# Is there a closed-form solution for $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(3n+m)}$?

I am seeking a closed-form solution for this double sum:

$$\begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(\color{blue}{3}n+m)}= ?. \end{eqnarray*}$$

I will turn it into $$3$$ tough integrals in a moment. But first I will state some similar results:

$$\begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(n+m)} &=& 2 \zeta(3) \\ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(\color{blue}{2}n+m)} &=& \frac{11}{8} \zeta(3) \\ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(\color{blue}{4}n+m)} &=& \frac{67}{32} \zeta(3) -\frac{G \pi}{2}. \\ \end{eqnarray*}$$

where $$G$$ is the Catalan constant. The last result took some effort ...

Now I know most of you folks prefer integrals to sums, so lets turn this into an integral. Using

$$\begin{eqnarray*} \frac{1}{n} &=& \int_0^1 x^{n-1} dx\\ \frac{1}{m} &=& \int_0^1 y^{m-1} dy\\ \frac{1}{3n+m} &=& \int_0^1 z^{3n+m-1} dz \\ \end{eqnarray*}$$ and summing the geometric series, we have the following triple integral $$\begin{eqnarray*} \int_0^1 \int_0^1 \int_0^1 \frac{z^3 dx dy dz}{(1-xz^3)(1-yz)}. \end{eqnarray*}$$

Now doing the $$x$$ and $$y$$ integrations we have $$\begin{eqnarray*} I=\int_0^1 \frac{\ln(1-z) \ln(1-z^3)}{z} dz. \end{eqnarray*}$$

Factorize the argument of the second logarithm ...

$$\begin{eqnarray*} I= \underbrace{\int_0^1 \frac{\ln(1-z) \ln(1-z)}{z} dz}_{=2\zeta(3)} + \int_0^1 \frac{\ln(1-z) \ln(1+z+z^2)}{z} dz. \end{eqnarray*}$$

So if you prefer my question is ... find a closed form for:

$$\begin{eqnarray*} I_1 = - \int_0^1 \frac{\ln(1-z) \ln(1+z+z^2)}{z} dz. \end{eqnarray*}$$

Integrating by parts gives:

$$\begin{eqnarray*} I_1 = - \int_0^1 \frac{\ln(z) \ln(1+z+z^2)}{1-z} dz + \int_0^1 \frac{(1+2z)\ln(z) \ln(1-z)}{1+z+z^2} dz. \end{eqnarray*}$$

and let us call these integrals $$I_2$$ and $$I_3$$ respectively.

All $$3$$ of these integrals are not easy for me to evaluate and any help with their resolution will be gratefully received.

• When I tried Wolfram ... it just told me $I_1 =1.10925$ , $I_2=0.713975$ and $I_3=0.395273$ ... at least it adds up ! – Donald Splutterwit Jul 15 at 21:19
• Mathematica gives the closed-form solution: $\frac{1}{54} \left(-6 \Gamma \left(\frac{4}{3}\right) \text{HypergeometricPFQRegularized}^{(\{0,0,0,0\},\{0,0,1\},0)}\left(\left\{1,1,1,\frac{4}{3}\right\},\left\{2,2,\frac{4}{3}\right\},1\right)-6 \Gamma \left(\frac{5}{3}\right) \text{HypergeometricPFQRegularized}^{(\{0,0,0,0\},\{0,0,1\},0)}\left(\left\{1,1,1,\frac{5}{3}\right\},\left\{2,2,\frac{5}{3}\right\},1\right)+12 \zeta (3)+2 \gamma \pi ^2+\pi ^2 \log (27)\right)$. – David G. Stork Jul 15 at 21:19
• The sequence double sums are of the form $$\frac{1}{a}\sum _{m=1}^{\infty } \frac{H_{a m}}{n^2}$$ where $H_n$ is the Harmonic Number and $n=am$; $a$ being equal to $1$,$2$,$3$ or $4$. – James Arathoon Jul 15 at 21:43
• Mathematica gives for the integral $$I=\frac{1}{216} \left(-216 \left(\text{Li}_3\left(\frac{\sqrt{-1}}{\sqrt{3}}\right)+\text{Li}_3\left(-\frac{(-1)^{5/6}}{\sqrt{3}}\right)\right)+672 \zeta (3)+9 \log ^3(3)-15 \pi ^2 \log (3)+4 \pi \sqrt{3} \left(\psi ^{(1)}\left(\frac{2}{3}\right)-\psi ^{(1)}\left(\frac{1}{3}\right)\right)\right)$$ – Yuriy S Jul 15 at 21:57
• @DonaldSplutterwit: This generalization may be of interest. – Tito Piezas III Jul 23 at 4:30

## 4 Answers

A slightly different approach where I will make use of a particular Euler sum.

