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Prove that:

$$ I=\int_0^1\frac{\log^2(1+x)\log(x)\log(1-x)}{1-x}dx=\frac{7}{2}\zeta(3){\log^22}-\frac{\pi^2}{6}{\log^32}-\frac{\pi^2}{2}\zeta(3)+{6}\zeta(5)-\frac{\pi^4}{48}\ln2 $$ Using integration by parts:

$$u=\log^2(1+x) \log x$$ thus $$du=\left[\frac{\log^2(1+x)}{x}+2\frac{\log x\log(1+x)}{1+x}\right]\,dx, v=\log^2(1-x)$$ We have: $$I= \left[-\frac{1}{2} \log^2(1-x) \log x \log^2(1+x)\right]^1_0+\frac{1}{2} \int^1_0 \log^2(1-x) \log^2(1+x) \frac{dx}{x} + \int^1_0 \log x \log(1+x) \log^2(1-x) \frac{dx}{1+x}$$

How to calculate these last two integrals?

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    $\begingroup$ can see math.stackexchange.com/questions/503405/… $\endgroup$ – math110 Jan 8 '15 at 15:22
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    $\begingroup$ One cannot help but wonder if one can show all these integrals are in the algebra generated by $\pi,\log 2,\zeta(n)$ and $e$ over $\Bbb Q$. $\endgroup$ – Pedro Tamaroff Jan 8 '15 at 15:42
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    $\begingroup$ By writing $\log(1+x)=\log(1-x^2)-\log(1-x)$ and using the technique of differentiation under the integral sign, we can see that such integral depends on a linear combination of derivatives of the beta function in particular points. Despite that, why this particular integral should be interesting? $\endgroup$ – Jack D'Aurizio Jan 8 '15 at 15:51
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    $\begingroup$ Ah, you again!? $\endgroup$ – Venus Jan 8 '15 at 16:48
  • $\begingroup$ @JackD'Aurizio Could you elaborate? How does one evaluate an integral such as $\int_0^1 \ln(1-x^2)\ln(1-x) \ln x\, dx/(1-x)$ in terms of derivatives of the Beta function? $\endgroup$ – user111187 Jan 8 '15 at 17:34
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Denote the integral as $I$. With some algebra, $I$ can be simplified like so. \begin{align} \small \small{I} &\small{=\int^1_0\frac{\log{x}\log(1-x)\log^2(1+x)}{1-x}}\ dx\\ &\small{=\frac{1}{6}\int^1_0\frac{\log{x}\log^3(1-x^2)}{1-x}\ dx+\frac{1}{6}\int^1_0\frac{\log{x}\log^3\left(\frac{1-x}{1+x}\right)}{1-x}\ dx-\frac{1}{3}\int^1_0\frac{\log{x}\log^3(1-x)}{1-x}\ dx}\\ &\small{=\frac{1}{6}\int^1_0\frac{\log{x}\log^3(1-x^2)}{1-x^2}\ dx+\frac{1}{6}\int^1_0\frac{\log{x}\log^3\left(\frac{1-x}{1+x}\right)}{1-x}\ dx-\frac{7}{24}\int^1_0\frac{\log{x}\log^3(1-x)}{1-x}\ dx}\\ &\small{=\frac{1}{6}\int^1_0\frac{\log{x}\log^3(1-x^2)}{1-x^2}\ dx+\frac{1}{6}\int^1_0\frac{\log^3{x}\log\left(\frac{1-x}{1+x}\right)}{x(1+x)}\ dx-\frac{7}{24}\int^1_0\frac{\log^3{x}\log(1-x)}{x}dx}\\ &\small{=\frac{1}{6}\int^1_0\frac{\log{x}\log^3(1-x^2)}{1-x^2}\ dx-\frac{1}{6}\int^1_0\frac{\log^3{x}\log\left(1-x^2\right)}{x}dx+\frac{1}{24}\int^1_0\frac{\log^3{x}\log(1-x)}{x}dx}\\ &\small{\ -\frac{1}{6}\int^1_0\frac{\log^3{x}\log(1-x)}{1+x}}dx+\frac{1}{6}\int^1_0\frac{\log^3{x}\log(1+x)}{1+x}dx\\ &\small{=\frac{1}{6}\int^1_0\frac{\log{x}\log^3(1-x^2)}{1-x^2}\ dx+\frac{1}{32}\int^1_0\frac{\log^3{x}\log(1-x)}{x}dx-\frac{1}{6}\int^1_0\frac{\log^3{x}\log(1-x^2)}{1+x}dx}\\ &\small{\ +\frac{1}{3}\int^1_0\frac{\log^3{x}\log(1+x)}{1+x}dx}\\ &\small{=\color{red}{\frac{1}{6}\int^1_0\frac{\log{x}\log^3(1-x^2)}{1-x^2}\ dx}+\color{darkorange}{\frac{1}{32}\int^1_0\frac{\log^3{x}\log(1-x)}{x}dx}-\color{darkgreen}{\frac{1}{6}\int^1_0\frac{\log^3{x}\log(1-x^2)}{1-x^2}dx}}\\ &\small{\ +\color{blue}{\frac{1}{96}\int^1_0\frac{\log^3{x}\log(1-x)}{1-x}dx}+\color{purple}{\frac{1}{3}\int^1_0\frac{\log^3{x}\log(1+x)}{1+x}dx}}\\ &=\color{red}{I_1}+\color{darkorange}{I_2}-\color{darkgreen}{I_3}+\color{blue}{I_4}+\color{purple}{I_5} \end{align} Clearly $I_2=\dfrac{3\zeta(5)}{16}$ is trivial.

