A Challenging Logarithmic Integral $\int_0^1 \frac{\log(x)\log(1-x)\log^2(1+x)}{x}dx$

Question

How can we prove that:

$$\int_0^1 \frac{\log(x)\log(1-x)\log^2(1+x)}{x}dx=\frac{7\pi^2}{48}\zeta(3)-\frac{25}{16}\zeta(5)$$ where $\zeta(z)$ is the Riemann Zeta Function.

The best I could do was to express it in terms of Euler Sums. Let $I$ denote the integral.

$$I=-\frac{\pi^2}{24}\zeta(3)+2\sum_{r=2}^\infty \frac{(-1)^r (H_r)^2}{r^3}-2\sum_{r=2}^\infty \frac{(-1)^r H_r}{r^4}+2 \sum_{r=2}^\infty \frac{(-1)^r H_r^{(2)}H_r}{r^2}-2\sum_{r=2}^\infty \frac{(-1)^r H_r^{(2)}}{r^3}$$

where $\displaystyle H_r^{(n)}=\sum_{n=1}^r \frac{1}{k^n}$. I am unable to simplify these sums further. Does anyone have any idea on how to solve this integral?

Please see here for more details.

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Some time ago I was able to solve the simpler integral:

\begin{align*} \int_0^1 \frac{\log(1-x)\log(x)\log(1+x)}{x}dx &=-\frac{3 \pi^4}{160}+\frac{7\log(2)}{4}\zeta(3)-\frac{\pi^2 \log^2(2)}{12} +\frac{\log^4(2)}{12} \\ &\quad+ 2 \text{Li}_4 \left(\frac{1}{2} \right) \end{align*} where $\text{Li}_n(z)$ is the Polylogarithm.

Answer 1

Let the considered integral be denoted by $I$. Our starting point is to reduce the number of logarithms of different arguments in the integrand. Thus, using the fact that $6ab^2=(a+b)^3-2a^3+(a-b)^3$ we obtain \begin{align*} 6I&=\underbrace{\int_0^1\frac{\log x}{x}\log^3(1-x^2)dx}_{x\leftarrow \sqrt{u}} -2\int_0^1\frac{\log x}{x}\log^3(1-x)dx+ \underbrace{\int_0^1\frac{\log x}{x}\log^3\left(\frac{1-x}{1+x}\right)dx}_{x\leftarrow\frac{1-u}{1+u}} \\ &= \frac{1}{4}\int_0^1\frac{\log u}{u}\log^3(1-u)du -2\int_0^1\frac{\log x}{x}\log^3(1-x)dx+ 2\int_0^1\frac{\log^3 u}{1-u^2}\log\left(\frac{1-u}{1+u}\right)du \\ &=\frac{-7}{4}\int_0^1\frac{\log x}{x}\log^3(1-x)dx+ 2\int_0^1\frac{\log^3 x}{1-x^2}\log\left(\frac{1-x}{1+x}\right)dx \\ \end{align*} Thus, $$ I=-\frac{7}{24}J+\frac{1}{3}K \tag{1} $$ with $$ J=\int_0^1\frac{\log(1- x)}{1-x}\log^3x\,dx,\quad K=\int_0^1\frac{1}{1-x^2}\log\left(\frac{1-x}{1+x}\right)\log^3 x\,dx \tag{2} $$ Now, for $x\in(-1,1)$, we have $$ \frac{\log(1-x)}{1-x}=-\left(\sum_{k=0}^\infty x^k\right)\left(\sum_{k=1}^\infty \frac{x^k}{k}\right) =-\sum_{n=1}^\infty H_nx^n $$ where $H_n=\sum_{k=1}^n1/k$ is the $n$-th Harmonic number. Similarly, $$ \frac{1}{1-x^2}\log\left(\frac{1-x}{1+x}\right) =-\left(\sum_{k=0}^\infty x^{2k}\right)\left(\sum_{k=1}^\infty \frac{-2x^{2k-1}}{2k-1}\right) =-2\sum_{n=1}^\infty \widetilde{H}_nx^{2n-1} $$ where $$\widetilde{H}_n=\sum_{k=1}^n\frac{1}{2k-1} =H_{2n}-\frac{1}{2}H_n$$ Thus, using the fact that $\int_0^1x^n(-\log x)^3\,dx=-6/(n+1)^4$ we conclude that \begin{align*} J&=\sum_{n=1}^\infty H_n\int_0^1x^n(-\log x)^3\,dx=6\sum_{n=1}^\infty \frac{H_n}{(n+1)^4}\\ &=6\sum_{n=0}^\infty \frac{1}{(n+1)^4}\left(H_{n+1}-\frac{1}{n+1}\right)=6A-6\zeta(5)\tag{3} \end{align*} with \begin{equation*} A=\sum_{n=1}^\infty\frac{H_n}{n^4}\tag{4} \end{equation*} Similarly, \begin{align*} K&=\sum_{n=1}^\infty (2H_{2n}-H_n)\int_0^1x^{2n-1}(-\log x)^3\,dx=6\sum_{n=1}^\infty\frac{2H_{2n}-H_n}{(2n)^4}\\ &=6\sum_{n=1}^\infty\frac{(1+(-1)^{2n})H_{2n}}{(2n)^4} -\frac{3}{8}\sum_{n=1}^\infty\frac{H_n}{n^4}\\ &=6\sum_{n=1}^\infty\frac{(1+(-1)^{n})H_{n}}{n^4} -\frac{3}{8}\sum_{n=1}^\infty\frac{H_n}{n^4}\\ &=\frac{45}{8}\sum_{n=1}^\infty\frac{H_{n}}{n^4} +6\sum_{n=1}^\infty\frac{(-1)^{n}H_{n}}{n^4}=\frac{45}{8}A+6B\tag{5} \end{align*} with \begin{equation*} B=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}\tag{6} \end{equation*} Combining (1) with (3) and (5) we find that \begin{equation*} I=\frac{1}{8}A+2B+\frac{7}{4}\zeta(5)\tag{7} \end{equation*} Now, sums $A$ and $B$ are known (see here ) (in a more general setting), and we have $$ A=3\zeta(5)-\zeta(2)\zeta(3),\qquad B=-\frac{59}{32}\zeta(5)+\frac{1}{2}\zeta(2)\zeta(3) $$ Thus $$ I=\frac{7}{8}\zeta(2)\zeta(3)-\frac{25}{16}\zeta(5). $$ which is the announced result.$\qquad\square$

