I ran across this Euler sum while trying to evaluate an integral. I mentioned it in another thread, but though perhaps asking about it separate may be a good idea.

Is there a closed form for this Euler sum? $$\sum_{n=1}^{\infty}\frac{H_{2n}}{n(6n+1)}.$$

Numerically, it converges to around $0.502788$

I found that sum while trying to evaluate


I managed to obtain two of the sums. This is the only one giving me trouble.

It comes from the identity:


Note that:




I thought maybe the residue method would work, but I am not so sure.

By considering $f(z)=\frac{\pi cot(\pi z)(\gamma+\psi(-2z))}{z(6z+1)}$

The residue at the origin is $18-\frac{\pi^{2}}{2}$

The residue at $-1/6$ is $3\pi (\gamma+\psi(1/3))$

But, there does not appear to be a residue for the positive half integers.

The series for the positive integers is:

$$\frac{1}{2}\cdot \frac{1}{(z-n)^{2}}+\frac{H_{2n}}{z-n}+...$$

which gives a residue of :


At the negative integers:


giving a residue of:


I do not think it quite adds up though. I could have easily went astray in all of that, though.

  • $\begingroup$ What is your question? $\endgroup$ – Wojowu Dec 6 '15 at 14:34
  • $\begingroup$ What is your question exactly? $\endgroup$ – Jan Dec 6 '15 at 14:45
  • 2
    $\begingroup$ The question is (I guess) : is there a closed form for the sum ? $\endgroup$ – Claude Leibovici Dec 6 '15 at 14:47
  • $\begingroup$ Yes, of course. Can anyone find a closed form for said Euler sum?. $\endgroup$ – Cody Dec 6 '15 at 16:15
  • $\begingroup$ I emended the heading to be more specific about what I was asking. $\endgroup$ – Cody Dec 6 '15 at 16:19

This is a partial answer but I believe I am making progress. Let $S$ denote the sum. Then $$ \begin{align*} S=\sum_{n=1}^\infty \frac{H_{2n}}{n(6n+1)} &= \sum_{n=1}^\infty\frac{H_{2n}}{n}\int_0^1 x^{6n}\; dx \\ &= \int_0^1\left( \sum_{n=1}^\infty\frac{H_{2n}}{n}x^{6n}\right)dx \tag{1}\end{align*} $$

Let $f(x)=\sum_{n=1}^\infty \frac{H_n}{n}x^n$ where $|x|<1$. It can be shown that $$f(x)=\text{Li}_2(x)+\frac{1}{2}\log^2(1-x) \tag{2}$$ Then, we can write $$\begin{align*} \sum_{n=1}^\infty\frac{H_{2n}}{n}x^{6n} &= f(x^3)+f(-x^3) \\ &= \text{Li}_2(x^3)+\text{Li}_2(-x^3)+\frac{\log^2(1-x^3)+\log^2(1+x^3)}{2}\tag{3} \end{align*} $$ Substitute (3) into (1) to get $$ \begin{align*} S=\int_0^1 \left(\text{Li}_2(x^3)+\text{Li}_2(-x^3)+\frac{\log^2(1-x^3)+\log^2(1+x^3)}{2} \right) dx \tag{4} \end{align*} $$ Note that $$\begin{align*} \int_0^1\left( \text{Li}_2(x^3)+\text{Li}_2(-x^3)\right)dx &= \frac{1}{2}\int_0^1\sum_{n=1}^\infty\frac{x^{6n}}{n^2} \; dx \\ &=\frac{1}{2}\sum_{n=1}^\infty\frac{1}{n^2(6n+1)} \\ &= \frac{1}{2}\sum_{n=1}^\infty \left(\frac{1}{n^2}-\frac{6}{n}+\frac{36}{1+6n} \right) \\ &= \frac{1}{2}\left(\frac{\pi^2}{6} -6\psi_0\left(\frac{1}{6} \right)-6\gamma_0-36\right) \\ &= \frac{\pi^2}{12}+\frac{3\pi\sqrt{3}}{2}+6\log 2+\frac{9}{2}\log 3-18 \tag{5} \end{align*} $$