$I_1$ and $I_3$ can be evaluated by differentiating the Beta function and expanding at $0$. \begin{align} I_1 &=\frac{1}{24}\frac{\partial^4}{\partial a\partial b^3}B\left(\frac{1}{2},0^+\right)=\frac{31\zeta(5)}{4}-\frac{\pi^4}{24}\log{2}-\frac{13\pi^2}{24}\zeta(3)+\frac{7\zeta(3)}{2}\log^2{2}-\frac{\pi^2}{6}\log^3{2}\tag1\\ I_3 &=\frac{1}{96}\frac{\partial^4}{\partial a^3\partial b}B\left(\frac{1}{2},0^+\right)=\frac{31\zeta(5)}{8}-\frac{\pi^4}{48}\log{2}-\frac{7\pi^2}{32}\zeta(3)\tag2 \end{align} $I_4$ and $I_5$ can be expressed in terms of well-known Euler sums. \begin{align} I_4 &=\frac{1}{16}\sum^\infty_{n=1}\frac{H_{n-1}}{n^4}=\frac{\zeta(5)}{8}-\frac{\pi^2}{96}\zeta(3)\tag3\\ I_5 &=2\sum^\infty_{n=1}\frac{(-1)^{n-1}H_{n-1}}{n^4}=\frac{29\zeta(5)}{16}-\frac{\pi^2}{6}\zeta(3)\tag4\\ \end{align} Therefore, we obtain the final result $$\boxed{\displaystyle I=6\zeta(5)-\frac{\pi^4}{48}\log{2}-\frac{\pi^2}{2}\zeta(3)+\frac{7\zeta(3)}{2}\log^2{2}-\frac{\pi^2}{6}\log^3{2}}$$


Details on $(1)$

Differentiating $B(a,b)$ with respect to $a$ once, we have $$\frac{\partial}{\partial a}B\left(\frac{1}{2},b\right)=B\left(\frac{1}{2},b\right)\left(\psi_0\left(\frac{1}{2}\right)-\psi_0\left(b+\frac{1}{2}\right)\right)$$ By Legendre's Duplication Formula, \begin{align} B\left(\frac{1}{2},b\right) &=2^{2b-1}\frac{\Gamma^2(b)}{\Gamma(2b)}\\ &=\frac{2^{2b}}{b}\frac{\Gamma^2(1+b)}{\Gamma(1+2b)}\\ &=\frac{\exp\left(2b\log{2}+2\log\Gamma(1+b)-\log\Gamma(1+2b)\right)}{b}\\ &=\frac{1}{b}\exp\left(2b\log{2}-\frac{\pi^2}{6}b^2+2\zeta(3)b^3+\mathcal{O}(b^4)\right)\\ &=\frac{1}{b}+2\log{2}-\frac{\pi^2}{6}b+2\zeta(3)b^2+2b\log^2{2}-\frac{\pi^2}{3}b^2\log{2}+\frac{4\log^3{2}}{3}b^2+\mathcal{O}(b^3)\\ &=\frac{1}{b}+2\log{2}+\left(2\log^2{2}-\frac{\pi^2}{6}\right)b+\left(2\zeta(3)-\frac{\pi^2}{3}\log{2}+\frac{4\log^3{2}}{3}\right)b^2+\mathcal{O}(b^3) \end{align} Furthermore, one can compute \begin{align} \psi_0\left(\frac{1}{2}\right)-\psi_0\left(b+\frac{1}{2}\right) &=-\frac{\pi^2}{2}b+7\zeta(3)b^2-\frac{\pi^4}{6}b^3+31\zeta(5)b^4+\mathcal{O}(b^5) \end{align} simply by taking successive derivatives and evaluating them at $b=0$. It follows that \begin{align} I_1 &=\operatorname*{Res}_{b=0}\frac{B\left(\frac{1}{2},b\right)\left(\psi_0\left(\frac{1}{2}\right)-\psi_0\left(b+\frac{1}{2}\right)\right)}{4b^4}\\ &=\frac{31\zeta(5)}{4}-\frac{\pi^4}{12}\log{2}+\frac{7\zeta(3)}{4}\left(2\log^2{2}-\frac{\pi^2}{6}\right)-\frac{\pi^2}{8}\left(2\zeta(3)-\frac{\pi^2}{3}\log{2}+\frac{4\log^3{2}}{3}\right)\\ &=\frac{31\zeta(5)}{4}-\frac{\pi^4}{24}\log{2}-\frac{13\pi^2}{24}\zeta(3)+\frac{7\zeta(3)}{2}\log^2{2}-\frac{\pi^2}{6}\log^3{2} \end{align} More generally, if I had made no mistake, \begin{align} &\frac{4}{m!}\int^1_0\frac{\log{x}\log^m(1-x^2)}{1-x^2}dx+(-1)^{m}(2^{m+2}-1)\zeta(m+2)\\ &=\sum^m_{k=0}\sum^{k+1}_{j=1}\left[\frac{(-1)^{{m-k}}(2^{m-k+1}-1)\zeta(m-k+1)}{j!}\sum_{\substack{r_1,\cdots,r_j\geq 1 \\ r_1+\cdots+r_j=k+1}}\left(\prod^j_{i=1}\frac{(-1)^{r_i-1}2^{r_i}\eta(r_i)}{r_i}\right)\right] \end{align}


