(The results below extend this post.) Given the Clausen function $\operatorname{Cl}_n\left(z\right)$. And,

$$\begin{aligned} \operatorname{Cl}_2\left(\frac\pi2\right) &= \text{Catalan's constant}\\ \operatorname{Cl}_2\left(\frac\pi3\right) &= \text{Gieseking's constant}\\ \operatorname{Cl}_2\left(\frac\pi4\right) &= \text{unnamed}\\ \operatorname{Cl}_2\left(\frac\pi6\right) &= \tfrac23\,\operatorname{Cl}_2\left(\frac\pi2\right)+\tfrac14\,\operatorname{Cl}_2\left(\frac\pi3\right) \end{aligned}$$

Then we have the closed-forms,

\begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(n+m)} &=& 2 \zeta(3) \\ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(2n+m)} &=& \frac{11}{8} \zeta(3) \\ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(3n+m)} &=& \frac{5}{3} \zeta(3) -\frac{2}{9}\pi\,\operatorname{Cl}_2\left(\frac\pi{\color{blue}3}\right)\\ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(4n+m)} &=& \frac{67}{32} \zeta(3) -\frac{1}{2}\pi\, \operatorname{Cl}_2\left(\frac\pi2\right) \\ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(6n+m)} &=& \frac{73}{24} \zeta(3) -\frac{8}{9}\pi\,\operatorname{Cl}_2\left(\frac\pi{\color{blue}3}\right)\\ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(8n+m)} &=& \frac{515}{128} \zeta(3) -\frac{3}{8}\pi\,\operatorname{Cl}_2\left(\frac\pi2\right)-\pi\,\operatorname{Cl}_2\left(\frac\pi{\color{red}4}\right)\\ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(12n+m)} &=& \frac{577}{96} \zeta(3) -\frac{7}{6}\pi\,\operatorname{Cl}_2\left(\frac\pi2\right)-\frac{19}{18}\pi\,\operatorname{Cl}_2\left(\frac\pi{\color{blue}3}\right)\\ \end{eqnarray*}

where for $p=12$ we could have used $\operatorname{Cl}_2\left(\frac\pi2\right)$ and $\operatorname{Cl}_2\left(\frac\pi6\right)$. As the OP from the other post points out, note that,

$$I(p)=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(pn+m)} =\int_0^1 \frac{\ln(1-z) \ln(1-z^p)}{z} dz$$

Q: The results above suggest a family. Can we find the closed-form of the integral $I(p)$ for $p=5$ and others?

$\color{red}{\text{Update July 24}}$: Thanks to Zacky's answer which provided the clue that more than one Clausen function with argument $\frac{m\,\pi}p$ may be needed, after some tinkering, I managed to find a closed-form for $I(p)$, namely,

$$I(p)= \frac{p^3+3}{2p^2}\zeta(3)-\frac{\pi}p\sum_{k=1}^{\lfloor(p-1)/2\rfloor}(p-2k)\operatorname{Cl}_2\left(\frac{2k\pi}p\right)$$

with floor function $\lfloor x\rfloor$. I found this using odd $p$, but it seems to work for even $p$ as well. However, a rigorous proof is needed to show it holds true for all $p$.

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    $\begingroup$ The constant $\rm{Cl}\big(\frac\pi3\big)$ is heavily discussed in this post while $\rm{Cl}\big(\frac\pi4\big)$ figures here. It might be nice to slowly build a database of integrals where the latter constant appears. $\endgroup$ – Tito Piezas III Jul 23 '19 at 4:36
  • $\begingroup$ This could be a really good idea, for sure ! $\endgroup$ – Claude Leibovici Jul 23 '19 at 5:40
  • $\begingroup$ Can you give some info on how you found those values? In the other post (from the first link) OP said even the case $p=4$ took a lot of effort. $\endgroup$ – Zacky Jul 23 '19 at 8:04
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    $\begingroup$ Well, it is not difficult to express $$\int_{0}^{1}\frac{\log(1-x)\log(1-\alpha x)}{x}\,dx$$ as a combination of logarithms, dilogarithms and trilogarithms, so $I(p)$ can be computed through such expression evaluated at the $p$-th roots of unity. $\endgroup$ – Jack D'Aurizio Jul 23 '19 at 10:48
  • $\begingroup$ Simplifications arise from the fact that $\text{Im}\,\text{Li}_3(e^{i\theta})$ is a piecewise-polynomial function, like $\text{Re}\,\text{Li}_2(e^{i\theta})$. $\endgroup$ – Jack D'Aurizio Jul 23 '19 at 10:53

