An easy looking integral, not so easy...PART 2

In this previous post We have computed the following integral

$\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx=\frac{11}{8}\zeta(3)$ (1)

Today we are going to compute the third integral of the list, namely:

$\boxed{\int_{0}^{1}\frac{\log(1-x)\log(1-x^4)}{x}dx=\frac{67}{32}\zeta(3)-\frac{\pi}{2}G}$

Lets get to it!

Start with

$\int_{0}^{1}\frac{\log(1-x)\log(1-x^4)}{x}dx=\int_{0}^{1}\frac{\log(1-x)\log[(1-x^2)(1+x^2)]}{x}dx$

$=\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx+\int_{0}^{1}\frac{\log(1-x)\log(1+x^2)}{x}dx$

By (1) we have

$\underbrace{\int_{0}^{1}\frac{\log(1-x)\log(1-x^4)}{x}dx}_{J}=\frac{11}{8}\zeta(3)+\int_{0}^{1}\frac{\log(1-x)\log(1+x^2)}{x}dx$

$J=\frac{11}{8}\zeta(3)+I$ (2)

We now focus in

$I=\int_{0}^{1}\frac{\log(1-x)\log(1+x^2)}{x}dx=\sum_{k=1}^{\infty}\frac{(-1)^{n-1}}{k}\int_{0}^{1}x^{2k-1}\log(1-x)dx$

Integrating by parts

$I=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\bigg\{\frac{\log(1-x)(x^{2n}-1)}{2n}\Big|_{0}^{1}+\frac{1}{2n}\int_{0}^{1}\frac{x^{2n}-1}{1-x}dx \bigg\}$

$=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\bigg\{-\frac{1}{2n}\int_{0}^{1}\frac{1-x^{2n}}{1-x}dx \bigg\}$

$=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\bigg\{-\frac{1}{2n}H_{2n} \bigg\}$

$\boxed{I=\frac{1}{2}\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}}$ (3)

So now, the task is to evaluate the Sum (3)

To calculate this sum, consider the following generating function proved here eq. (10)

$\sum_{n=1}^\infty \frac{x^n}{n^2}H_n=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\operatorname{Li}_2(1-x)\ln(1-x)+\frac{1}{2}\ln x \ln^2(1-x)+\zeta(3)$

Letting x=i

The left hand side becomes

$\sum_{n=1}^\infty \frac{i^n}{n^2}H_n=-\frac{H_2}{2^2}+\frac{H_4}{4^2}-\frac{H_6}{6^2}+\frac{H_8}{8^2}- \cdots+i\Big(-\frac{H_1}{1^2}+\frac{H_3}{3^2}-\frac{H_5}{5^2}+\frac{H_7}{7^2}- \cdots \Big)$

$\sum_{n=1}^\infty \frac{i^n}{n^2}H_n=\frac{1}{4}\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}+i\sum_{n=1}^\infty \frac{(-1)^n}{(2n-1)^{2}}H_{2n-1}$

We therefore have that

$Re\Big\{ \sum_{n=1}^\infty \frac{i^n}{n^2}H_n\Big\}=\frac{1}{4}\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}$

$\boxed{\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}=4\Big\{ \sum_{n=1}^\infty \frac{i^n}{n^2}H_n\Big\}}$

$Re\Big\{ \sum_{n=1}^\infty \frac{i^n}{n^2}H_n\Big\}=Re\Big\{ \operatorname{Li}_3(i)-\operatorname{Li}_3(1-i)+\operatorname{Li}_2(1-i)\ln(1-i)+\frac{1}{2}\ln (i) \ln^2(1-i)+\zeta(3)\Big\} \quad \quad (4)$

Lets now compute the components of (4) and then, put all together to get the solution.

$\ln(i)=\ln(e^{i\frac{\pi}{2}})=i\frac{\pi}{2}\quad \quad$ (5)

$\ln(1-i)=\ln(\sqrt{2}e^{-i\frac{\pi}{4}})=\frac{1}{2}\ln(2)-i\frac{\pi}{4}\quad \quad$ (6)

and

$\ln^2(1-i)=\Big(\frac{1}{2}\ln(2)-i\frac{\pi}{4}\Big)^2=\frac{1}{4}\ln^2(2)-\frac{\pi^2}{16}-i\frac{ \pi}{4}\ln(2)\quad \quad \tag4$ (7)

$Li_{2}(i)=\sum_{n=1}^{\infty}\frac{i^n}{n^2}=\frac{i}{1^2}+\frac{(i)^2}{2^2}+\frac{(i)^3}{3^2}+\frac{(i)^4}{4^2}+\cdots$

$=\frac{i}{1^2}-\frac{1}{2^2}-\frac{i}{3^2}+\frac{1}{4^2}+\cdots$

$=\Big[-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{6^2}+\frac{1}{8^2}-\cdots \Big]+i\Big[\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\cdots \Big]$