Details on $(2)$

Differentiate $B(a.b)$ with respect to $a$ and $b$ once each and obtain the first term of the series expansion at $b=0$. \begin{align} \small{\frac{\partial^2}{\partial a\partial b}B(a,0^+)} &\small{=\lim_{b\to 0^+} B(a,b)\left[(\psi_0(a)-\psi_0(a+b))(\psi_0(b)-\psi_0(a+b))-\psi_1(a+b)\right]}\\ &\small{=\frac{1}{b}\left[\left(\psi_1(a)b+\frac{\psi_2(a)}{2}b^2+\mathcal{O}(b^3)\right)\left(\frac{1}{b}+\gamma+\psi_0(a)+\mathcal{O}(b)\right)-\psi_1(a)-\psi_2(a)b+\mathcal{O}(b^2)\right]_{b\to0^+}}\\ &\small{=(\gamma+\psi_0(a))\psi_1(a)-\frac{\psi_2(a)}{2}} \end{align} Therefore, differentiating the above expression twice with respect to $a$ yields \begin{align} I_3 &=\frac{1}{96}\left[(\gamma+\psi_0(a))\psi_3(a)+3\psi_1(a)\psi_2(a)-\frac{\psi_4(a)}{2}\right]_{a=\frac12}\\ &=\frac{1}{96}\left[(-2\log{2})\pi^4+3\left(\frac{\pi^2}{2}\right)(-14\zeta(3))+372\zeta(5)\right]\\ &=\frac{31\zeta(5)}{8}-\frac{\pi^4}{48}\log{2}-\frac{7\pi^2}{32}\zeta(3) \end{align} Applying Leibniz's product rule gives us a more general result. \begin{align} 2\sum^\infty_{n=0}\frac{H_n}{(n+a)^{m+1}} &=2\frac{(-1)^{m-1}}{m!}\int^1_0\frac{x^{a-1}\ln^m{x}\ln(1-x)}{1-x}dx\\ &=\frac{(-1)^{m-1}}{m!}\left[2(\gamma+\psi_0(a))\psi_{m}(a)+2\sum^{m-1}_{k=1}\binom{m-1}{k}\psi_{k}(a)\psi_{m-k}(a)-\psi_{m+1}(a)\right]\\ \end{align}