Let $$I = \int_0^1 \frac{\ln (1 - x) \ln (1 - x^3)}{x} \, dx.$$ Expanding the $$\ln (1 - x^3)$$ term gives \begin{align} I &= - \sum_{n = 1}^\infty \frac{1}{n} \int_0^1 x^{3n - 1} \ln (1 - x) \, dx\tag1 \end{align} Making use of the result (for a proof of this, see here) $$\int_0^1 x^{n - 1} \ln (1 - x) \, dx = -\frac{H_n}{n}.$$ Re-indexing, namely $$n \mapsto 3n$$ gives $$\int_0^1 x^{3n - 1} \ln (1 - x) \, dx = -\frac{H_{3n}}{3n}.$$ Substitution of this result into (1) reduces our integral $$I$$ to the following Euler sum $$I = \frac{1}{3} \sum_{n = 1}^\infty \frac{H_{3n}}{n^2} = 3 \sum_{n = 1}^\infty \frac{H_{3n}}{(3n)^2}.$$

For the Euler sum, since the series converges absolutely, terms in the sum may be rearranged. Doing so we have \begin{align} \sum_{n =1}^\infty \frac{H_{3n}}{(3n)^2} &= \frac{H_3}{3^2} + \frac{H_6}{6^2} + \frac{H_9}{9^2} + \cdots\\ &= \frac{2}{3} \left [\frac{3}{2} \frac{H_3}{3^2} + \frac{3}{2} \frac{H_6}{6^2} + \frac{3}{2} \frac{H_9}{9^2} + \cdots \right ]\\ &= \frac{2}{3} \left [\left (\frac{H_3}{3^2} + \frac{H_6}{6^2} + \frac{H_9}{9^2} + \cdots \right ) + \frac{1}{2} \left (\frac{H_3}{3^2} + \frac{H_6}{6^2} + \frac{H_9}{9^2} + \cdots \right ) \right ]\\ &= \frac{2}{3} \left [\left (-\frac{1}{2} \frac{H_1}{1^2} - \frac{1}{2} \frac{H_2}{2^2} + \frac{H_3}{3^2} - \frac{1}{2} \frac{H_4}{4^2} - \frac{1}{2} \frac{H_5}{5^2} + \frac{H_6}{6^2} - \cdots \right )\right.\\ & \qquad + \left. \frac{1}{2} \left (\frac{H_1}{1^2} + \frac{H_2}{2^2} + \frac{H_3}{3^2} + \frac{H_4}{4^2} + \cdots \right ) \right ]\\ &= \frac{2}{3} \sum_{n = 1}^\infty \frac{H_n}{n^2} \cos \left (\frac{2 \pi n}{3} \right ) + \frac{1}{3} \sum_{n = 1}^\infty \frac{H_n}{n^2} \tag2\\ &= \frac{2}{3} \zeta (3) + \frac{2}{3} \operatorname{Re} \sum_{n = 1}^\infty \frac{H_n}{n^2} \left (e^{\frac{2 \pi i}{3}} \right )^n \end{align} Note in (2) the well-known result of $$\sum_{n = 1}^\infty \frac{H_n}{n^2} = 2 \zeta (3)$$ has been used.