Answer 2

On the path of Omran Kouba,

\begin{align} K_1&=\int_0^1 \frac{\ln^3(x)\ln(1-x)}{1-x}dx\\ K_2&=\int_0^1 \frac{\ln^3(x)\ln(1+x)}{1+x}dx\\ U&=\int_0^1 \frac{\ln^3(x)\ln(1+x)}{1-x}dx\\ V&=\int_0^1 \frac{\ln^3(x)\ln(1-x)}{1+x}dx\\ Z&=\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln^3 x}{1-x^2}dx\\ &=\frac{1}{2}\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln^3 x}{1-x}dx+\frac{1}{2}\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln^3 x}{1+x}dx\\ &=\frac{1}{2}K_1-\frac{1}{2}U+\frac{1}{2}V-\frac{1}{2}K_2\\ J&=\int_0^1 \frac{\ln(x)\ln(1-x)\ln^2(1+x)}{x}dx&=\\ J&=-\frac{7}{24}K_1+\frac{1}{3}Z\\ &=-\frac{7}{24}K_1+\frac{1}{3}\left(\frac{1}{2}K_1-\frac{1}{2}U+\frac{1}{2}V-\frac{1}{2}K_2\right)\\ &=\frac{1}{6}V-\frac{1}{6}U-\frac{1}{8}K_1-\frac{1}{6}K_2 \end{align} From Integrating $\int_0^1 \frac{\ln(1+x)\ln^3 x}{1+x}\,dx$ with restricted techniques

One obtains, \begin{align*}K_1&=12\zeta(5)-6\zeta(2)\zeta(3)\\ K_2&=\frac{1}{2}K_1-\frac{9}{16}\zeta(5)\\ U&=-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)+12\zeta(5)\\ V&=\frac{273}{16}\zeta(5)-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)\\ \end{align*}

Therefore,

\begin{align*}J&=\frac{1}{6}\left(\frac{273}{16}\zeta(5)-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)\right)-\frac{1}{6}\left(-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)+12\zeta(5)\right)-\\ &\frac{1}{8}\Big(12\zeta(5)-6\zeta(2)\zeta(3)\Big)-\frac{1}{6}\left(\frac{87}{16}\zeta(5)-3\zeta(2)\zeta(3)\right)\\ &=\frac{7}{8}\zeta(2)\zeta(3)-\frac{25}{16}\zeta(5)\\ &=\boxed{\frac{7}{48}\pi^2\zeta(3)-\frac{25}{16}\zeta(5)} \end{align*}

Answer 3

Here is another solution. We start with @Omran Kouba's simplification:

$$ I = -\frac{7}{24}J + \frac{1}{3}K, $$

where

$$ J := \int_{0}^{1} \frac{\log(1-x)\log^3 x}{1-x} \, \mathrm{d}x \qquad\text{and}\qquad K := \int_{0}^{1} \frac{\log^3 x}{1-x^2} \log\left(\frac{1-x}{1+x}\right) \, \mathrm{d}x. $$

Step 1. To make use of this decomposition, we first introduce two auxiliary integrals:

$$ A := \int_{0}^{1} \frac{\log (1-x)\log^3 x}{(1-x)x} \, \mathrm{d}x \qquad\text{and}\qquad B := \int_{0}^{1} \frac{\log (1+x)\log^3 x}{(1+x)x} \, \mathrm{d}x. $$

Plugging the power series $\frac{\log(1-x)}{1-x} = -\sum_{n=1}^{\infty} H_n x^n$ into the integral and performing term-wise integration, we easily find that

$$ \color{red}{ A = 6 \sum_{n=1}^{\infty} \frac{H_n}{n^4} } \qquad\text{and}\qquad \color{red}{ B = 6 \sum_{n=1}^{\infty} (-1)^n \frac{H_n}{n^4} }. $$

We will demonstrate how to compute these sums in Step 3.

Step 2. Next, we relate $A$ and $B$ to $J$ and $K$. We achieve this by decomposing the respective integrand of $A$ and $B$ algebraically. Indeed, for $J$, we can write

$$\begin{align*} \color{red}{J} &= \int_{0}^{1} \left[ \color{blue}{ \frac{\log(1-x)}{1-x} } \right] \log^3 x \, \mathrm{d}x \\ &= \int_{0}^{1} \left[ \color{blue}{ \frac{\log(1-x)}{(1-x)x} - \frac{\log(1-x)}{x} } \right] \log^3 x \, \mathrm{d}x \\ &= \color{red}{ A - 6\zeta(5)}. \end{align*}$$

The blue part in the first and second line are equal algebraically. Similarly, for $K$, we have

$$\begin{align*} \color{red}{K} &= \int_{0}^{1} \left[ \color{blue}{ \frac{1}{1-x^2} \log\left(\frac{1-x}{1+x}\right) } \right] \log^3 x \, \mathrm{d}x \\ &= \int_{0}^{1} \left[ \color{blue}{ \frac{\log (1-x)}{(1-x)x} + \frac{\log (1+x)}{(1+x)x} - \frac{\log (1-x^2)}{(1-x^2)x} } \right] \log^3 x \, \mathrm{d}x \\ &= \int_{0}^{1} \frac{\log (1-x)\log^3 x}{(1-x)x} \, \mathrm{d}x + \int_{0}^{1} \frac{\log (1+x)\log^3 x}{(1+x)x} \, \mathrm{d}x - \smash[b]{ \underbrace{ \int_{0}^{1} \frac{\log (1-x^2)\log^3 x}{(1-x^2) x} \, \mathrm{d}x }_{x^2 \mapsto x} } \\ &= \color{red}{\frac{15}{16} A + B}. \end{align*}$$

Again, the blue part in the first and second line are identical as functions. Therefore we can express $I$ as a linear combination of $A$, $B$, and $\zeta(5)$:

$$ \color{red}{ I = \frac{1}{48} A + \frac{1}{3} B + \frac{7}{4}\zeta(5). } $$

Step 3. So it remains to compute $A$ and $B$. To this end, we extend the harmonic numbers $H_n$ to the meromorphic function on $\mathbb{C}$ via the identity $H_t = \psi(t+1) - \psi(1)$, where $\psi$ is the digamma function. Then by the Euler's reflection formula, we have

$$ H_t = H_{-1-t} - \frac{\pi}{\tan(\pi t)}. \tag{*} $$

This relation will be used critically in our computation.