$$\boxed{\int_0^1 \frac{\ln(1-x) \ln(1-x^5)}{x} dx=\\ 4\zeta(3)-\frac{\pi}{5}\operatorname{Cl}_2\left(\frac{4\pi}{5}\right)-\frac{3\pi}{5}\operatorname{Cl}_2\left(\frac{2\pi}{5}\right)+3\operatorname{Cl}_3\left(\frac{4\pi}{5}\right)+3\operatorname{Cl}_3\left(\frac{2\pi}{5}\right)}$$ $$\operatorname{Cl}_2\left(x\right)=\sum_{n=1}^\infty \frac{\sin(nx)}{n^2},\quad \operatorname{Cl}_3\left(x\right)=\sum_{n=1}^\infty \frac{\cos(nx)}{n^3}$$

(Added by OP.) But since,

$$\operatorname{Cl}_3\left(\frac{4\pi}{5}\right)+\operatorname{Cl}_3\left(\frac{2\pi}{5}\right) =-\frac{12}{25}\zeta(3)$$

then the above can be simplified as,

$$\boxed{\int_0^1 \frac{\ln(1-x) \ln(1-x^5)}{x} dx=\frac{64}{25}\zeta(3)-\frac{\pi}{5}\operatorname{Cl}_2\left(\frac{4\pi}{5}\right)-\frac{3\pi}{5}\operatorname{Cl}_2\left(\frac{2\pi}{5}\right)}$$

Tools used: $$(1-x^5)=(1-x)(1+\varphi x+x^2)(1-\frac{1}{\varphi}x+x^2), \quad \varphi =\frac{\sqrt 5+1}{2} $$ $$\ln(1+\varphi x+x^2)=-2\sum_{n=1}^\infty \frac{\cos\left(\frac{4n\pi}{5}\right)}{n}x^n$$ $$\ln(1-\frac{1}{\varphi} x+x^2)=-2\sum_{n=1}^\infty \frac{\cos\left(\frac{2n\pi}{5}\right)}{n}x^n$$ $$\int_0^1 x^{n-1}\ln(1-x)dx=-\frac1n\sum_{k=1}^n \frac{1}{k}=-\frac{H_n}{n}$$ $$S(x)=\sum_{n=1}^\infty \frac{x^n}{n^2}H_n=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\operatorname{Li}_2(1-x)\ln(1-x)+\frac{1}{2}\ln x \ln^2(1-x)+\zeta(3) $$

$$\small I(5)=\int_0^1 \frac{\ln^2(1-x)}{x}dx+\int_0^1 \frac{\ln(1-x)\ln(1+\varphi x+x^2)}{x}dx+\int_0^1\frac{\ln(1-x)\ln(1-\frac{1}{\varphi} x+x^2)}{x}dx$$ $$=\sum_{n=1}^\infty \int_0^1 x^{n-1} \ln^2 xdx-2\sum_{n=1}^\infty \frac{\cos\left(\frac{4n\pi}{5}\right)+\cos\left(\frac{2n\pi}{5}\right)}{n}\int_0^1 x^{n-1} \ln(1-x)dx$$ $$=2\sum_{n=1}^\infty \frac{1}{n^3}+2\sum_{n=1}^\infty \frac{\cos\left(\frac{4n\pi}{5}\right)+\cos\left(\frac{2n\pi}{5}\right)}{n^2}H_n=2\zeta(3)+2\Re \left(S\left(e^{4 i \pi/5}\right)+S\left(e^{2 i \pi/5}\right)\right)\tag 1$$