The sum can now be found by making use of the following generating function (for a simple proof of this result, see here) $$\sum_{n=1}^\infty\frac{H_n}{n^2}x^n=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3).$$ Setting $$x = e^{\frac{2 \pi i}{3}}$$ gives \begin{align} \sum_{n = 1}^\infty \frac{H_{3n}}{(3n)^2} &= \frac{2}{3} \zeta (3) + \frac{2}{3} \operatorname{Re} \left [\operatorname{Li}_3 (e^{\frac{2 \pi i}{3}}) - \operatorname{Li}_3 (1 - e^{\frac{2 \pi i}{3}}) + \ln (1 - e^{\frac{2 \pi i}{3}}) \operatorname{Li}_2 (1 - e^{\frac{2 \pi i}{3}}) \right.\\ & \qquad \left. + \frac{1}{2} \ln (e^{\frac{2 \pi i}{3}}) \ln^2 (1 - e^{\frac{2 \pi i}{3}}) + \zeta (3) \right ]\\ &= \frac{5}{3} \zeta (3) + \frac{2}{3} \operatorname{Re} \left [\operatorname{Li}_3 (e^{\frac{2 \pi i}{3}}) - \operatorname{Li}_3 (1 - e^{\frac{2 \pi i}{3}}) + \ln (1 - e^{\frac{2 \pi i}{3}}) \operatorname{Li}_2 (1 - e^{\frac{2 \pi i}{3}}) \right.\\ & \qquad \left. + \frac{\pi i}{3} \ln^2 (1 - e^{\frac{2 \pi i}{3}}) \right ]\tag3 \end{align} Now, since (this part is tedious, but readily doable) \begin{align} \operatorname{Re} \operatorname{Li}_3 (e^{\frac{2 \pi i}{3}}) &= -\frac{4}{9} \zeta (3)\\ \operatorname{Re} \operatorname{Li}_3 (1 - e^{\frac{2 \pi i}{3}}) &= \frac{\pi^2}{18} \ln 3 + \frac{13}{18} \zeta (3)\\ \operatorname{Re} \left [i \ln^2 (1 - e^{\frac{2 \pi i}{3}}) \right ] &= \frac{\pi}{6} \ln 3\\ \operatorname{Re} \left [\ln (1 - e^{\frac{2 \pi i}{3}}) \operatorname{Li}_2 (1 - e^{\frac{2 \pi i}{3}}) \right ] &= \frac{\pi^3}{27 \sqrt{3}} - \frac{\pi}{18 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right ) \end{align} The Euler sum in (3) thus becomes $$\sum_{n = 1}^\infty \frac{H_{3n}}{(3n)^2} = \frac{5}{9} \zeta (3) + \frac{2 \pi^3}{81 \sqrt{3}} - \frac{\pi}{27 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right ),$$ so that we finally arrive at the following value for the integral (and hence your double sum) of

$$\int_0^1 \frac{\ln (1 - x) \ln (1 - x^3)}{x} \, dx = \frac{5}{3} \zeta (3) + \frac{2 \pi^3}{27 \sqrt{3}} - \frac{\pi}{9 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right )$$

• (+1) Good. In general, as stated very often in the book, (Almost) Impossible Integrals, Sums, and Series, use that $\sum_{k=1}^{\infty} \frac{1}{k(k+n)}=\frac{H_n}{n}$. Thus, you may reduce much the size of the solution. You get directly $3 \sum_{n = 1}^\infty \frac{H_{3n}}{(3n)^2}$. – user97357329 Jul 16 at 13:27
• @omegadot would it be possible to tell me how to find (by hand) those real parts of the polylogarithms? – Zacky Jul 20 at 20:36
• You may be interested in this generalization. – Tito Piezas III Jul 23 at 4:29
• @Zacky - I calculate, by hand, the real parts of all those quantities I used in my answer in an addendum here. – omegadot Jul 23 at 8:29

$$\boxed{I=\int_0^1 \frac{\ln(1-x) \ln(1-x^3)}{x}dx=\frac53\zeta(3) +\frac{2\pi^3}{27\sqrt 3} -\frac{\pi}{9\sqrt 3}\psi_1\left(\frac13\right)}$$ The value of the integral can be extracted from here, since using $$2ab=a^2+b^2-(a-b)^2$$ we get: $$2I=\int_0^1 \frac{\ln^2(1-x)}{x}dx+\int_0^1 \frac{\ln^2(1-x^3)}{x}dx-\int_0^1 \frac{\ln^2(1+x+x^2)}{x}dx$$ $$\int_0^1 \frac{\ln^2(1-x^3)}{x}dx\overset{x^3\to x}=\frac13\int_0^1 \frac{\ln^2(1-x)}{x} dx=\frac13\sum_{n=1}^\infty\int_0^1 x^n \ln^2 x dx=\frac23 \zeta(3)$$ $$\int_0^1\frac{\ln^2(1+x+x^2)}{x}dx=\frac{2\pi}{9\sqrt3}\psi_1\left(\frac13\right)-\frac{4\pi^3}{27\sqrt3}-\frac23\zeta(3)$$ And the result from above follows.