Now fix a number $0 < \sigma < 1$. Then by the residue calculation followed by the identity $\text{(*)}$, we obtain

$$\begin{align*} A &= \frac{6}{2\pi i} \int_{\sigma+i\infty}^{\sigma-i\infty} \frac{H_t}{t^4} \frac{\pi}{\tan(\pi t)} \, \mathrm{d}t \\ &= \frac{6}{2\pi i} \int_{\sigma+i\infty}^{\sigma-i\infty} \frac{H_{-1-t}}{t^4} \frac{\pi}{\tan(\pi t)} \, \mathrm{d}t - \frac{6}{2\pi i} \int_{\sigma+i\infty}^{\sigma-i\infty} \frac{1}{t^4} \frac{\pi^2}{\tan^2(\pi t)} \, \mathrm{d}t \end{align*}$$

To proceed, we apply the substitution $t \mapsto -t$ to the first term. This leads to:

$$\begin{align*} &\frac{6}{2\pi i}\int_{\sigma+i\infty}^{\sigma-i\infty} \frac{H_{-1-t}}{t^4} \frac{\pi}{\tan(\pi t)} \, \mathrm{d}t \\ &= - \frac{6}{2\pi i}\int_{-\sigma+i\infty}^{-\sigma-i\infty} \frac{H_{t-1}}{t^4} \frac{\pi}{\tan(\pi t)} \, \mathrm{d}t \\ &= -6 \, \underset{t=0}{\mathrm{Res}} \, \left[ \frac{H_{t-1}}{t^4} \frac{\pi}{\tan(\pi t)} \right] - \frac{6}{2\pi i}\int_{\sigma+i\infty}^{\sigma-i\infty} \frac{H_t - t^{-1}}{t^4} \frac{\pi}{\tan(\pi t)} \, \mathrm{d}t \\ &= -6 \left[ 2\zeta(2)\zeta(3) - \zeta(5) \right] - A + 6\zeta(5). \end{align*}$$

For the second term, note that we have $\int_{\sigma+i\infty}^{\sigma-i\infty} t^{-4} \, \mathrm{d}t = 0$. So we can add a constant to $\frac{\pi^2}{\tan^2(\pi t)}$ without changing the value of the second integral. In particular, we can use $\frac{\pi^2}{\sin^2(\pi t)} = \pi^2 + \frac{\pi^2}{\tan^2(\pi t)}$ instead. Since $(\frac{\pi}{\tan(\pi t)})' = -\frac{\pi^2}{\sin^2(\pi t)}$, we can then perform integration by parts to obtain

$$\begin{align*} - \frac{6}{2\pi i} \int_{\sigma+i\infty}^{\sigma-i\infty} \frac{1}{t^4} \frac{\pi^2}{\tan^2(\pi t)} \, \mathrm{d}t &= - \frac{6}{2\pi i} \int_{\sigma+i\infty}^{\sigma-i\infty} \frac{1}{t^4} \frac{\pi^2}{\sin^2(\pi t)} \, \mathrm{d}t \\ &= \frac{6}{2\pi i} \int_{\sigma+i\infty}^{\sigma-i\infty} \frac{4}{t^5} \frac{\pi}{\tan(\pi t)} \, \mathrm{d}t \\ &= 24\zeta(5). \end{align*}$$

Combining altogether and solving the resulting equation for $A$, we obtain

$$ \color{red}{ A = -6\zeta(2)\zeta(3) + 18\zeta(5) } $$

A similar technique shows that we also have

$$\begin{align*} \color{red}{B} &= \frac{6}{2\pi i} \int_{\sigma+i\infty}^{\sigma-i\infty} \frac{H_t}{t^4} \frac{\pi}{\sin(\pi t)} \, \mathrm{d}t = \color{red}{ 3\zeta(2)\zeta(3) - \frac{177}{16}\zeta(5) }. \end{align*}$$

Plugging these back into $I$, the desired identity is established.

Answer 4

Okay, I simplified my solution. My original solution was about 3 times longer than this, so it is so lucky for me to find a shortcut like this :)

Step 1. Let $I$ be the integral in question: \begin{align*} I &= \int_{0}^{1} \frac{\log x \log (1 - x) \log^{2} (1 + x)}{x} \, dx. \end{align*} By the simple algebraic formula $(a + b)^{3} + (a - b)^{3} - 2 a^{3} = 6 a b^{2}$, it follows that \begin{align*} I &= \frac{1}{6} \int_{0}^{1} \frac{\log x}{x} \left\{ \log^{3} (1-x^{2}) + \log^{3} \left( \frac{1-x}{1+x} \right) - 2\log^{3}(1-x) \right\} \, dx \\ &= \frac{1}{6} \int_{0}^{1} \frac{\log x \log^{3} (1-x^{2})}{x} \, dx + \frac{1}{6} \int_{0}^{1} \frac{\log x \log^{3} \left( \frac{1-x}{1+x} \right)}{x} \, dx - \frac{1}{3} \int_{0}^{1} \frac{\log x \log^{3} (1-x)}{x} \, dx \tag{1} \end{align*} Applying the substitution $x^{2} \mapsto x$, the first integral reduces to \begin{align*} \frac{1}{6} \int_{0}^{1} \frac{\log x \log^{3} (1-x^{2})}{x} \, dx &= \frac{1}{24} \int_{0}^{1} \frac{\log x \log^{3} (1-x)}{x} \, dx = \frac{1}{24} \frac{\partial^{4}\beta}{\partial z \partial w^{3}}(0^{+}, 1) \\ &= \frac{1}{2}\zeta(5) - \frac{1}{4}\zeta(2)\zeta(3). \end{align*} So $\text{(1)}$ can be written as \begin{align*} I &= \frac{7}{4} \zeta(2)\zeta(3) - \frac{7}{2} \zeta(5) + I_{2}, \tag{2} \end{align*} where ( I_{2} ) is given by \begin{align*} I_{2} = \frac{1}{6} \int_{0}^{1} \frac{\log x \log^{3} \left( \frac{1-x}{1+x} \right)}{x} \, dx \end{align*}