$$ \begin{align*} \frac{1}{2}\int_0^1\log^3(1-x^3)dx &= \frac{1}{6}\int_0^1t^{-2/3}\log^2(1-t)dt \quad (t=x^3)\\ &= \frac{1}{6}\left[\frac{\partial^2}{\partial y^2} B(x,y)\right]_{x=1/3,y=1} \\ &= \frac{\pi ^2}{8}-\frac{\sqrt{3} \pi }{2}+\frac{9}{2}+\frac{9}{8} \log^2 3-\frac{9 \log 3}{2}+\frac{1}{4} \sqrt{3} \pi \log 3-\frac{\psi_1\left(\frac{4}{3}\right)}{2}\tag{6} \end{align*} $$ Substitute (5) and (6) into equation (4) to get $$S=-\frac{27}{2}+\frac{5\pi^2}{24}+\frac{9}{8}\log^2 3+\frac{\pi\sqrt{3}}{4}(4+\log 3)+6\log 2-\frac{1}{2}\psi_1\left(\frac{4}{3} \right)+\frac{1}{2}\int_0^1 \log^2(1+x^3)dx \tag{7}$$

Now, it remains to calculate $\int_0^1\log^2(1+x^3)dx$.


Use integration by parts, $$ \begin{align*} \int_0^1\log^2(1+x^3) dx &= \left[x\log^2(1+x^3) \right]_0^1-6\int_0^1 \frac{x^3\log(1+x^3)}{1+x^3}dx \\ &= \log^2 2-6 \int_0^1 \log(1+x^3)dx +6\int_0^1\frac{\log(1+x^3)}{1+x^3}dx\tag{8} \end{align*} $$ The first integral is "elementary": $$\int_0^1\log(1+x^3)dx = -3+\frac{\pi}{\sqrt{3}}+2\log 2 \tag{9}$$

Evaluating $\int_0^1\frac{\log(1+x^3)}{1+x^3}$ is more tedious. Let $\omega=e^{i\pi /3}$. Then note that $1+x^3=(x+1)(x-\omega)(x-\bar{\omega})$ and


Expanding the logarithms and using the above partial fraction decomposition in $\int_0^1\frac{\log(1+x^3)}{1+x^3}$ will result in 9 pieces which can be evaluated in terms of dilogarithms. Ultimately, all this will give a closed form for $\int_0^1 \log^2(1+x^3)dx$ and $\sum_{n=1}^\infty\frac{H_{2n}}{n(6n+1)}$.

  • 1
    $\begingroup$ according to Mathematica, that last integral equals $18-\frac{5}{36}\pi^2+\frac{\log^23}{4}+(\log2-4)\log8+\frac{\log\frac{2187}{16}-12}{2\sqrt{3}}\pi+\text{Li}_2(-1/3)-(1+i\sqrt{3})\text{Li}_2\left(\frac{3-i \sqrt{3}}{6}\right)$ $+(1-i\sqrt{3})\text{Li}_2\left(\frac{3-i\sqrt{3}}{4}\right)-(1-i\sqrt{3})\text{Li}_2\left(\frac{3+i\sqrt{3}}{6}\right)+(1+i\sqrt{3})\text{Li}_2\left(\frac{3+ i \sqrt{3}}{4}\right)=0.0837342\dots$ $\endgroup$ – AccidentalFourierTransform Feb 12 '16 at 16:22
  • $\begingroup$ @AccidentalFourierTransform: Great! Now we have the closed form for the sum. $\endgroup$ – Shobhit Bhatnagar Feb 12 '16 at 16:28
  • $\begingroup$ Wow, look at you I&S!. It is surprising and cool to see a reply to this problem after all this time. I had forgotten about this sum. Nice indeed. btw...Shobit...is that you?. That is one wacky result Mathematica gave. Maybe it can be simplified. $\endgroup$ – Cody Feb 13 '16 at 14:40
  • $\begingroup$ @Cody: Yes, that is me. $\endgroup$ – Shobhit Bhatnagar Feb 14 '16 at 11:12
  • $\begingroup$ Wow, dude, where ya' been?. Good to hear from you. Good work! $\endgroup$ – Cody Feb 14 '16 at 14:33

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