Euler Sums系列（六）

\[\Large\displaystyle \sum_{n=1}^{\infty}\frac{H_{2n}}{n(6n+1)}\]

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\(\Large\mathbf{Solution:}\)
Let \(S\) denote the sum. Then
\[\begin{align*}
S=\sum_{n=1}^\infty \frac{H_{2n}}{n(6n+1)} &= \sum_{n=1}^\infty\frac{H_{2n}}{n}\int_0^1 x^{6n}\mathrm dx \\
&= \int_0^1\left( \sum_{n=1}^\infty\frac{H_{2n}}{n}x^{6n}\right)\mathrm dx \tag{1}\end{align*}\]
Let \(\displaystyle f(x)=\sum_{n=1}^\infty \frac{H_n}{n}x^n\) where \(|x|<1\). It can be shown that
\[\begin{align*}
f(x)=\text{Li}_2(x)+\frac{1}{2}\ln^2(1-x) \tag{2}
\end{align*}\]
Then, we can write
\[\begin{align*}
\sum_{n=1}^\infty\frac{H_{2n}}{n}x^{6n} &= f\left(x^3\right)+f\left(-x^3\right) \\
&= \text{Li}_2\left(x^3\right)+\text{Li}_2\left(-x^3\right)+\frac{\ln^2\left(1-x^3\right)+\ln^2\left(1+x^3\right)}{2}\tag{3}
\end{align*}\]
Substitute (3) into (1) to get
\[\begin{align*}
S=\int_0^1 \left(\text{Li}_2\left(x^3\right)+\text{Li}_2\left(-x^3\right)+\frac{\ln^2\left(1-x^3\right)+\ln^2\left(1+x^3\right)}{2} \right)\mathrm dx \tag{4}
\end{align*}\]
Note that
\[\begin{align*}
\int_0^1\left( \text{Li}_2\left(x^3\right)+\text{Li}_2\left(-x^3\right)\right)\mathrm dx &= \frac{1}{2}\int_0^1\sum_{n=1}^\infty\frac{x^{6n}}{n^2} \mathrm dx \\
&=\frac{1}{2}\sum_{n=1}^\infty\frac{1}{n^2(6n+1)} \\
&= \frac{1}{2}\sum_{n=1}^\infty \left(\frac{1}{n^2}-\frac{6}{n}+\frac{36}{1+6n} \right) \\
&= \frac{1}{2}\left(\frac{\pi^2}{6} -6\psi_0\left(\frac{1}{6} \right)-6\gamma_0-36\right) \\
&= \frac{\pi^2}{12}+\frac{3\pi\sqrt{3}}{2}+6\ln 2+\frac{9}{2}\ln 3-18 \tag{5}
\end{align*}\]
\[\begin{align*}
\frac{1}{2}\int_0^1\ln^3\left(1-x^3\right)\mathrm dx &= \frac{1}{6}\int_0^1t^{-2/3}\ln^2(1-t)\mathrm dt \quad (t=x^3)\\
&= \frac{1}{6}\left[\frac{\partial^2}{\partial y^2} \mathrm{B}(x,y)\right]_{x=1/3,y=1} \\
&= \frac{\pi ^2}{8}-\frac{\sqrt{3} \pi }{2}+\frac{9}{2}+\frac{9}{8} \ln^2 3-\frac{9 \ln 3}{2}+\frac{1}{4} \sqrt{3} \pi \ln 3-\frac{\psi_1\left(\dfrac{4}{3}\right)}{2}\tag{6}
\end{align*}\]
Substitute (5) and (6) into equation (4) to get
\[S=-\frac{27}{2}+\frac{5\pi^2}{24}+\frac{9}{8}\ln^2 3+\frac{\pi\sqrt{3}}{4}(4+\ln 3)+6\ln 2-\frac{1}{2}\psi_1\left(\frac{4}{3} \right)+\frac{1}{2}\int_0^1 \ln^2(1+x^3)\mathrm{d}x\]
Now, it remains to calculate \(\displaystyle \int_0^1 \ln^2(1+x^3)\mathrm{d}x\).
According to Mathematica, it equals
\[\begin{align*}
\int_0^1 \ln^2(1+x^3)\mathrm{d}x&=18-\frac{5}{36}\pi ^{2}+\frac{\ln^{2}3}{4}+3\ln^{2}2-12\ln 2+\frac{\ln\dfrac{2187}{16}-12}{2\sqrt{3}}\pi +\mathrm{Li}_{2}\left ( -\frac{1}{3} \right )\\
&~~~-\left ( 1+i\sqrt{3} \right )\mathrm{Li}_{2}\left ( \frac{3-i\sqrt{3}}{6} \right )+\left ( 1-i\sqrt{3} \right )\mathrm{Li}_{2}\left ( \frac{3-i\sqrt{3}}{4} \right )\\
&~~~-\left ( 1-i\sqrt{3} \right )\mathrm{Li}_{2}\left ( \frac{3+i\sqrt{3}}{6} \right )+\left ( 1+i\sqrt{3} \right )\mathrm{Li}_{2}\left ( \frac{3+i\sqrt{3}}{4} \right )
\end{align*}\]
Hence, the final result is
\[\boxed{\displaystyle \begin{align*}
\sum_{n=1}^{\infty}\frac{H_{2n}}{n(6n+1)}&=\color{blue}{-\frac{9}{2}+\frac{5}{36}\pi ^{2} +\frac{5}{4}\ln^23+\frac{3}{2}\ln^22-\frac{\ln 2}{\sqrt{3}}+\left ( \frac{7}{4\sqrt{3}}+\frac{\sqrt{3}}{4} \right )\ln 3}\\
&~~~\color{blue}{+\frac{1}{2}\Bigg\{\mathrm{Li}_{2}\left ( -\frac{1}{3} \right )-\psi _{1}\left ( \frac{4}{3} \right )-\left ( 1+i\sqrt{3} \right )\mathrm{Li}_{2}\left ( \frac{3-i\sqrt{3}}{6} \right )}\\
&~~~\color{blue}{+\left ( 1-i\sqrt{3} \right )\mathrm{Li}_{2}\left ( \frac{3-i\sqrt{3}}{4} \right )-\left ( 1-i\sqrt{3} \right )\mathrm{Li}_{2}\left ( \frac{3+i\sqrt{3}}{6} \right )}\\
&~~~\color{blue}{+\left ( 1+i\sqrt{3} \right )\mathrm{Li}_{2}\left ( \frac{3+i\sqrt{3}}{4} \right )\Bigg\}}
\end{align*}}\]