In order to calculate the real parts of the polylogs I used this approach to find: $$\Re \operatorname{Li}_3(e^{4i\pi/5})=\operatorname{Cl}_3\left(\frac{4\pi}{5}\right)$$ $$\Re \operatorname{Li}_3(1-e^{4i\pi/5})=\frac{\zeta(3)}{2}-\frac12 \operatorname{Cl}_3\left(\frac{4\pi}{5}\right)+\frac{2\pi^2}{25}\ln\left(\frac{5+\sqrt 5}{2}\right)$$ $$\Re \operatorname{Li}_2(1-e^{4i\pi/5})\ln(1-e^{i4\pi/5})=\frac{\pi^2}{25}\ln\left(\frac{5+\sqrt 5}{2}\right)-\frac{\pi}{10}\operatorname{Cl}_2\left(\frac{4\pi}{5}\right)$$ $$\Re \ln(e^{i4\pi/5})\ln^2(1-e^{i4\pi/5})=\frac{2\pi^2}{25}\ln\left(\frac{5+\sqrt 5}{2}\right)$$

$$\Re \operatorname{Li}_3(e^{2i\pi/5})=\operatorname{Cl}_3\left(\frac{2\pi}{5}\right)$$ $$\Re \operatorname{Li}_3(1-e^{2i\pi/5})=\frac{\zeta(3)}{2}-\frac12 \operatorname{Cl}_3\left(\frac{2\pi}{5}\right)+\frac{\pi^2}{50}\ln\left(\frac{5-\sqrt 5}{2}\right)$$ $$\Re \operatorname{Li}_2(1-e^{4i\pi/5})\ln(1-e^{i4\pi/5})=-\frac{\pi^2}{25}\ln\left(\frac{5-\sqrt 5}{2}\right)-\frac{3\pi}{10}\operatorname{Cl}_2\left(\frac{2\pi}{5}\right)$$ $$\Re \ln(e^{i4\pi/5})\ln^2(1-e^{i4\pi/5})=\frac{3\pi^2}{25}\ln\left(\frac{5-\sqrt 5}{2}\right)$$

And plugging those values in $(1)$ yields the announced result.

  • 1
    $\begingroup$ Very nice! +1. Do you think any closed form is possible for $p=\pi n$? The asymptotic expression (6) from my answer makes me think something is going on there. Of course, the factoring wouldn't work in this case $\endgroup$ – Yuriy S Jul 23 '19 at 17:56
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    $\begingroup$ I made edits to fix a typo (it should be $3$ not $\frac32$) and added a simplification to get rid of the $\operatorname{Cl}_3(z)$. I hope it is ok. $\endgroup$ – Tito Piezas III Jul 24 '19 at 4:43
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    $\begingroup$ @Zacky: It really wasn't a wasted 4 hours. :) Your particular result for $p=5$ was enough clue for me to find the result for general $p$. Kindly see the update in the main post. Of course, it still needs a rigorous proof though. $\endgroup$ – Tito Piezas III Jul 24 '19 at 13:13
  • $\begingroup$ Wolfram gives you 0 since you are derivating a constant (try to take a derivative then plug in the value). // The values of cos(...) can be obtained separately. Also separately we can find the sum of the derivative of Hurwitz zeta using the last eq from here: dlmf.nist.gov/25.11#vi. It doesn't look nice. $\endgroup$ – Zacky Jul 26 '19 at 13:43
  • $\begingroup$ I'm afraid you should take a derivative with respect to the first parameter (then plug in $2$). $\endgroup$ – Zacky Jul 26 '19 at 13:50

We may apply a discrete Fourier transform to the following generating function $$\sum_{n=1}^\infty \frac{x^n}{n^2}H_n=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\operatorname{Li}_2(1-x)\ln(1-x)+\frac{1}{2}\ln x \ln^2(1-x)+\zeta(3)$$ since $$ I(p) = \sum_{n\geq 1}\frac{H_{p n}}{pn^2}. $$ The only term leading to a non-elementary contribution is the sum of $\operatorname{Li}_3(1-x)$ over the $p$-th roots of unity.