$Li_{2}(i)=\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)^2}+i\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^2}$

$Li_{2}(i)=\frac{1}{4}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}+i\beta(2)$

$Li_{2}(i)=-\frac{1}{4}\eta(2)+i G$

$Li_{2}(i)=-\frac{1}{4} \cdot\frac{\pi^2}{12}+i G$

$\boxed{Li_{2}(i)=-\frac{1}{8}\zeta(2)+iG}\quad \quad \tag5$ (8)

Recall the Dilogarithm reflection formula (proof here, eq. (4))

$Li_{2}(x)+Li_{2}(1-x)=\zeta(2)-\ln(x)\ln(1-x)$

Plug x=i

$Li_{2}(1-i)=\zeta(2)-\ln(i)\ln(1-i)-Li_{2}(i)$

$Li_{2}(1-i)=\zeta(2)-\left(i\frac{\pi}{2}\right) \cdot\left( \frac{\ln(2)}{2}-i\frac{\pi}{4}\right)+\frac{1}{8}\zeta(2)-iG$

$Li_{2}(1-i)=\zeta(2)+\frac{1}{8}\zeta(2)-\frac{\pi^2}{8}-i\left(\frac{\pi ln(2)}{4}+G\right)$

$Li_{2}(1-i)=\zeta(2)+\frac{1}{8}\frac{\pi^2}{6}-\frac{\pi^2}{8}-i\left(\frac{\pi ln(2)}{4}+G\right)$

$Li_{2}(1-i)=\zeta(2)-\frac{5}{8}\frac{\pi^2}{6}-i\left(\frac{\pi ln(2)}{4}+G\right)$

$\boxed{Li_{2}(1-i)=\frac{3}{8}\zeta(2)-i\left(\frac{\pi ln(2)}{4}+G\right)}\quad \quad \tag6$ (9)

$Li_{3}(x)=\sum_{k=1}^{\infty}\frac{x^k}{k^3}$

$Li_{3}(i)=\sum_{k=1}^{\infty}\frac{i^k}{k^3}=\frac{i}{1^3}+\frac{i^2}{2^3}+\frac{i^3}{3^3}+\frac{i^4}{4^3}+\cdots$

$=\frac{i}{1^3}-\frac{1}{2^3}-\frac{i}{3^3}+\frac{1}{4^3}+\cdots$

$=-\left[\frac{1}{2^3}-\frac{1}{4^3}+\frac{1}{6^3}-\cdots \right]+i\left[\frac{1}{1^3}-\frac{1}{3^3}+\frac{1}{5^3}-\cdots \right]$

$=-\sum_{k=1}^{\infty}\frac{(-1)^{n-1}}{(2k)^3}+i\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^3}$

$=-\frac{1}{8}\eta(3)+i\beta(3)$

$\boxed{Li_{3}(i)=-\frac{3}{32}\zeta(3)+i\beta(3)}\quad \quad \tag7$ (10)

Recall the identity proved here equation (5)

$Li_{3}(x)+Li_{3}(1-x)+Li_{3}\Big(1-\frac{1}{x}\Big)=\zeta(3)+\frac{\ln^3(x)}{6}+\frac{\pi^2\ln(x)}{6}-\frac{\ln^2(x)\ln(1-x)}{2}$

$x=i=\frac{i \pi}{2}$

$1-\frac{1}{i}=1+i=(1-i)^{*}$

therefore

$Li_{3}(x)+Li_{3}(1-i)+Li_{3}\big((1-i)^{*}\big)=\zeta(3)+\frac{\ln^3(i)}{6}+\frac{\pi^2\ln(i)}{6}-\frac{\ln^2(i)\ln(1-i)}{2}\quad \quad (11)$

but

$1-i=\sqrt{2}e^{-\frac{i\pi}{4}}$

and

$1+i=\sqrt{2}e^{\frac{i\pi}{4}}$

$Li_{3}(1-i)+Li_{3}\big((1-i)^{*}\big)$

$Li_{3}(1-i)=\sum_{n=1}^{\infty}\frac{2^{k/2}e^{-\frac{ik \pi}{4}}}{k^3}$

$Li_{3}\big((1-i)^{*}\big)=\sum_{n=1}^{\infty}\frac{2^{k/2}e^{\frac{ik \pi}{4}}}{k^3}$

$Li_{3}(1-i)+Li_{3}\big((1-i)^{*}\big)=\sum_{n=1}^{\infty}\frac{2^{k/2}e^{\frac{ik \pi}{4}}}{k^3}+\sum_{n=1}^{\infty}\frac{2^{k/2}e^{-\frac{ik \pi}{4}}}{k^3}$

