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How to evaluate

$$I=\int_0^1\frac{\ln(1-x^2)}{x}\operatorname{Li}_2\left(\frac{1-x}{2}\right)\ dx\ ?$$

This integral was mentioned by @nospoon in the comments of this problem.

What I tried is integration by parts which gives

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

Now if we use the following identity that can be found on page $95$ Eq $(4)$ of this paper

$$\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)x^n=-\ln(2)+\sum_{n=1}^\infty(-1)^n(\overline{H}_n-\ln2)x^n=\frac{\ln\left(\frac{1+x}{2}\right)}{1-x}$$

and multiply both sides by $\large \frac{\operatorname{Li}_2(x^2)}{x}$ then $\int_0^1$, we obtain

$$\int_0^1\frac{\operatorname{Li}_2(x^2)}{x(1-x)}\ln\left(\frac{1+x}{2}\right)\ dx=-\frac12\ln2\zeta(3)+\sum_{n=1}^\infty(-1)^n(\overline{H}_n-\ln2)\int_0^1 x^{n-1}\operatorname{Li}_2(x^2)\ dx$$

where

$$\int_0^1 x^{n-1}\operatorname{Li}_2(x^2)\ dx\overset{x^2\to x}{=}\frac12\int_0^1 x^{\frac n2-1}\operatorname{Li}_2(x)\ dx=\frac12\left(\frac{2\zeta(2)}{n}-\frac{4H_{n/2}}{n^2}\right)$$

so

$$\int_0^1\frac{\operatorname{Li}_2(x^2)}{x(1-x)}\ln\left(\frac{1+x}{2}\right)\ dx=-\frac12\ln2\zeta(3)+\sum_{n=1}^\infty(-1)^n(\overline{H}_n-\ln2)\left(\frac{\zeta(2)}{n}-\frac{2H_{n/2}}{n^2}\right)$$

and since

$$\int_0^1\frac{\operatorname{Li}_2(x^2)}{x(1-x)}\ln\left(\frac{1+x}{2}\right)\ dx=\int_0^1\frac{\operatorname{Li}_2(x^2)}{x}\ln\left(\frac{1+x}{2}\right)\ dx+2I$$

therefore

$$I=-\frac14\ln2\zeta(3)+\frac12\color{blue}{\sum_{n=1}^\infty(-1)^n(\overline{H}_n-\ln2)\left(\frac{\zeta(2)}{n}-\frac{2H_{n/2}}{n^2}\right)}-\frac12\underbrace{\int_0^1\frac{\operatorname{Li}_2(x^2)}{x}\ln\left(\frac{1+x}{2}\right)\ dx}_{\text{manageable}}$$

any idea how to evalute the blue sum? I think I made it more complicated. Any other ideas?

Thank you.

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    $\begingroup$ (+1) Interesting question. $\endgroup$ Jan 18 '20 at 21:47
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Here is a reduction proposed by Cornel:

We may start with the representation (derive and integrate back $\operatorname{Li}_2((1-x)/2)$), $\displaystyle \int_0^x \frac{\log(1+y)}{1-y}\textrm{d}y=\underbrace{\operatorname{Li}_2\left(\frac{1-x}{2}\right)}_{\text{Main part}}-\operatorname{Li}_2\left(\frac{1}{2}\right)-\log (2) \log (1-x)$, which if we plug in the original integral, reverse the integration order in the double integral, then carefully split the whole integral to avoid the divergence issue and apply integration by parts where needed, we arrive at $$\mathcal{I}=\left(\frac{\pi^2}{12}-\frac{\log^2(2)}{2}\right)\int_0^1\frac{\log \left(1-x^2\right)}{x}\textrm{d}x-\frac{\pi ^2}{12}\int_0^1\frac{\log (1-x)}{1+x}\textrm{d}x$$ $$-\int_0^1\frac{\log (1-x) \log ^2(1+x)}{x}\textrm{d}x-\int_0^1\frac{\log ^2(1-x) \log (1+x)}{x}\textrm{d}x$$ $$+\log(2)\int_0^1\frac{\log (1-x) \log \left(1-x^2\right)}{x}\textrm{d}x-\int_0^1\frac{\log (1-x)\operatorname{Li}_2(-x) }{1+x}\textrm{d}x$$ $$+\underbrace{\int_0^1\frac{ \log (1-x)\operatorname{Li}_2(x)}{1+x}\textrm{d}x}_{\text{The challenging part}}.$$

All integrals are known and flow natually, except the last one which is the challenging part of the problem. After we apply Landen's Identity, we want to use the following result,

$$ \int_0^1 \frac{\displaystyle \log(1-x)\operatorname{Li}_2\left(\frac{x}{x-1}\right)}{1+x} \textrm{d}x=\frac{29}{16} \zeta (4)+\frac{1}{4}\log ^2(2) \zeta (2) -\frac{1}{8} \log ^4(2),$$ which is presented and calculated in (Almost) Impossible Integrals, Sums, and Series, see page $17$ (also you may see a different approach here).

