Prove $\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^3=-\frac5{16}\zeta(3)$

Question

How to prove that

$$S=\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^3=-\frac5{16}\zeta(3)$$

where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number.

I came up with this problem after I solved a similar one here. I managed to prove the equality above but I am not happy with my solution as I used Mathematica for calculating the blue integral below, so any better ideas and without using softwares? Thank you,

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My solution:

In page $105$ of this paper we have $$\overline{H}_n-\ln2=(-1)^{n-1}\int_0^1\frac{x^n}{1+x}dx$$

$$\Longrightarrow S=-\int_0^1\int_0^1\int_0^1\frac{1}{(1+x)(1+y)(1+z)}\sum_{n=0}^\infty(xyz)^n\ dx\ dy\ dz$$

$$=-\int_0^1\int_0^1\int_0^1\frac{dx\ dy\ dz}{(1+x)(1+y)(1+z)(1-xyz)}$$

$$=-\int_0^1\int_0^1\frac{dx\ dy}{(1+x)(1+y)}\left(\int_0^1\frac{dz}{(1+z)(1-xyz)}\right)$$

$$=-\int_0^1\int_0^1\frac{dx\ dy}{(1+x)(1+y)}\left(-\frac{\ln(1-xy)-\ln2}{1+xy}\right)$$

$$=\int_0^1\frac{dx}{1+x}\left(\int_0^1\frac{\ln(1-xy)-\ln2}{(1+y)(1+xy)}dy\right)$$

Mathematica gives

$$\color{blue}{\int_0^1\frac{\ln(1-xy)-\ln2}{(1+y)(1+xy)}dy}$$

$$\small{=\frac{1}{x-1}\left[\frac{\pi^2}{12}+\frac12\ln^22-\ln(1-x)\ln\left(\frac{2x}{1+x}\right)+\operatorname{Li}_2\left(\frac{1}{1+x}\right)-\operatorname{Li}_2\left(\frac{1+x}{2}\right)-\operatorname{Li}_2\left(\frac{1-x}{1+x}\right)\right]}$$

giving us

$$ S=-\int_0^1\frac{\frac{\pi^2}{12}+\frac12\ln^22-\ln(1-x)\ln\left(\frac{2x}{1+x}\right)+\operatorname{Li}_2\left(\frac{1}{1+x}\right)-\operatorname{Li}_2\left(\frac{1+x}{2}\right)-\operatorname{Li}_2\left(\frac{1-x}{1+x}\right)}{1-x^2}dx$$

By integration by parts and some simplifications, we get

$$S=\underbrace{2\int_0^1\tanh^{-1}x\frac{\ln(1-x)-\ln2}{1+x}dx}_{\Large\mathcal{I}_1}-\underbrace{\int_0^1\frac{\tanh^{-1}x\ln(1-x)}{x}dx}_{\Large\mathcal{I}_2}$$

For $\mathcal{I}_1$ we know that $\tanh^{-1}x=-\frac12\ln\left(\frac{1-x}{1+x}\right)$, so set $\frac{1-x}{1+x}=u$

$$\Longrightarrow \mathcal{I}_1=\int_0^1\ln u\frac{\ln(1+u)-\ln u}{1+u}du=\boxed{-\frac{13}{8}\zeta(3)}$$

For $\mathcal{I}_2$ use $\tanh^{-1}x=\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}$

$$\Longrightarrow \mathcal{I}_2=\sum_{n=0}^\infty\frac1{2n+1}\int_0^1 x^{2n}\ln(1-x)\ dx=-\sum_{n=0}^\infty\frac{H_{2n+1}}{(2n+1)^2}$$

$$=-\sum_{n=0}^\infty\frac{H_{n+1}}{(n+1)^2}\left(\frac{1+(1)^n}{2}\right)=-\sum_{n=1}^\infty\frac{H_{n}}{n^2}\left(\frac{1-(1)^n}{2}\right)$$

$$=-\frac12\sum_{n=1}^\infty\frac{H_n}{n^2}+\frac12\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}=\boxed{-\frac{21}{16}\zeta(3)}$$

where we used $\sum_{n=1}^\infty\frac{H_n}{n^2}=2\zeta(3)$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}=-\frac58\zeta(3)$

Combine the boxed results, we get the claimed closed form of $S$.

