) = 0 \\ \implies \boxed{\sum_{n=1}^\infty \frac{(-1)^{n+1}H_{2n}}{(2n+1)^3} = -\beta(4) + \frac{35\pi \zeta(3)}{128}} \color{blue}{\cdots (5)} \end{aligned} where β(s)=n=0(1)n(2n+1)s\beta(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s} is the Dirichlet beta function. Finally, plugging equation (5) into (4) gives us: K=2β(4)35πζ(3)64 \boxed{K = 2\beta(4) - \frac{35\pi \zeta(3)}{64}}

Following are some more examples of Euler sums that can be evaluated using the contour integration method: n=0(1)nψ2(n+1)2n+1=8β(4)+π23G7π2ζ(3)n=0(1)nψ1(n+1)(2n+1)2=6β(4)+π24G7π4ζ(3)n=1(1)n+1Hn(2n+1)3=3β(4)+7π16ζ(3)+π316log(2) \begin{aligned} \sum_{n=0}^\infty\frac{(-1)^n\psi_2(n+1)}{2n+1} &= 8\beta(4)+\frac{\pi^2}{3}G-\frac{7\pi}{2}\zeta(3) \\ \sum_{n=0}^\infty\frac{(-1)^n\psi_1(n+1)}{(2n+1)^2} &= 6\beta(4)+\frac{\pi^2}{4}G-\frac{7\pi}{4}\zeta(3) \\ \sum_{n=1}^\infty \frac{(-1)^{n+1} H_n}{(2n+1)^3} &= -3\beta(4)+\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\log(2) \end{aligned} where G=β(2)G=\beta(2) denotes the Catalan’s constant. Some integrals that can be evaluated with the above Euler sums are: 01log2(x)arctan(x)x(1x2)dx=β(4)+7πζ(3)64+π3log(2)3201log(x)arctan(x)arctanh(x)xdx=π216G7πζ(3)320π2xsinxlog2(1+cosxsinx1+cosx+sinx)dx=π26G+4β(4) \begin{aligned} \int_0^1 \frac{\log^2(x)\arctan(x)}{x\left(1-x^2 \right)}dx &=\beta(4)+\frac{7\pi \zeta(3)}{64}+\frac{\pi^3 \log(2)}{32} \\ \int_0^1\frac{\log(x)\arctan(x)\text{arctanh}(x)}{x}dx &= \frac{\pi^2}{16}G-\frac{7\pi\zeta(3)}{32} \\ \int_0^{\frac{\pi}{2}}\frac{x}{\sin x}\log^2\left(\frac{1+\cos x-\sin x}{1+\cos x+\sin x}\right)dx &= \frac{\pi^2}{6}G +4\beta(4) \end{aligned}