Step 2. Now we calculate $I_{2}$. Applying integrating by parts, followed by the substitution $x \mapsto \displaystyle \frac{1-y}{1+y}$, we have \begin{align*} I_{2} &= \left[ \frac{1}{12} \log^{2} x \log^{3} \left( \frac{1-x}{1+x} \right) \right]_{0}^{1} + \frac{1}{2} \int_{0}^{1} \frac{\log^{2} x \log^{2} \left( \frac{1-x}{1+x} \right)}{1 - x^{2}} \, dx \\ &= \frac{1}{4} \int_{0}^{1} \frac{\log^{2} y \log^{2} \left( \frac{1-y}{1+y} \right)}{y} \, dy \\ &= \int_{0}^{1} \frac{\log^{2} y}{y} \left\{ \frac{1}{2} \log^{2} \left( \frac{1+y}{1-y} \right) \right\}^{2} \, dy \\ &= \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{1}{(2m+1)(2n+1)} \int_{0}^{1} y^{2m+2n+1} \log^{2} y \, dy \\ &= \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{2}{(2m+1)(2n+1)(2m+2n+2)^{3}}. \end{align*} By the following partial fraction decomposition \begin{align*} \frac{1}{ab(a+b)^{3}} = \frac{a+b}{a b(a+b)^{4}} = \frac{1}{a(a+b)^{4}} + \frac{1}{b(a+b)^{4}}, \end{align*} together with the symmetry in the role of $m$ and $n$, it follows that \begin{align*} I_{2} &= 4 \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{1}{(2n+1)(2m+2n+2)^{4}} = \frac{1}{4} \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{1}{(2n+1)(m+n+1)^{4}} \\ &= \frac{1}{4} \sum_{n=0}^{\infty} \sum_{m=n+1}^{\infty} \frac{1}{2n+1} \cdot \frac{1}{m^{4}} = \frac{1}{4} \sum_{n=1}^{\infty} \sum_{m=n}^{\infty} \frac{1}{2n-1} \cdot \frac{1}{m^{4}} \\ &= \frac{1}{4} \sum_{m=1}^{\infty} \left( \sum_{n=1}^{m} \frac{1}{2n-1} \right) \frac{1}{m^{4}}. \end{align*} Now by exploiting the following identity \begin{align*} &\int_{0}^{1} \frac{1 - x^{k}}{1 - x} \, dx = 1 + \frac{1}{2} + \cdots + \frac{1}{k} \\ \Longrightarrow & \quad \int_{0}^{1} \frac{1 + x^{k} - 2x^{2k}}{2(1 - x)} \, dx = 1 + \frac{1}{3} + \cdots + \frac{1}{2k-1}, \end{align*} we can rephrase $I_{2}$ as follows: \begin{align*} I_{2} &= \frac{1}{8} \sum_{m=1}^{\infty} \frac{1}{m^{4}} \int_{0}^{1} \frac{1 + x^{m} - 2x^{2m}}{1 - x} \, dx \\ &= \frac{1}{8} \int_{0}^{1} \frac{1}{1-x} \left( \zeta(4) + \mathrm{Li}_{4}(x) - 2 \mathrm{Li}_{4}(x^{2}) \right) \, dx. \end{align*} Writing ( \displaystyle \frac{1}{1-x} = \frac{\mathrm{Li}{0}(x)}{x} ) and applying the simple differentiation rule $$\displaystyle \frac{d}{dx} \mathrm{Li}_{s} (x^{p}) = \frac{p \mathrm{Li}_{s-1}(x^{p})}{x}$$, by applying integrating parts 3 times, \begin{align*} I_{2} &= \frac{1}{8} \left[ \mathrm{Li}_{1}(x) \left( \zeta(4) + \mathrm{Li}_{4}(x) - 2 \mathrm{Li}_{4}(x^{2}) \right) \right]_{0}^{1} - \frac{1}{8} \int_{0}^{1} \frac{\mathrm{Li}_{1}(x)}{x} \left( \mathrm{Li}_{3}(x) - 4 \mathrm{Li}_{3}(x^{2}) \right) \, dx \\ &= -\frac{1}{8} \left[ \mathrm{Li}_{2}(x) \left( \mathrm{Li}_{3}(x) - 4 \mathrm{Li}_{3}(x^{2}) \right) \right]_{0}^{1} + \frac{1}{8} \int_{0}^{1} \frac{\mathrm{Li}_{2}(x)}{x} \left( \mathrm{Li}_{2}(x) - 8 \mathrm{Li}_{2}(x^{2}) \right) \, dx \\ &= \frac{3}{8} \zeta(2)\zeta(3) + \frac{1}{8} \left[ \mathrm{Li}_{3}(x) \left( \mathrm{Li}_{2}(x) - 8 \mathrm{Li}_{2}(x^{2}) \right) \right]_{0}^{1} - \frac{1}{8} \int_{0}^{1} \frac{\mathrm{Li}_{3}(x)}{x} \left( \mathrm{Li}_{1}(x) - 16 \mathrm{Li}_{1}(x^{2}) \right) \, dx \\ &= -\frac{1}{2}\zeta(2)\zeta(3) + \frac{1}{8} \int_{0}^{1} \frac{\mathrm{Li}_{3}(x)}{x} \left( \log(1 - x) - 16 \log(1 - x^{2}) \right) \, dx \\ &= -\frac{1}{2}\zeta(2)\zeta(3) + \frac{1}{8} \int_{0}^{1} \frac{\mathrm{Li}_{3}(x) \log(1 - x)}{x} \, dx - 2 \int_{0}^{1} \frac{\mathrm{Li}_{3}(x) \log(1 - x^{2}) }{x}\, dx. \end{align*} Now applying the substitution $x \mapsto x^{2}$ to the first integral, from the identity $\mathrm{Li}_{3}(x^{2}) = 4 \mathrm{Li}_{3}(x) + 4 \mathrm{Li}_{3}(-x)$ it follows that \begin{align*} \frac{1}{8} \int_{0}^{1} \frac{\mathrm{Li}_{3}(x) \log(1 - x)}{x} \, dx &= \frac{1}{4} \int_{0}^{1} \frac{\mathrm{Li}_{3}(x^{2}) \log(1 - x^{2})}{x} \, dx \\ &= \int_{0}^{1} \frac{ \{ \mathrm{Li}_{3}(x) + \mathrm{Li}_{3}(-x) \} \log(1 - x^{2})}{x} \, dx. \end{align*} This finally gives \begin{align*} I_{2} = -\frac{1}{2}\zeta(2)\zeta(3) - \int_{-1}^{1} \frac{\mathrm{Li}_{3}(x) \log(1 - x^{2}) }{x}\, dx. \end{align*} Plugging this back to ( \text{(2)} ), we obtain \begin{align*} I_{1} = \frac{5}{4}\zeta(2)\zeta(3) - \frac{7}{2}\zeta(5) + I_{3}, \tag{3} \end{align*} where ( I{3} ) is given by \begin{align*} I_{3} = - \int_{-1}^{1} \frac{\mathrm{Li}_{3}(x) \log(1 - x^{2})}{x} \, dx. \end{align*}

