Harmonic sum with Dirichlet eta tail

Question

The following problem is proposed by Cornel Valean:

$$\sum_{n=1}^{\infty} \frac{H_n}{2n+1}\left(\eta(2)- \overline{H}_n^{(2)}\right)$$ $$=2 G^2-2\ln(2) \pi G+\ln^2(2)\frac{\pi ^2}{6} +\frac{53}{1440}\pi ^4-\frac{1}{6} \ln ^4(2)-4 \operatorname{Li}_4\left(\frac{1}{2}\right),$$ where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}, \ m\ge1$, is the $n$th generalized harmonic number of order $m$, $\overline{H}_n^{(m)}=1-\frac{1}{2^m}+\cdots+(-1)^{n-1}\frac{1}{n^m}, \ m\ge1$, represents the $n$th generalized skew-harmonic number of order $m$, $G$ denotes the Catalan's constant, $\zeta$ designates the Riemann zeta function, $\eta$ designates the Dirichlet eta function, and $\operatorname{Li}_n$ signifies the Polylogarithm.

Here is my approach: Using $$\eta(2)-\overline{H}_n^{(2)}=-\int_0^1\frac{(-x)^n\ln(x)}{1+x}dx$$

and $$\sum_{n=1}^\infty\frac{H_n}{2n+1}x^n=-\int_0^1\frac{\ln(1-y^2x)}{1-y^2x}dy,$$

the sum converts to

$$\int_0^1\int_0^1\frac{\ln(x)\ln(1+y^2x)}{(1+x)(1+y^2x)}dydx.$$ Any idea how to crack this integral or a different approach than mine?

Thank you in advance.

Answer 1

Here is my route in large steps

Consider the integral found in the OP $$\eta \left(2\right)-\overline{H}_k^{\left(2\right)}=-4\int _0^1\left(-x^2\right)^k\frac{x\ln \left(x\right)}{1+x^2}\:dx,$$ as well as the generating function $$\sum _{k=1}^{\infty }\frac{H_k}{2k+1}x^{2k}=\frac{1}{2x}\left(\ln \left(2\right)+\frac{1}{2}\ln \left(1-x^2\right)\right)\ln \left(\frac{1-x}{1+x}\right)+\frac{1}{2x}\left(\operatorname{Li}_2\left(\frac{1+x}{2}\right)-\operatorname{Li}_2\left(\frac{1-x}{2}\right)\right).$$ Replacing $x\rightarrow ix$, multiplying both sides by $-4\frac{x\ln \left(x\right)}{1+x^2}$ and integrating from $0$ to $1$ with respect to $x$ yields $$\sum _{k=1}^{\infty }\frac{H_k}{2k+1}\left(\eta \left(2\right)-\overline{H}_k^{\left(2\right)}\right)=4\int _0^1\frac{\left(\ln \left(2\right)+\frac{1}{2}\ln \left(1+x^2\right)\right)\arctan \left(x\right)\ln \left(x\right)}{1+x^2}\:dx$$ $$+2i\int _0^1\left(\operatorname{Li}_2\left(\frac{1+ix}{2}\right)-\operatorname{Li}_2\left(\frac{1-ix}{2}\right)\right)\frac{\ln \left(x\right)}{1+x^2}\:dx$$ $$=4\int _0^1\frac{\left(\ln \left(2\right)+\frac{1}{2}\ln \left(1+x^2\right)\right)\arctan \left(x\right)\ln \left(x\right)}{1+x^2}\:dx+4\Re \left\{i\int _0^1\frac{\ln \left(x\right)\operatorname{Li}_2\left(\frac{1+ix}{2}\right)}{1+x^2}\:dx\right\}.$$ Integrating by parts the $2$nd integral and taking the real part leads to $$\Re \left\{i\int _0^1\frac{\ln \left(x\right)\operatorname{Li}_2\left(\frac{1+ix}{2}\right)}{1+x^2}\:dx\right\}$$ $$=G^2-\frac{\pi }{8}\ln \left(2\right)G-\int _0^1\frac{\arctan \left(x\right)\ln \left(x\right)\left(\frac{1}{2}\ln \left(1+x^2\right)-x\arctan \left(x\right)-\ln \left(2\right)\right)}{1+x^2}\:dx$$ $$+\int _0^1\frac{\left(\frac{1}{2}\ln \left(1+x^2\right)-x\arctan \left(x\right)-\ln \left(2\right)\right)\operatorname{Ti}_2\left(x\right)}{1+x^2}\:dx,$$ and this means that $$\sum _{k=1}^{\infty }\frac{H_k}{2k+1}\left(\eta \left(2\right)-\overline{H}_k^{\left(2\right)}\right)$$ $$=4G^2-\frac{\pi }{2}\ln \left(2\right)G+8\ln \left(2\right)\int _0^1\frac{\arctan \left(x\right)\ln \left(x\right)}{1+x^2}\:dx+4\int _0^1\frac{x\arctan ^2\left(x\right)\ln \left(x\right)}{1+x^2}\:dx$$ $$+4\int _0^1\frac{\left(\frac{1}{2}\ln \left(1+x^2\right)-x\arctan \left(x\right)-\ln \left(2\right)\right)\operatorname{Ti}_2\left(x\right)}{1+x^2}\:dx$$ $$=4G^2-\pi \ln \left(2\right)G+8\ln \left(2\right)\int _0^{\frac{\pi }{4}}x\ln \left(\tan \left(x\right)\right)\:dx+8\int _0^{\frac{\pi }{4}}x\ln \left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)\:dx$$ $$-8\int _0^{\frac{\pi }{4}}x\ln ^2\left(\cos \left(x\right)\right)\:dx-8\underbrace{\int _0^{\frac{\pi }{4}}\operatorname{Ti}_2\left(\tan \left(x\right)\right)\ln \left(\cos \left(x\right)\right)\:dx}_{I_1}-4\ln \left(2\right)\underbrace{\int _0^1\frac{\operatorname{Ti}_2\left(x\right)}{1+x^2}\:dx}_{I_2}.$$