Details on $(3)$

It is possible to prove $(3)$ using only algebraic manipulations. \begin{align} 80I_4 &=-\zeta(5)+\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{1}{n^4}\left(\frac{1}{k}-\frac{1}{k+n}\right)+2\sum^\infty_{n=1}\sum^{n-1}_{k=1}\frac{1}{n^4}\left(\frac{1}{k}+\frac{1}{n-k}\right)\\ &=-\zeta(5)+\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{1}{kn^3(k+n)}+2\sum^\infty_{k=1}\sum^\infty_{n=k+1}\frac{1}{kn^3(n-k)}\\ &=-\zeta(5)+\frac{1}{2}\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(k+n)^2-2kn}{k^3n^3(k+n)}+2\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{1}{kn(k+n)^3}\\ &=-\zeta(5)+\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{1}{k^2n^3}-\sum^\infty_{n=1}\sum^\infty_{k=1}\left(\frac{1}{k^2n^2(k+n)}-\frac{2}{kn(k+n)^3}\right)\\ &=-\zeta(5)+\zeta(2)\zeta(3)-\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{k^2+n^2}{k^2n^2(k+n)^3}\\ &=-\zeta(5)+\zeta(2)\zeta(3)-2\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{1}{n^2(k+n)^3}\\ &=-\zeta(5)+\zeta(2)\zeta(3)-2\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{1}{k^3n^2}+2\sum_{k=n}\frac{1}{k^3n^2}+\color{brown}{2\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{1}{k^3(k+n)^2}}\\ &=-\zeta(5)+\zeta(2)\zeta(3)-2\zeta(2)\zeta(3)+2\zeta(5)-\color{brown}{\zeta(2)\zeta(3)+48I_4+3\zeta(5)}\\ &=4\zeta(5)-2\zeta(2)\zeta(3)+48I_4\\ \\ \implies I_4 &=\frac{\zeta(5)}{8}-\frac{\zeta(2)\zeta(3)}{16} \end{align} I used the identity $$2\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{1}{k^3(k+n)^2}=-\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{1}{k^3n^2}+3\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{1}{k^4}\left(\frac{1}{n}-\frac{1}{n+k}\right)$$ which follows from the partial fraction decomposition of $\frac{1}{n^2(k+n)^3}$ and the obvious fact that $\sum_{n,k}\frac{1}{n^2(k+n)^3}=\sum_{n,k}\frac{1}{k^2(k+n)^3}$.


Details on $(4)$

Here is an elementary proof of $(4)$ which uses a similar approach as the one robjohn used in his answer here.

$\underline{\textbf{5 Different Representations of S}}$

Denote $S=\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^4}$. Then \begin{align} S &=\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^n}{kn^3(k+n)}\tag{a}\\ 2S+\frac{15\zeta(5)}{8} &=\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{kn(k+n)^3}\tag{b}\\ -S-\frac{3\zeta(2)\zeta(3)}{4} &=\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^k}{k^2n^2(k+n)}\tag{c}\\\ 2S+\frac{21\zeta(5)}{8}-\frac{\zeta(2)\zeta(3)}{4} &=\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{k^2n^2(k+n)}\tag{d}\\\ -S-\frac{43\zeta(5)}{16}+\frac{7\zeta(2)\zeta(3)}{8} &=\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^k}{kn(k+n)^3}\tag{e} \end{align} $\underline{\text{Proof of (a) and (c)}}$ \begin{align} S &=\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^n}{n^4}\left(\frac{1}{k}-\frac{1}{k+n}\right) =\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^n}{kn^3(k+n)}\\ &=\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^n[(k+n)-n]}{k^2n^3(k+n)} =\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^n}{k^2n^3}-\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^n}{k^2n^2(k+n)}\\ &=-\frac{3\zeta(2)\zeta(3)}{4}-\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^k}{k^2n^2(k+n)} \end{align} $\underline{\text{Proof of (b)}}$ \begin{align} S &=-\frac{15\zeta(5)}{16}+\frac{1}{2}\sum^\infty_{n=1}\sum^{n-1}_{k=1}\frac{(-1)^n}{n^4}\left(\frac{1}{k}+\frac{1}{n-k}\right) =-\frac{15\zeta(5)}{16}+\sum^\infty_{k=1}\sum^\infty_{n=k+1}\frac{(-1)^n}{kn^3(n-k)}\\ &=-\frac{15\zeta(5)}{16}+\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{kn(k+n)^3} \end{align} $\underline{\text{Proof of (d) and (e)}}$

Since $1+(-1)^k+(-1)^n+(-1)^{k+n}$ equals $4$ if and only if both $k,n$ are even and $0$ otherwise, \begin{align} 0 &=7\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{1}{k^2n^2(k+n)}+16\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^k}{k^2n^2(k+n)}+8\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{k^2n^2(k+n)}\\ &=7\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{1}{kn(k+n)^3} +16\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^k}{kn(k+n)^3}+8\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{kn(k+n)^3} \end{align} Using the results $\sum_{k,n}\frac{1}{k^2n^2(k+n)}=-3\zeta(5)+2\zeta(2)\zeta(3)$ and $\sum_{k,n}\frac{1}{kn(k+n)^3}=4\zeta(5)-2\zeta(2)\zeta(3)$, which were obtained in the proof of $(3)$, in tandem with $\text{(b)}$ and $\text{(c)}$ yields $\text{(d)}$ and $\text{(e)}$.