Step 3. Finally, we evaluate $I_{3}$ into a closed form and thus complete the calculation. Since the integrand is holomorphic on the unit disc, we can shift up the contour of integration to the clockwise semicircular arc: \begin{align*} I_{3} &= -i \int_{\pi}^{0} \mathrm{Li}_{3}(e^{i\theta}) \log(1 - e^{2i\theta}) \, d\theta = - \frac{1}{i} \int_{0}^{\pi} \mathrm{Li}_{3}(e^{i\theta}) \log(1 - e^{2i\theta}) \, d\theta \end{align*} Now expanding with Taylor series and taking advantage of the fact that $I_{3}$ is real, we have \begin{align*} I_{3} &= \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n^{3}} \frac{1}{m} \int_{0}^{\pi} \{ \cos (n\theta) \sin (2m\theta) + \sin(n\theta) \cos(2m\theta) \} \, d\theta \\ &= \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n^{3}} \frac{1}{m} \int_{0}^{\pi} \cos (n\theta) \sin (2m\theta) \, d\theta \\ &\qquad + \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n^{3}} \frac{1}{m} \int_{0}^{\pi} \sin(n\theta) \cos(2m\theta) \, d\theta \\ &=: S_{1} + S_{2}. \end{align*} To evaluate $S_{1}$, we note the following simple identity \begin{align*} \int_{0}^{\pi} \cos (n\theta) \sin (2m\theta) \, d\theta = \frac{1 + (-1)^{n-1}}{2} \int_{0}^{\pi} \cos (\tfrac{1}{2} n\theta) \sin (m\theta) \, d\theta. \end{align*} Then we can write \begin{align*} S_{1} &= \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{3}} \int_{0}^{\pi} \left\{ \sum_{m=1}^{\infty} \frac{\sin (m\theta)}{m} \right\} \cos((n+\tfrac{1}{2})\theta) \, d\theta, \\ S_{2} &= \sum_{m=1}^{\infty} \frac{1}{m} \int_{0}^{\pi} \left\{ \sum_{n=1}^{\infty} \frac{\sin(n\theta)}{n^{3}} \right\} \cos(2m\theta) \, d\theta. \end{align*} But it is not hard to check that the following Fourier expansion holds: \begin{align*} \sum_{n=1}^{\infty} \frac{\sin n\theta}{n} = \Im \sum_{n=1}^{\infty} \frac{e^{i\theta}}{n} = - \Im \log(1 - e^{i\theta}) = \frac{\pi-\theta}{2}, \quad 0 < \theta < \pi, \end{align*} \begin{align*} \sum_{n=1}^{\infty} \frac{\sin n\theta}{n^{3}} = \frac{\theta ^3}{12}-\frac{\pi \theta ^2}{4}+\frac{\pi ^2 \theta }{6}, \quad 0 < \theta < \pi. \end{align*} Plugging these to $S_{1}$, $S_{2}$ and performing some tedious calculation, we obtain \begin{align*} S_{1} = \sum_{n=0}^{\infty} \frac{2}{(2n+1)^{5}} = \frac{31}{16} \zeta(5) \quad \text{and} \quad S_{2} = -\frac{\pi^{2}}{16} \sum_{m=1}^{\infty} \frac{1}{m^{3}} = -\frac{3}{8}\zeta(2)\zeta(3). \end{align*} Therefore we have \begin{align*} I = \frac{7}{8}\zeta(2)\zeta(3) - \frac{25}{16} \zeta(5). \end{align*}

Answer 5

Some work of professor omran kouba

Let the considered integral be denoted by $I$. Our starting point is to reduce the number of logarithms of different arguments in the integrand. Thus, using the fact that $6ab^2=(a+b)^3-2a^3+(a-b)^3$ we obtain \begin{align*} 6I&=\underbrace{\int_0^1\frac{\log x}{x}\log^3(1-x^2)dx}_{x\leftarrow \sqrt{u}} -2\int_0^1\frac{\log x}{x}\log^3(1-x)dx+ \underbrace{\int_0^1\frac{\log x}{x}\log^3\left(\frac{1-x}{1+x}\right)dx}_{x\leftarrow\frac{1-u}{1+u}} \\ &= \frac{1}{4}\int_0^1\frac{\log u}{u}\log^3(1-u)du -2\int_0^1\frac{\log x}{x}\log^3(1-x)dx+ 2\int_0^1\frac{\log^3 u}{1-u^2}\log\left(\frac{1-u}{1+u}\right)du \\ &=\frac{-7}{4}\int_0^1\frac{\log x}{x}\log^3(1-x)dx+ 2\int_0^1\frac{\log^3 x}{1-x^2}\log\left(\frac{1-x}{1+x}\right)dx \\ \end{align*} Thus, $$ I=-\frac{7}{24}J+\frac{1}{3}K \tag{1} $$ with $$ J=\int_0^1\frac{\log(1- x)}{1-x}\log^3x\,dx,\quad K=\int_0^1\frac{1}{1-x^2}\log\left(\frac{1-x}{1+x}\right)\log^3 x\,dx \tag{2} $$