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Next, it's time for $2$ observations.

In order to reduce $I_1$ we must first consider $$\int _0^{\frac{\pi }{4}}\sin \left(2\left(2k+1\right)x\right)\ln \left(\cos \left(x\right)\right)\:dx=-\frac{1}{4}\left(\frac{-\overline{H}_k+2H_{2k}-2H_k-\ln \left(2\right)}{2k+1}+\frac{1}{\left(2k+1\right)^2}\right),$$ with this it becomes $$\int _0^{\frac{\pi }{4}}\operatorname{Ti}_2\left(\tan \left(x\right)\right)\ln \left(\cos \left(x\right)\right)\:dx=\sum _{k=0}^{\infty }\frac{1}{\left(2k+1\right)^2}\int _0^{\frac{\pi }{4}}\sin \left(2\left(2k+1\right)x\right)\ln \left(\cos \left(x\right)\right)\:dx$$ $$+\int _0^{\frac{\pi }{4}}x\ln \left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)\:dx-\int _0^{\frac{\pi }{4}}x\ln ^2\left(\cos \left(x\right)\right)\:dx$$ $$=\frac{1}{4}\sum _{k=1}^{\infty }\frac{\overline{H}_k-2H_{2k}+2H_k}{\left(2k+1\right)^3}-\frac{15}{64}\zeta \left(4\right)+\frac{7}{32}\ln \left(2\right)\zeta \left(3\right)+\int _0^{\frac{\pi }{4}}x\ln \left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)\:dx$$ $$-\int _0^{\frac{\pi }{4}}x\ln ^2\left(\cos \left(x\right)\right)\:dx.$$ Secondly, $I_2$ becomes $$\int _0^1\frac{\operatorname{Ti}_2\left(x\right)}{1+x^2}\:dx=\frac{\pi }{4}G+2\int _0^{\frac{\pi }{4}}x\ln \left(\tan \left(x\right)\right)\:dx.$$

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Replacing everything from before we obtain $$\sum _{k=1}^{\infty }\frac{H_k}{2k+1}\left(\eta \left(2\right)-\overline{H}_k^{\left(2\right)}\right)=4G^2-2\pi \ln \left(2\right)G+\frac{5}{8}\zeta \left(4\right)-\frac{7}{4}\ln \left(2\right)\zeta \left(3\right)-2\sum _{k=1}^{\infty }\frac{\overline{H}_k}{\left(2k+1\right)^3}$$ $$-2\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{k^3}-4\sum _{k=1}^{\infty }\frac{H_k}{\left(2k+1\right)^3},$$ and given that the resulting series are either known or easily relatable to simple integrals, we conclude with $$\sum _{k=1}^{\infty }\frac{H_k}{2k+1}\left(\eta \left(2\right)-\overline{H}_k^{\left(2\right)}\right)$$ $$=2G^2-2\pi \ln \left(2\right)G+\frac{53}{16}\zeta \left(4\right)-4\operatorname{Li}_4\left(\frac{1}{2}\right)+\ln ^2\left(2\right)\zeta \left(2\right)-\frac{1}{6}\ln ^4\left(2\right).$$ Unfortunately, the imaginary unit was briefly involved and therefore everything started with pretty complicated terms. Interestingly enough, all the rough integrals cancelled and the last relation only featured very simple or known series.

Perhaps there's a way to use some of the ideas presented here to further simplify the already given approach.

Answer 2

A solution may be found in this short presentation, Exploiting The Master Theorem of Series to Evaluate a Series with the Tail of the Dirichlet Eta Function Artistically, by C.I. Valean.

Another solution, as emphasized at the given link, may be obtained by exploiting \begin{equation*} \int_0^1 x^{2n-1}\operatorname{arctanh}^2(x)\textrm{d}x=\log(2)\frac{H_{2n}}{n}-\frac{1}{2}\log(2)\frac{H_n}{n}+\frac{1}{2}\frac{1}{n}\sum_{k=1}^{n-1}\frac{H_k}{2k+1}, \end{equation*} which is given and derived in More (Almost) Impossible Integrals, Sums, and Series, Springer, Cham, First Edition (2023), more precisely in Ch. $1$, Sect. $1.22$, p. $29$.

End of story (apart from this, I would love to see more real solutions using very different ideas)