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    $\begingroup$ @Zacky: because it is? $\endgroup$ – Jack D'Aurizio Jul 23 '19 at 18:05

Some work in progress on the general series. No closed form, sorry, but I think this may be interesting anyway.

Let's study the function $I(p)$. Obviously:

$$I \left( \frac{1}{p} \right)= p I(p)$$

Thus we are only interested in the case $p \geq 1$.

Let's sum over $m$. This gives us:

$$I(p)=\frac{\pi^2}{6}\frac{\gamma}{p}+\frac{1}{p} \sum_{n=1}^\infty \frac{\psi(pn+1)}{n^2} \tag{1}$$

There's a lot of various identities for polygamma which could be useful here.

1) Consider the following identity:

$$\psi(pn+1)=\log (pn+1)-\sum_{k=1}^\infty \frac{|G_k| (k-1)!}{(pn+1)_k}$$

Where $G_k$ are so called Gregory coefficients. $G_k= \int_0^1 \binom{x}{k} dx$ and $|G_k| \asymp \frac{1}{k \log^2 k}$ if $k \to \infty$.

$$I(p)=\frac{\pi^2}{6}\frac{\gamma+\log p}{p}+\frac{1}{p} \sum_{n=1}^\infty \frac{\log(n+1/p)}{n^2}-\frac{1}{p} \sum_{k=1}^\infty \frac{|G_k| k!}{k} \sum_{n=0}^\infty \frac{1}{(n+1)^2 (pn+p+1)_k} $$

The second series doesn't have a closed form as far as I know but it's elementary at least.

Third double series should be small in value and you might notice I changed the order of summation.

$$\sum_{n=0}^\infty \frac{1}{(n+1)^2 (pn+p+1)_k}= \frac{p!}{(p+k)!} {_{k+3} F_{k+2}} \left( \begin{array}(1,1,1, \frac{1}{p}+1, \ldots, \frac{1}{p}+k \\ 2,2,\frac{1}{p}+2, \ldots, \frac{1}{p}+k+1 \end{array};1 \right)$$

So we have:

$$pI(p)=\frac{\pi^2}{6}(\gamma+\log p)+\sum_{n=1}^\infty \frac{\log(n+1/p)}{n^2}- \\ -\sum_{k=1}^\infty \frac{|G_k|}{k \binom{p+k}{k}} {_{k+3} F_{k+2}} \left( \begin{array}(1,1,1, \frac{1}{p}+1, \ldots, \frac{1}{p}+k \\ 2,2,\frac{1}{p}+2, \ldots, \frac{1}{p}+k+1 \end{array};1 \right) \tag{2}$$

For $p>1$ the first terms and the log series give the most important contribution. The last series is complicated, but we can easily compute any finite number of terms to get more digits.

Further expanding the logarithm and using:

$$\sum_{n=1}^\infty \frac{\log(n)}{n^2}=- \frac{\pi^2}{6} (\gamma+ \log(2 \pi))+2 \pi^2 \log A $$

Whee A is http://mathworld.wolfram.com/Glaisher-KinkelinConstant.html.

We have:

$$pI(p)=\frac{\pi^2}{6}(\log p+12 \log A-\log 2 \pi)+\sum_{n=1}^\infty \frac{1}{n^2} \log \left(1+\frac{1}{pn} \right)- \\ -\sum_{k=1}^\infty \frac{|G_k|}{k \binom{p+k}{k}} {_{k+3} F_{k+2}} \left( \begin{array}(1,1,1, \frac{1}{p}+1, \ldots, \frac{1}{p}+k \\ 2,2,\frac{1}{p}+2, \ldots, \frac{1}{p}+k+1 \end{array};1 \right) \tag{3}$$

For $p \to \infty$ the asymptotic expansion will then be:

$$p I(p) \asymp \frac{\pi^2}{6}(\log p+12 \log A-\log 2 \pi)+ \frac{\zeta(3)}{2p} \tag{4}$$

Where an additional $-\zeta(3)/(2p)$ comes from the third series as the first term in asymptotic expansion for large $p$.