evaluation of. $J$

$$\begin{aligned} & \text { We know, } \sum_{n \in \mathbb{N}} H_n x^n=-\frac{\ln (1-x)}{1-x},|x|<1 \\ & \Longrightarrow \int_0^1 \frac{\ln (1-x) \ln ^3(x)}{1-x} d x=-\sum_{n \in \mathbb{N}} H_n \int_0^1 x^n \ln ^3(x) d x \\ & \left(\because \int_0^1 t^m \ln ^n(t) d t=\frac{(-1)^n \cdot n!}{(m+1)^{n+1}}, n>-1 \wedge m \neq-1\right) \\ & \Longrightarrow 6 \sum_{n \in \mathbb{N}} \frac{1}{(n+1)^4}\left(H_{n+1}-\frac{1}{n+1}\right) \xlongequal{n \rightarrow n-1} 6 \sum_{n \in \mathbb{N}} \frac{H_n}{n^4}-6 \underbrace{\sum_{n \in \mathbb{N}} \frac{1}{n^5}}_{=\zeta(5)} \\ & \text { Consider, } \sum_{j=1}^{j=q-2} \zeta(q-j) \zeta(j+1)=\sum_{n \in \mathbb{N}} \sum_{m \in \mathbb{N}} \sum_{j=1}^{j=q-2} \frac{1}{m^{q-j} n^{j+1}} \Longrightarrow \sum_{m \in \mathbb{N}}\left(\sum_{n=1}^{n=m-1}+\sum_{\substack{n \geq m \\ m \in \mathbb{N}}}\right) \sum_{j=1}^{j=q-2} \frac{1}{m^{q-j} n^{j+1}} \\ & \Longrightarrow \sum_{m \in \mathbb{N}} \sum_{j=1}^{j=q-2} \frac{1}{m^{q+1}}+\sum_{m \in \mathbb{N}}\left(\sum_{n=1}^{n=m-1}+\sum_{\substack{n \geq m+1 \\ m \in \mathbb{N}}}\right) \sum_{j=1}^{j=q-2} \frac{1}{m^{q-j} n^{j+1}} \Longrightarrow \sum_{j=1}^{j=q-2}\left(\sum_{m \in \mathbb{N}} \frac{1}{m^{q+1}}\right) \\ & +\sum_{m \in \mathbb{N}}\left(\sum_{n=1}^{n=m-1}+\sum_{n \geq m+1}\right) \frac{1}{n m^q} \underbrace{\sum_{j=1}^{j=q-2}\left(\frac{m}{n}\right)^j}_{\text {Geometric Sum }} \Longrightarrow \sum_{j=1}^{j=q-2} \zeta(q+1) \\ & +\sum_{m \in \mathbb{N}}\left(\sum_{n=1}^{n=m-1}+\sum_{\substack{n \geq m+1 \\ m \in \mathbb{N}}}\right)\left(\frac{1}{n m^{q-1}(n-m)}-\frac{1}{m n^{q-1}(n-m)}\right) \Longrightarrow(q-2) \zeta(q+1) \\ & +\sum_{m \in \mathbb{N}} \sum_{n=1}^{n=m-1}\left(\frac{1}{n m^{q-1}(n-m)}-\frac{1}{m n^{q-1}(n-m)}\right)+\sum_{m \in \mathbb{N}} \sum_{n \geq m+1}\left(\frac{1}{n m^{q-1}(n-m)}-\frac{1}{m n^{q-1}(n-m)}\right) \\ & \end{aligned}$$

$$ \begin{gathered} \Longrightarrow \sum_{j=1}^{j=q-2} \zeta(q-j) \zeta(j+1)=(q+2) \zeta(q+1)-2 \sum_{m \in \mathbb{N}} \frac{H_m}{m^q} \\ \Longrightarrow \sum_{m \in \mathbb{N}} \frac{H_m}{m^q}=\left(1+\frac{q}{2}\right) \zeta(q+1)-\frac{1}{2} \sum_{j=1}^{j=q-2} \zeta(q-j) \zeta(j+1) \\ \xlongequal[\because \zeta(2)=\pi^2 / 6]{\text { Set } q=4} \sum_{m \in \mathbb{N}} \frac{H_m}{m^4}=3 \zeta(5)-\frac{1}{2}\left(\frac{\pi^2}{6} \zeta(3)+\frac{\pi^2}{6} \zeta(3)\right) \Longrightarrow 3 \zeta(5)-\frac{\pi^2}{6} \zeta(3) \end{gathered} $$

Put the value of above harmonic sum in equation-(1). We get, $$ \begin{gathered} \Longrightarrow 6 \sum_{n \in \mathbb{N}} \frac{H_n}{n^4}-6 \zeta(5)=6\left(3 \zeta(5)-\frac{\pi^2}{6} \zeta(3)\right)-6 \zeta(5) \\ \Longrightarrow 18 \zeta(5)-\pi^2 \zeta(3)-6 \zeta(5) \\ \Longrightarrow \int_0^1 \frac{\ln (1-x) \ln ^3(x)}{1-x} d x=12 \zeta(5)-\pi^2 \zeta(3) \end{gathered} $$

$$\begin{aligned} & \left(\because \sum_{m \in \mathbb{N}} \sum_{n=1}^{n=m} a_m b_n=\sum_{n \in \mathbb{N}} \sum_{m \geq n} a_m b_n\right) \\ & \begin{array}{l}\Longrightarrow(q-2) \zeta(q+1)+\underbrace{\sum_{m \geq n+1} \sum_{n \in m^2}\left(\frac{1}{n m^{q-1}(n-m)}-\frac{1}{m n^{q-1}(n-m)}\right)}_{n \in \mathbb{N}} \\ +\sum_{m \in \mathbb{N}} \sum_{n \geq m+1}\left(\frac{1}{n m^{q-1}(n-m)}-\frac{1}{m n^{q-1}(n-m)}\right)^m \Longrightarrow(q-2) \zeta(q+1)\end{array} \\ & +2 \underbrace{\sum_{m \in \mathbb{N}} \sum_{n \geq m+1}\left(\frac{1}{n m^{q-1}(n-m)}-\frac{1}{m n^{q-1}(n-m)}\right)}_{n \rightarrow n+m} \Longrightarrow(q-2) \zeta(q+1) \\ & +2 \sum_{m \in \mathbb{N}} \sum_{n \in \mathbb{N}}\left(\frac{1}{n m^{q-1}(n+m)}-\frac{1}{n m(n+m)^{q-1}}\right) \Longrightarrow(q-2) \zeta(q+1)+2 \sum_{n \in \mathbb{N}} \sum_{m \in \mathbb{N}} \frac{1}{n m^{q-1}(n+m)} \\ & -2 \sum_{n \in \mathbb{N}}^{m \in \mathbb{N}} \sum_{m \in \mathbb{N}} \frac{1}{n m(n+m)^{q-1}} \Longrightarrow(q-2) \zeta(q+1)+2 \sum_{m \in \mathbb{N}} \frac{1}{m^q} \sum_{n \in \mathbb{N}} \frac{m+n-n}{n(n+m)}-2 \sum_{n \in \mathbb{N}} \sum_{m \in \mathbb{N}} \frac{1}{n(n+m)^q} \\ & -2 \underbrace{\sum_{n \in \mathbb{N}} \sum_{m \in \mathbb{N}} \frac{1}{m(n+m)^q}}_{n \leftrightarrow m} \Longrightarrow(q-2) \zeta(q+1)+2 \sum_{m \in \mathbb{N}} \frac{1}{m^q} \underbrace{\sum_{n \in \mathbb{N}}\left(\frac{1}{n}-\frac{1}{n+m}\right)}_{=H_m}-4 \underbrace{\sum_{n \in \mathbb{N}} \sum_{m \in \mathbb{N}} \frac{1}{n(n+m)^q}}_{m \rightarrow m-n} \\ & \Longrightarrow(q-2) \zeta(q+1)+2 \sum_{m \in \mathbb{N}} \frac{H_m}{n^q}-4 \sum_{n \in \mathbb{N}}\left[-\frac{1}{n^{q+1}}+\sum_{m \geq n+1}^{=H_m}\left(\frac{1}{n m^q}\right)+\frac{1}{n^{q+1}}\right] \\ & \Longrightarrow(q-2) \zeta(q+1)+2 \sum_{m \in \mathbb{N}} \frac{H_m}{m^q}-4 \sum_{n \in \mathbb{N}} \sum_{m \geq n} \frac{1}{n m^q}+4 \sum_{n \in \mathbb{N}} \frac{1}{n^{q+1}} \\ & \left(\because \sum_{m \in \mathbb{N}} \sum_{n=1}^{n=m} a_m b_n=\sum_{n \in \mathbb{N}} \sum_{m \geq n} a_m b_n\right) \\ & \Longrightarrow(q-2) \zeta(q+1)+4 \zeta(q+1)+2 \sum_{m \in \mathbb{N}} \frac{H_m}{m^q}-4 \sum_{m \in \mathbb{N}} \frac{1}{m^q} \underbrace{\sum_{n=1}^{n=m} \frac{1}{n}}_{=H_m} \\ & \end{aligned}$$