Alternatively we can make use of the following series: $$-\frac12 \ln(1-2x\cos t+x^2)=\sum_{n=1}^\infty \frac{\cos(nt)}{n} x^n,\quad |x|<1, t\in \mathbb R$$ $$\Rightarrow I_1 =-\int_0^1 \frac{\ln(1-x)\ln(1+x+x^2)}{x}dx=2\sum_{n=1}^\infty \frac{\cos\left(\frac{2n \pi}{3}\right)}{n}\int_0^1 \ln(1-x) x^{n-1}dx$$ $$=2\sum_{n=1}^\infty \frac{\cos\left(\frac{2n \pi}{3}\right)}{n^2}H_n=2\Re \left(\sum_{n=1}^\infty \frac{z^n}{n^2}H_n\right),\quad z=e^{\frac{2\pi i}{3}}$$ Using the following generating function: $$\sum_{n=1}^\infty \frac{x^n}{n^2}H_n=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\operatorname{Li}_2(1-x)\ln(1-x)+\frac{1}{2}\ln x \ln^2(1-x)+\zeta(3)$$ And by plugging in the values found in this post yields the announced result.

• One word: yikes (+1) – clathratus Jul 17 at 14:51

Not a full answer, but another interesting expression for the series.

Let's introduce a function:

$$S(x,y)=\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n y^m}{ n m (3n +m)}$$

Assuming $$|x|<1$$ and $$|y|<1$$ we avoid any convergence problems, and can use partial fractions:

$$S(x,y)=\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n y^m}{ n m^2}-\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n y^m}{ m^2(n+ \frac{1}{3} m)}$$

$$S(x,y)=-\log(1-x) \text{Li}_2(y) -x \sum_{m=1}^\infty \frac{y^m}{ m^2} \Phi \left(x,1,\frac13 m+1 \right)$$

Let's use the integral representation of the Lerch transcendent:

$$\Phi \left(x,1,\frac13 m+1 \right)= \int_0^\infty \frac{e^{-(1+\frac13 m)t} ~dt}{1-x e^{-t}}$$

Summation under the integral gives us:

$$S(x,y)=-\log(1-x) \text{Li}_2(y) -x \int_0^\infty \text{Li}_2 \left(y e^{-t/3} \right) \frac{dt}{e^t-x}$$

So we can assume:

$$S=\lim_{x \to 1} \left[-\log(1-x) \text{Li}_2(x) -x \int_0^\infty \text{Li}_2 \left(x e^{-t/3} \right) \frac{dt}{e^t-x} \right]$$

Which does seem to work numerically, though of course is quite hard to evaluate symbolically.

Numerical check:

In:= x=9999999/10000000;
y=9999999/10000000;
N[-Log[1-x]PolyLog[2,y],10]-x NIntegrate[PolyLog[2,y Exp[-t/3]]/(Exp[t]-x),{t,0,Infinity},WorkingPrecision->10]
Out= 1.29484017


Compare to exact expression:

In:= N[(1/216)*(-15*Pi^2*Log + 9*Log^3 + 4*Sqrt*Pi*(-PolyGamma[1, 1/3] +
PolyGamma[1, 2/3]) - 216*(PolyLog[3, (-1)^(1/6)/Sqrt] +
PolyLog[3, -((-1)^(5/6)/Sqrt)]) + 672*Zeta), 10]
Out= 1.2948652620+0.*10^-11 I


It may be that $$x=y$$ is not the best choice for the limit. For example, we can assume $$x=y^a$$ where $$a$$ is some real number. A good choice may lead to a better numerical convergence or even a closed form.