An exapmle:

$$100 I(100)=9.4682325532367113866$$

$$\frac{\pi^2}{6}(\log 100+12 \log A-\log 2 \pi)+ \frac{\zeta(3)}{2 \cdot 100}=9.4682415725122177074876$$

As you can see the asymptotic expansion works well enough, though some further correction terms are needed.

From (1), expanding the logarithm as we did, and using the well known asymptotic expansion of harmonic numbers we can make a full asymptotic series:

$$p I(p) \asymp \frac{\pi^2}{6}(\log p+12 \log A-\log 2 \pi)+\frac{\zeta(3)}{2p} -\sum_{k=1}^\infty \frac{B_{2k}}{2k p^{2k}} \zeta(2k+2) \tag{5}$$

I'll check it numerically later, but I'm pretty sure it doesn't converge. Still, for large $p$ a few first terms should give a lot of correct digits.

Using the explicit form for even zetas, we have:

$$p I(p) \asymp \frac{\pi^2}{6}\log \frac{p}{2\pi}+2\pi^2 \log A+\frac{\zeta(3)}{2p} -\frac{\pi^2}{2} \sum_{k=1}^\infty \frac{(-1)^k B_{2k}B_{2k+2}}{k(k+1) (2k+1)!} \frac{(2\pi)^{2k}}{p^{2k}} \tag{6}$$

The logarithmic terms and the series make me think that $p=2\pi$ is some special value.

  • $\begingroup$ Interesting that $\log A$ also can be expressed as a zeta series $\endgroup$ – Yuriy S Jul 23 '19 at 16:43
  • $\begingroup$ Got confused for a moment, but the current edit should be correct. Aside from $\zeta(3)$ only even zeta values figure in the series (5). Which explains the closed forms I think. Let me look further into this $\endgroup$ – Yuriy S Jul 23 '19 at 16:49
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    $\begingroup$ It turns out there is a general closed-form in terms of Clausen functions. Kindly see the July 24 update in the main post. $\endgroup$ – Tito Piezas III Jul 24 '19 at 13:08
  • $\begingroup$ @TitoPiezasIII, a beautiful closed form, and indeed $2 \pi/p$ is there $\endgroup$ – Yuriy S Jul 24 '19 at 13:24

An additional note on how to derive the digamma (or harmonic numbers) series from the integral:

$$ p I(p) = \sum_{n=1}^\infty \frac{H_{p n}}{n^2}$$

$$I(p)= \int_0^1 x^{-1} \log (1-x) \log (1-x^p) dx= \\ = - \sum_{n=1}^\infty \frac{1}{n} \int_0^1 x^{pn-1} \log (1-x) dx$$

Now consider the following integral:

$$J(s)=-\int_0^1 x^s \log (1-x) dx$$

Let's integrate by parts with: $$u=x^s, \qquad du=s x^{s-1} dx \\ dv=- \log(1-x) dx, \qquad v=x+(1-x) \log(1-x)$$

We get:

$$J(s)=1-s\int_0^1 x^s dx-s\int_0^1 x^{s-1} \log (1-x) dx+s \int_0^1 x^s \log (1-x) dx$$

$$(s+1)J(s)=\frac{1}{s+1}+s J(s-1)$$

It's easy to check that $J(0)=1$.

Introducing a new function:

$$Y(s+1)=(s+1) J(s)$$

We see that:

$$Y(s+1)=\frac{1}{s+1}+Y(s) \\ Y(1)=1$$

But this is exactly the definition of harmonic numbers.

So we have:

$$I(p)= \sum_{n=1}^\infty \frac{1}{n} J(pn-1)=\sum_{n=1}^\infty \frac{1}{n} \frac{Y(pn)}{pn}=\frac{1}{p} \sum_{n=1}^\infty \frac{H_{pn}}{n^2}$$


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