$$ \begin{gathered} \Longrightarrow \sum_{j=1}^{j=q-2} \zeta(q-j) \zeta(j+1)=(q+2) \zeta(q+1)-2 \sum_{m \in \mathbb{N}} \frac{H_m}{m^q} \\ \Longrightarrow \sum_{m \in \mathbb{N}} \frac{H_m}{m^q}=\left(1+\frac{q}{2}\right) \zeta(q+1)-\frac{1}{2} \sum_{j=1}^{j=q-2} \zeta(q-j) \zeta(j+1) \\ \xlongequal[\because \zeta(2)=\pi^2 / 6]{\text { Set } q=4} \sum_{m \in \mathbb{N}} \frac{H_m}{m^4}=3 \zeta(5)-\frac{1}{2}\left(\frac{\pi^2}{6} \zeta(3)+\frac{\pi^2}{6} \zeta(3)\right) \Longrightarrow 3 \zeta(5)-\frac{\pi^2}{6} \zeta(3) \end{gathered} $$

Put the value of above harmonic sum in equation-(1). We get, $$ \begin{gathered} \Longrightarrow 6 \sum_{n \in \mathbb{N}} \frac{H_n}{n^4}-6 \zeta(5)=6\left(3 \zeta(5)-\frac{\pi^2}{6} \zeta(3)\right)-6 \zeta(5) \\ \Longrightarrow 18 \zeta(5)-\pi^2 \zeta(3)-6 \zeta(5) \\ \Longrightarrow \int_0^1 \frac{\ln (1-x) \ln ^3(x)}{1-x} d x=12 \zeta(5)-\pi^2 \zeta(3) \end{gathered} $$

Evaluation of $k$.