Using the dilogarithm properties, we have:

$$F(x,y)=-x \int_0^\infty \text{Li}_2 \left(y e^{-t/3} \right) \frac{dt}{e^t-x}=x \int_0^\infty \int_0^1 \frac{\log(1-e^{-t/3} y u) du dt}{u (e^t-x)}$$

Let's change the variable:

$$e^{-t}=v \\ t=- \log v$$

$$F(x,y)=x \int_0^1 \int_0^1 \frac{\log(1- y u v^{1/3}) du dv}{u (1-x v)}$$

Let's take:

$$y=x^{1/3}$$

We have:

$$F(x,y)=x \int_0^1 \int_0^1 \frac{\log(1- u (xv)^{1/3}) du dv}{u (1-x v)}$$

$$v=w/x$$

$$F(x,y)=\int_0^x \int_0^1 \frac{\log(1- u w^{1/3}) du dw}{u (1-w)}$$

$$F(x,y)=-\int_0^x \frac{\text{Li}_2 (w^{1/3}) dw}{1-w}$$

So, there's a neater expression for the limit:

$$\color{blue}{S=\lim_{x \to 1} \left[-\log(1-x) \text{Li}_2(x^{1/3}) -\int_0^x \frac{\text{Li}_2 (w^{1/3}) dw}{1-w} \right]}$$

This allows a simple generalization:

$$\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n y^m}{ n m (an +m)}=\lim_{x \to 1} \left[-\log(1-x) \text{Li}_2(x^{1/a}) -\int_0^x \frac{\text{Li}_2 (w^{1/a}) dw}{1-w} \right]$$

Which checks out numerically with the examples from the OP.

I wonder if we can somehow use L'Hospital here to deal with the integral and get a closed form for the limit.

Integration by parts could also work.

• In fact, integration by parts will immediately give us the same integral as in the OP – Yuriy S Jul 15 at 23:46

Addendum

This is an addendum to the answer I gave here. As requested by @Zacky, I will show how the real parts of those quantities considered in my solution, particularly those containing polylogarithms, can be extracted by hand. The quantities in question are: \begin{align} \operatorname{Re} \left [i \ln^2 (1 - e^{\frac{2 \pi i}{3}}) \right ] &= \frac{\pi}{6} \ln 3\tag1\\ \operatorname{Re} \operatorname{Li}_3 (e^{\frac{2 \pi i}{3}}) &= -\frac{4}{9} \zeta (3)\tag2\\ \operatorname{Re} \operatorname{Li}_3 (1 - e^{\frac{2 \pi i}{3}}) &= \frac{\pi^2}{18} \ln 3 + \frac{13}{18} \zeta (3)\tag3\\ \operatorname{Re} \left [\ln (1 - e^{\frac{2 \pi i}{3}}) \operatorname{Li}_2 (1 - e^{\frac{2 \pi i}{3}}) \right ] &= \frac{\pi^3}{27 \sqrt{3}} - \frac{\pi}{18 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right )\tag4 \end{align}

As we are dealing with complex valued functions, in all cases the principal value is taken.

Preliminaries

$$\ln \left (e^{\frac{2 \pi i}{3}} \right ) = \frac{2 \pi i}{3}.$$

The first quantity

It is routine to show that: $$\ln \left (1 - e^{\frac{2 \pi i}{3}} \right ) = \frac{1}{2} \ln 3 - i \frac{\pi}{6}.$$ Thus \begin{align} i \ln^2 \left (1 - e^{\frac{2 \pi i}{3}} \right ) &= i \left (\frac{1}{2} \ln 3 - i \frac{\pi}{6} \right )^2\\ &= i \left (\frac{1}{4} \ln^2 3 - \frac{\pi^2}{36} - i\frac{\pi}{6} \ln 3 \right )\\ &=\frac{\pi}{6} \ln 3 + \frac{i}{4} \ln^2 3 - \frac{i \pi^2}{36}, \end{align} and we immediately see that $$\boxed{\operatorname{Re} \left [i \ln^2 (1 - e^{\frac{2 \pi i}{3}}) \right ] = \frac{\pi}{6} \ln 3}$$