$\underline{\textbf{Main Proof of (4)}}$ \begin{align} -2S &=\frac{29\zeta(5)}{16}+\frac{5\zeta(2)\zeta(3)}{8}+\color{red}{S}-\color{blue}{\left(-S-\frac{43\zeta(5)}{16}+\frac{7\zeta(2)\zeta(3)}{8}\right)}-\color{#00A000}{\left(2S+\frac{15\zeta(5)}{8}\right)}\\ &\ \ \ \ -\color{purple}{\left(2S+\frac{21\zeta(5)}{8}-\frac{\zeta(2)\zeta(3)}{4}\right)}\\ &=\frac{29\zeta(5)}{16}+\frac{5\zeta(2)\zeta(3)}{8}+\sum^\infty_{n=1}\sum^\infty_{k=1}\left[\color{red}{\frac{(-1)^n}{kn^3(k+n)}\frac{k(k+n)+n^2+kn}{(k+n)^2}}-\color{blue}{\frac{(-1)^n}{kn(k+n)^3}}\right]\\ &\ \ \ \ -\color{#00A000}{\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{kn(k+n)^3}}-\color{purple}{\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{k^2n^2(k+n)}}\\ &=\frac{29\zeta(5)}{16}+\frac{5\zeta(2)\zeta(3)}{8}+\sum^\infty_{n=1}\sum^\infty_{k=1}\left[\color{magenta}{\frac{(-1)^n}{n^3(k+n)^2}}+\color{darkorange}{\frac{(-1)^n}{n^2(k+n)^3}}\right]-\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{kn(k+n)^3}\\ &\ \ \ \ -\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{k^2n^2(k+n)}\\ &=\frac{29\zeta(5)}{16}+\frac{5\zeta(2)\zeta(3)}{8}\color{magenta}{-\frac{3\zeta(2)\zeta(3)}{4}+\frac{15\zeta(5)}{16}-\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{k^2(k+n)^3}}\color{darkorange}{-\frac{\zeta(2)\zeta(3)}{2}+\frac{15\zeta(5)}{16}}\\ &\ \ \ \ \color{darkorange}{-\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{k^3(k+n)^2}}-\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{kn(k+n)^3}-\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{k^2n^2(k+n)}\\ &=\frac{59\zeta(5)}{16}-\frac{5\zeta(2)\zeta(3)}{8}-\frac{1}{2}\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}[\color{#C00000}{1}(k+n)^2\color{#C0A000}{-2}kn]}{k^2n^2(k+n)^3}\\ &-\color{salmon}{\frac{1}{2}\sum^\infty_{n=1}\sum^\infty_{k=1}}\frac{\color{salmon}{(-1)^{k+n}[(k+n)^3}\color{limegreen}{-3}kn(k+n)]}{\color{salmon}{k^3n^3(k+n)^2}}-\color{darkcyan}{\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{kn(k+n)^3}}-\color{orchid}{\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{k^2n^2(k+n)}}\\ &=\frac{59\zeta(5)}{16}-\frac{5\zeta(2)\zeta(3)}{8}+\left(-\frac{1}{2}\cdot\color{#C00000}{(1)}-\frac{1}{2}\color{limegreen}{\cdot(-3)}-1\cdot \color{orchid}{(1)}\right)\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{k^2n^2(k+n)}\\ &\ \ \ \ +\left(-\frac{1}{2}\color{#C0A000}{\cdot(-2)}-1\cdot\color{darkcyan}{(1)}\right)\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^{k+n}}{kn(k+n)^3}-\color{salmon}{\frac{3 \zeta(2)\zeta(3)}{8}}\\ &=\frac{59\zeta(5)}{16}-\zeta(2)\zeta(3) \end{align} Thus $S=-\frac{59\zeta(5)}{32}+\frac{\zeta(2)\zeta(3)}{2}$ and $I_5=\frac{29\zeta(5)}{16}-\zeta(2)\zeta(3)$.

The more general identity $$2 \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{n^{2k}} = (2k+1) \eta(2k+1) - \zeta(2k+1) - 2 \sum_{m=1}^{k-1} \zeta(2k+1-2m) \eta(2k)$$ was proven here using contour integration.

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    $\begingroup$ +1) It's answers like this that make Math.SE such an invaluable resource. I think you used every technique in the book for solving these log integrals! $\endgroup$ – David H Oct 1 '16 at 12:28

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