$$\begin{aligned} &\left.\int_0^1 \frac{\ln ^3(x)}{1-x^2} \ln \left(\frac{1-x}{1+x}\right) d x \Longrightarrow \int_0^1 \frac{\ln ^3(x) \ln (1-x y)}{1-x^2}\right|_{y=-1} ^{y=+1} d x \\ &\Longrightarrow \int_{-1}^{+1} \int_0^1 \frac{x \ln ^3(x)}{\left(1-x^2\right)(1-x y)} d x d y \\ &\xlongequal[\text { fraction }]{\text { Partial }} \int_{-1}^{+1} \int_0^1 \frac{y \ln ^3(x)}{\left(1-y^2\right)(1-x y)} d x d y-\int_{-1}^{+1} \int_0^1 \frac{x \ln ^3(x)}{\left(1-y^2\right)\left(1-x^2\right)} d x d y \\& \stackrel{=}{\text { fraction }} \int_{-1}^{+1} \int_0^1 \frac{y \ln ^3(x)}{\left(1-y^2\right)(1-x y)} d x d y-\int_{-1}^{+1} \int_0^1 \frac{x \ln ^3(x)}{\left(1-y^2\right)\left(1-x^2\right)} d x d y \\ & -\int_{-1}^{+1} \int_0^1 \frac{y \ln ^3(x)}{\left(1-y^2\right)\left(1-x^2\right)} d x d y = \int_{-1}^{+1} \frac{y}{1-y^2} \sum_{n \in \mathbb{N}} y^{n-1} \int_0^1 x^{n-1} \ln ^3(x) d x d y \\ & -\int_{-1}^{+1} \frac{1}{1-y^2} \sum_{n \in \mathbb{N}} \int_0^1 x^{2 n-1} \ln ^3(x) d x d y-\int_{-1}^{+1} \frac{y}{1-y^2} \sum_{n \in \mathbb{Z}_0^{+}} \int_0^1 x^{2 n} \ln ^3(x) d x d y = (1) \\ & -6 \int_{-1}^{+1} \frac{1}{1-y^2} \sum_{n \in \mathbb{N}} \frac{y^n}{n^4} d y \end{aligned}$$ $$\begin{aligned} & =[\underbrace{\left[\left(\frac{\pi^4}{120}-\frac{3}{4} \operatorname{Li}_4\left(x^2\right)\right) \frac{\tanh ^{-1}(x)}{2}\right]_{-1}^{+1}}_{=0}+\frac{3}{4} \int_{-1}^{+1} \frac{\operatorname{Li}_3\left(x^2\right)}{x} \tanh ^{-1}(x) d x \\ & \rightarrow\left(\frac{3 \operatorname{Li}_3\left(x^2\right)\left(\operatorname{Li}_2(x)-\mathrm{Li}_2(-x)\right)}{8}\right)_{-1}^{+1}-\frac{3}{4} \int_{-1}^{+1} \frac{\operatorname{Li}_2\left(x^2\right)\left(\operatorname{Li}_2(x)-\mathrm{Li}_2(-x)\right)}{x} d x \\ & =[\because \operatorname{Li}_2(-1)=-\pi^2 / 12]{\because \operatorname{Li}_3(1)=\zeta(3) ; \because \operatorname{Li}_2(1)=\pi^2 / 6} \frac{3 \pi^2}{16} \zeta(3)-\frac{3}{4} \int_{-1}^{+1} \frac{\operatorname{Li}_2\left(x^2\right)\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)\right)}{x} d x \\ & = \frac{3 \pi^2}{16} \zeta(3) - \frac{3}{2} \underbrace{\int_{-1}^{+1} \frac{\mathrm{Li}_2^2(x)}{x} d x}_{x \rightarrow -x} + \frac{3}{2} \int_{-1}^{+1} \frac{\mathrm{Li}_2^2(-x)}{x} d x \Longrightarrow \frac{3 \pi^2}{16} \zeta(3) + 3 \underbrace{\int_{-1}^0 \frac{\mathrm{Li}_2^2(-x)}{x} d x}_{x \rightarrow -x} + 3 \int_0^1 \frac{\mathrm{Li}_2^2(-x)}{x} d x \\ & \Longrightarrow \frac{3 \pi^2}{16} \zeta(3) - 3 \int_0^1 \frac{\mathrm{Li}_2^2(x)}{x} d x + 3 \int_0^1 \frac{\mathrm{Li}_2^2(-x)}{x} d x \\ & = \frac{3 \pi^2}{16} \zeta(3) - 3 \int_0^1 \frac{\mathrm{Li}_2^2(x)}{x} d x + 3 \int_0^1\left(\frac{\mathrm{Li}_2\left(x^2\right)}{2}-\mathrm{Li}_2(x)\right)^2 \frac{d x}{x} \\ & \Longrightarrow \frac{3 \pi^2}{16} \zeta(3) - 3 \int_0^1 \frac{\mathrm{Li}_2^2(x)}{x} d x + \frac{3}{4} \underbrace{\int_0^1 \frac{\mathrm{Li}_2^2\left(x^2\right)}{x} d x}_{x^2 \rightarrow x} - 3 \int_0^1 \frac{\operatorname{Li}_2(x) \mathrm{Li}_2\left(x^2\right)}{x} d x + 3 \int_0^1 \frac{\mathrm{Li}_2^2(x)}{x} d x \\ & \Longrightarrow \frac{3 \pi^2}{16} \zeta(3) + \frac{3}{8} \int_0^1 \frac{\mathrm{Li}_2^2(x)}{x} d x - 3 \int_0^1 \frac{\mathrm{Li}_2(x) \mathrm{Li}_2\left(x^2\right)}{x} d x \\ & \Longrightarrow \frac{3 \pi^2}{16} \zeta(3) + \frac{3}{8} \sum_{n \in \mathbb{N}} \frac{1}{n^2} \underbrace{\int_0^1 x^{n-1} \operatorname{Li}_2(x) d x}_{\text {I.B.P. }} - 3 \sum_{n \in \mathbb{N}} \frac{1}{n^2} \underbrace{\int_0^1 x^{2 n-1} \operatorname{Li}_2(x) d x}_{\text {I.B.P. }} \\ & \Longrightarrow \frac{3 \pi^2}{16} \zeta(3) + \frac{3}{8} \sum_{n \in \mathbb{N}} \frac{1}{n^2}\left[\left(\frac{x^n \operatorname{Li}_2(x)}{n}\right)_0^1+\frac{1}{n} \int_0^1 x^{n-1} \ln (1-x) d x\right] \\ & \end{aligned}$$ $$\begin{align*} & -3 \sum_{n \in \mathbb{N}} \frac{1}{n^2}\left[\left(\frac{x^{2 n} \operatorname{Li}_2(x)}{2 n}\right)_0^1+\frac{1}{2 n} \int_0^1 x^{2 n-1} \ln (1-x) d x\right] = \frac{3 \pi^2}{16} +\frac{\pi^2}{16} \sum_{n \in \mathbb{N}} \sum_{n(3)}^{Y^3}-\frac{3}{8} \sum_{n \in \mathbb{N}} \frac{H_n}{n^4}-\frac{\pi^2}{4} \sum_{=\zeta(3)}^{\sum_{n \in \mathbb{N}} \sum_n}+\frac{3}{4} \sum_{n \in \mathbb{N}} \frac{H_{2 n}}{n^4} \\ & \Longrightarrow-\frac{3}{8} \sum_{n \in \mathbb{N}} \frac{H_n}{n^4}+6 \sum_{n \in \mathbb{N}} \frac{H_n}{n^4} +6 \sum_{n \in \mathbb{N}} \frac{(-1)^n H_n}{n^4} \Longrightarrow \frac{45}{8} \sum_{n \in \mathbb{N}} \frac{H_n}{n^4}+6 \sum_{n \in \mathbb{N}} \frac{(-1)^n H_n}{n^4} \\ & \Longrightarrow \frac{135}{8} \zeta(5)-\frac{15 \pi^2}{16} \zeta(3)+\frac{\pi^2}{2} \zeta(3)-\frac{177}{16} \zeta(5) \\ & \Longrightarrow \int_0^1 \frac{\ln ^3(x)}{1-x^2} \ln \left(\frac{1-x}{1+x}\right) d x=\frac{93}{16} \zeta(5)-\frac{7 \pi^2}{16} \zeta(3) \end{align*}$$

Finally

$$ I=-\frac{7}{24}J+\frac{1}{3}K \tag{1} $$ With $$\int_0^1\frac{\log(1- x)}{1-x}\log^3x\,dx= \left(12\zeta(5)-\pi^2\zeta(3)\right)$$ And $$\int_0^1\frac{1}{1-x^2}\log\left(\frac{1-x}{1+x}\right)\log^3 x\,dx= \left(\frac{93}{16}\right)\zeta(5)-\frac{7\pi^2}{16} \zeta(3)$$ Finally $$-\frac{7}{24}\left(12\zeta(5)-\pi^2\zeta(3)\right) + \frac{1}{3}\left(\frac{93}{16}\right)\zeta(5)-\frac{7\pi^2}{16} \zeta(3) = \frac{7 \pi^2 \zeta(3)}{48} - \frac{25 \zeta(5)}{16}$$