Answer 1

Let's continue from where you showed that: $$S=\int_0^1\int_0^1\frac{1}{(1+x)(1+y)}\left(\frac{\ln(1-xy)-\ln2}{1+xy}\right)dydx$$ $$\overset{xy=t}=\int_0^1\int_0^x \frac{1}{(1+x)(x+t)}\frac{\ln\left(\frac{1-t}{2}\right)}{1+t}dtdx=\int_0^1\color{blue}{\int_t^1\frac{1}{(1+x)(x+t)}}\frac{\ln\left(\frac{1-t}{2}\right)}{1+t}\color{blue}{dx}dt $$ $$=\int_0^1 \frac{\color{blue}{\ln\left(\frac{(1+t)^2}{4t}\right)}\ln\left(\frac{1-t}{2}\right)}{\color{blue}{(1-t)}(1+t)}dt\overset{\large t\to\frac{1-x}{1+x}}=\frac12 \int_0^1 \frac{\ln(1-x^2)\ln\left(\frac{1+x}{x}\right)}{x}dx$$ $$=\frac12\int_0^1 \frac{\ln(1-x^2)\ln(1+x)}{x}dx-\frac12\int_0^1 \frac{\ln(1-x^2)\ln x}{x}dx$$ The first integral equals $-\frac{3\zeta(3)}{8}$ and can be found by plugging $m,n,q=1$ and $p=0$ in this general result: $$\small \int_0^1 \frac{[m\ln(1+x)+n\ln(1-x)][q\ln(1+x)+p\ln(1-x)]}{x}dx=\left(\frac{mq}{4}-\frac{5}{8}(mp+nq)+2np\right)\zeta(3)$$ The second integral can be expanded into power series: $$\int_0^1 \frac{\ln(1-x^2)\ln x}{x}dx\overset{x^2=t}=\frac14\int_0^1 \frac{\ln(1-t)\ln t}{t}dt$$ $$=-\frac14\sum_{n=1}^\infty \frac{1}{n}\int_0^1t^{n-1}\ln t\, dt=\frac14\sum_{n=1}^\infty \frac{1}{n^3}=\frac{\zeta(3)}{4}$$ $$\Rightarrow \sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^3=\frac12\left(-\frac{3\zeta(3)}{8}-\frac{\zeta(3)}{4}\right)=-\frac{5\zeta(3)}{16}$$

Answer 2

Different way to calculate $\mathcal{I}_1-\mathcal{I}_2$:

We have from the previous solution

$$\mathcal{I}_1=\int_0^1\ln x\frac{\ln(1+x)-\ln x}{1+x}dx$$

$$=\underbrace{\int_0^1\frac{\ln x\ln(1+x)}{1+x}dx}_{IBP}-\int_0^1\frac{\ln^2x}{1+x}dx$$

$$=-\frac12\int_0^1\frac{\ln^2(1+x)}{x}dx-\int_0^1\frac{\ln^2x}{1+x}dx$$

and

$$\mathcal{I}_2=\int_0^1\frac{\tanh^{-1}x\ln(1-x)}{x}dx$$

$$=-\frac12\int_0^1\frac{\ln^2(1-x)}{x}dx+\frac12\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}dx$$

so

$$\mathcal{I}_1-\mathcal{I}_2=\frac12\int_0^1\frac{\color{blue}{\ln^2(1-x)-\ln^2(1+x)-\ln(1-x)\ln(1+x)}}{x}\ dx-\int_0^1\frac{\ln^2x}{1+x}dx$$

now use the algebraic identity

$$a^2-b^2-ab=2a^2-\frac34(a+b)^2-\frac14(a-b)^2$$

with $a=\ln(1-x)$ and $b=\ln(1+x)$ we get

$$\mathcal{I}_1-\mathcal{I}_2=\frac12\int_0^1\frac{\color{blue}{\ln^2(1-x)-\frac34\ln^2(1-x^2)-\frac14\ln^2\left(\frac{1-x}{1+x}\right)}}{x}\ dx-\int_0^1\frac{\ln^2x}{1+x}dx$$

$$=\int_0^1\frac{\ln^2(1-x)}{x}dx-\frac38\underbrace{\int_0^1\frac{\ln^2(1-x^2)}{x}dx}_{\large x^2\mapsto x}-\frac18\underbrace{\int_0^1\frac{\ln^2\left(\frac{1-x}{1+x}\right)}{x}dx}_{\large\frac{1-x}{1+x}\mapsto x}-\int_0^1\frac{\ln^2x}{1+x}dx$$

$$=\frac{13}{16}\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}dx}_{1-x\mapsto x}-\frac14\int_0^1\frac{\ln^2x}{1-x^2}dx-\int_0^1\frac{\ln^2x}{1+x}dx$$

$$=\frac{11}{16}\int_0^1\frac{\ln^2x}{1-x}dx-\frac98\int_0^1\frac{\ln^2x}{1+x}dx=-\frac{5}{16}\zeta(3)$$