$\boxed{Li_{3}(1-i)+Li_{3}\big((1-i)^{*}\big)=2\sum_{n=1}^{\infty}\frac{2^{k/2} \cos\big( \frac{k\pi}{4}\big)}{k^3}}\quad \quad \tag9$ (12)

on the other hand

$Re\{ Li_{3}(1-i)\}= Re\left \{\sum_{n=1}^{\infty}\frac{2^{k/2} e^{-\frac{i \pi}{4}}}{k^3} \right\}$

$\boxed{Re\{ Li_{3}(1-i)\}=\sum_{n=1}^{\infty}\frac{2^{k/2} \cos\big( \frac{k\pi}{4}\big)}{k^3}}\quad \quad \tag{10}$ (13)

From (12) and (13) we conclude that

$\boxed{Li_{3}(1-i)+Li_{3}\big((1-i)^{*}\big)=2Re\{ Li_{3}(1-i)\}}\quad \quad \tag{11}$ (14)

We can now rewrite (11) as

$Li_{3}(x)+2Re\{ Li_{3}(1-i)\}=\zeta(3)+\frac{\ln^3(i)}{6}+\frac{\pi^2\ln(i)}{6}-\frac{\ln^2(i)\ln(1-i)}{2}$

$2Re\{ Li_{3}(1-i)\}=\zeta(3)+\frac{\ln^3(i)}{6}+\frac{\pi^2\ln(i)}{6}-\frac{\ln^2(i)\ln(1-i)}{2}-Li_{3}(x)$

$Re\{ Li_{3}(1-i)\}=\frac{\zeta(3)}{2}+\frac{\ln^3(i)}{12}+\frac{\pi^2\ln(i)}{12}-\frac{\ln^2(i)\ln(1-i)}{4}-\frac{Li_{3}(x)}{2}$

$Re\{ Li_{3}(1-i)\}=\frac{\zeta(3)}{2}+\frac{1}{12}\frac{-i\pi^3}{48}+\frac{\pi^2}{12}\frac{i \pi}{2}+\frac{1}{4}\frac{\pi^2}{4}\Big( \frac{\ln(2)}{2}+i\frac{\pi}{4}\Big)-\frac{Li_{3}(x)}{2}$

$Re\left\{ Li_{3}(1-i)\right\}=\frac{35}{64}\zeta(3)-\frac{\pi^2}{32}\ln(2)$

$\boxed{Re\left\{ Li_{3}(1-i)\right\}=\frac{35}{64}\zeta(3)-\frac{3}{16}\zeta(2)\ln(2)}\quad \quad \tag{12}$ (15)

From (4) and (6) we have

$\ln(i)\ln^2(1-i)=\left(i\frac{\pi}{2} \right)\left(\frac{1}{4}\ln^2(2)-\frac{\pi^2}{16}-i\frac{ \pi}{4}\ln(2) \right)$

$\boxed{\ln(i)\ln^2(1-i)=\frac{\pi^2}{8}\ln(2)-i\left[\frac{\pi^3}{32}-\frac{\pi}{8}\ln^2(2) \right]}\quad \quad \tag{13}$ (16)

$\ln(1-i)Li_{2}(1-i)=\left(\frac{1}{2}\ln(2)-i\frac{\pi}{4} \right)\left(\frac{3}{8}\zeta(2)-i\left(\frac{\pi ln(2)}{4}+G\right) \right)$

$\ln(1-i)Li_{2}(1-i)=\frac{3}{16}\zeta(2)\ln(2)-\frac{\pi^2 \ln(2)}{16}-\frac{\pi }{4}G-i\frac{1}{2}\ln(2)\left(\frac{\pi ln(2)}{4}+G\right)-i\frac{3 \pi}{32}\zeta(2)$

$\ln(1-i)Li_{2}(1-i)=\frac{\pi^2}{32}\ln(2)-\frac{\pi^2 \ln(2)}{16}-\frac{\pi }{4}G-i\frac{1}{2}\ln(2)\left(\frac{\pi ln(2)}{4}+G\right)-i\frac{3 \pi}{32}\zeta(2)$

$\boxed{\ln(1-i)Li_{2}(1-i)=-\frac{\pi^2}{32}\ln(2)-\frac{\pi }{4}G-i\left(\frac{\pi ln^2(2)}{8}+\frac{1}{2}\ln(2)G+\frac{ \pi^3}{64}\right)}\quad \quad \tag{14}$ (17)

Plugging the real part of (10), (15), (16) and (17) in (4) we get

$Re\left\{ \sum_{n=1}^\infty \frac{i^n}{n^2}H_n \right \}=-\frac{3}{32}\zeta(3)-\frac{35}{64}\zeta(3)-\frac{\pi^2}{32}\ln(2)-\frac{\pi^2}{32}\ln(2)-\frac{\pi}{4}G+\frac{\pi^2}{16}\ln(2)+\zeta(3)$