End of story.

A good note: If you want to calculate the blue sum you would like first to split the sum and calculate the easier part, and for the hard part you would like to split the sum based on parity. Then you arrive at known sums. More specifically, I would consider writing $H_{n/2}$ in terms of Digamma and then use the known identity $\psi(n+1/2)=2H_{2n}-H_n-\gamma-2\log(2)$, which is also given/mentioned in the same book, page $248$. The type of advanced sums you have to deal with are see here and see here. Also, see this one.

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    $\begingroup$ Impressive (+1) . Thank you. I'll follow your note to calculate the blue sum. I've already did the easy part of it. $\endgroup$ Jan 18 '20 at 23:15
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    $\begingroup$ @AliShather Welcome. Let me know if you have any difficulty with the blue sum which is possibly not covered by the comment. $\endgroup$ Jan 18 '20 at 23:17
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    $\begingroup$ @AliShather For the difficult part of the sum, after splitting according to parity you'll have no trouble with alternating series. $\endgroup$ Jan 18 '20 at 23:20
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    $\begingroup$ Thank you . I'll work on it. $\endgroup$ Jan 18 '20 at 23:22
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    $\begingroup$ yep thats what i did :) $\endgroup$ Jan 18 '20 at 23:35
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Starting with the following identity that can be found on page $95$ Eq $(5)$ of this paper

$$\sum_{n=1}^\infty \overline{H}_n\frac{x^n}{n}=\operatorname{Li}_2\left(\frac{1-x}{2}\right)-\operatorname{Li}_2(-x)-\ln2\ln(1-x)-\operatorname{Li}_2\left(\frac12\right)$$

Multiply both sides by $\large \frac{\ln(1-x^2)}{x}$ then integrate from $x=0$ to $x=1$ we get

$$\underbrace{\sum_{n=1}^\infty \frac{\overline{H}_n}{n}\int_0^1x^{n-1}\ln(1-x^2)\ dx}_{\large S}$$ $$\small{=I-\underbrace{\int_0^1\frac{\operatorname{Li}_2(-x)\ln(1-x^2)}{x}\ dx}_{\large J}-\ln2\underbrace{\int_0^1\frac{\ln(1-x)\ln(1-x^2)}{x}\ dx}_{\large K}-\operatorname{Li}_2\left(\frac12\right)\underbrace{\int_0^1\frac{\ln(1-x^2)}{x}\ dx}_{\large -\frac12\zeta(2)}}$$

or

$$I=S+J+\ln2\ K-\frac12\zeta(2)\operatorname{Li}_2\left(\frac12\right)\tag1$$


Evaluating $S$

Notice that

$$\int_0^1 x^{n-1}\ln(1-x^2)\ dx\overset{x^2\to x}{=}\frac12\int_0^1 x^{n/2-1}\ln(1-x)\ dx=-\frac{H_{n/2}}{n}$$

$$\Longrightarrow S=\boxed{-\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}}$$


Evaluating $J$

Writing $\ln(1-x^2)=\ln(1-x)+\ln(1+x)$ gives

$$J=\int_0^1\frac{\operatorname{Li}_2(-x)\ln(1-x)}{x}dx+\int_0^1\frac{\operatorname{Li}_2(-x)\ln(1+x)}{x}dx$$

$$=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 x^{n-1}\ln(1-x)\ dx-\frac12\operatorname{Li}_2^2(-x)|_0^1$$

$$=-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-\frac{5}{16}\zeta(4)$$

$$=\boxed{-2\operatorname{Li_4}\left(\frac12\right)+\frac{39}{16}\zeta(4)-\frac74\ln2\zeta(3)+\frac12\ln^22\zeta(2)-\frac{1}{12}\ln^42}$$

where we used $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}$$=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$


Evaluating $K$

Similarly

$$K=\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}\ dx}_{\large 2\zeta(3)}+\underbrace{\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}\ dx}_{\large -\frac{5}{8}\zeta(3)}=\boxed{\frac{11}8\zeta(3)}$$

where the second integral is evaluated here.

Plug the boxed results in $(1)$ we obtain that

$$I=-2\operatorname{Li}_4\left(\frac12\right)+\frac{29}{16}\zeta(4)-\frac38\ln2\zeta(3)+\frac34\ln^22\zeta(2)-\frac1{12}\ln^42-\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}$$

In here we proved

$$\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}=\frac1{24}\ln^42-\frac14\ln^22\zeta(2)+\frac{21}{8}\ln2\zeta(3)-\frac{9}{8}\zeta(4)+\operatorname{Li}_4\left(\frac12\right)$$

$$\Longrightarrow I=-\frac1{8}\ln^42+\ln^22\zeta(2)-3\ln2\zeta(3)+\frac{47}{16}\zeta(4)-3\operatorname{Li}_4\left(\frac12\right)$$

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