The closed-form solution of the family $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(pn+m)}$?

Question

(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?

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$\color{red}{\text{Update July 24}}$: 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 $p$.

Answer 1

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

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(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,

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

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

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

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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.

Answer 2

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.

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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.

Answer 3

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.

Answer 4

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:

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