The second quantity

\begin{align} \operatorname{Re} \operatorname{Li}_3 (e^{\frac{2 \pi i}{3}}) &= \operatorname{Re} \sum_{n = 1}^\infty \frac{1}{n^3} e^{\frac{2 \pi i n}{3}}\\ &= \sum_{n = 1}^\infty \frac{1}{n^3} \cos \left (\frac{2 \pi n}{3} \right )\\ &= -\frac{1}{2} \cdot \frac{1}{1^3} - \frac{1}{2} \cdot \frac{1}{2^3} + \frac{1}{3^3} - \frac{1}{2} \cdot \frac{1}{4^3} - \frac{1}{2} \cdot \frac{1}{5^3} + \frac{1}{6^3} - \cdots\\ &= \frac{3}{2} \left (\frac{1}{3^3} + \frac{1}{6^3} + \frac{1}{9^3} + \cdots \right ) - \frac{1}{2} \left (\frac{1}{1^3} + \frac{1}{2^3} + \frac{1}{3^3} + \cdots \right )\\ &= \frac{3}{2 \cdot 3^3} \left (\frac{1}{1^3} + \frac{1}{2^3} + \frac{1}{3^3} + \cdots \right ) - \frac{1}{2} \left (\frac{1}{1^3} + \frac{1}{2^3} + \frac{1}{3^3} + \cdots \right )\\ &= \frac{1}{18} \sum_{n = 1}^\infty \frac{1}{n^3} - \frac{1}{2} \sum_{n = 1}^\infty \frac{1}{n^3}\\ &= -\frac{4}{9} \sum_{n = 1}^\infty \frac{1}{n^3} \end{align} giving $$\boxed{\operatorname{Re} \operatorname{Li}_3 (e^{\frac{2 \pi i}{3}}) = -\frac{4}{9} \zeta (3)}$$

Later we will have a need for its imaginary part so let's calculate it now.

\begin{align} \operatorname{Im} \operatorname{Li}_3 (e^{\frac{2 \pi i}{3}}) &= \operatorname{Im} \sum_{n = 1} \frac{1}{n^3} e^{\frac{2 \pi i n}{3}}\\ &= \sum_{n = 1}^\infty \frac{1}{n^3} \sin \left (\frac{2 \pi n}{3} \right )\\ &= \frac{\sqrt{3}}{2} \underbrace{\sum_{\substack{n = 1\\n \in 1,4,7,\ldots}}^\infty \frac{1}{n^3}}_{n \, \mapsto \, 3n + 1} - \frac{\sqrt{3}}{2} \underbrace{\sum_{\substack{n = 1\\n \in 2,5,8,\ldots}}^\infty \frac{1}{n^3}}_{n \, \mapsto \, 3n + 2}\\ &= \frac{\sqrt{3}}{2} \sum_{n = 1}^\infty \frac{1}{(3n + 1)^3} - \frac{\sqrt{3}}{2} \sum_{n = 1}^\infty \frac{1}{(3n + 2)^3}\\ &= \frac{\sqrt{3}}{54} \sum_{n = 1}^\infty \frac{1}{(n + \frac{1}{3})^3} - \frac{\sqrt{3}}{54} \sum_{n = 1}^\infty \frac{1}{(n + \frac{2}{3})^3}\\ &= -\frac{\sqrt{3}}{54} \cdot \frac{1}{2!} \psi^{(2)} \left (\frac{1}{3} \right ) + \frac{\sqrt{3}}{54} \cdot \frac{1}{2!} \psi^{(2)} \left (\frac{2}{3} \right )\\ &= \frac{\sqrt{3}}{108} \left [\psi^{(2)} \left (1 - \frac{1}{3} \right ) - \psi^{(2)} \left (\frac{1}{3} \right ) \right ]\\ &= \frac{\sqrt{3}}{108} \cdot \pi \left. \frac{d^2}{dz^2} \cot (\pi z) \right |_{z = \frac{1}{3}}\\ &= \frac{\sqrt{3}}{108} \cdot \frac{8 \pi^2}{3 \sqrt{3}}\\ &= \frac{2 \pi^2}{81}. \end{align} Thus we see that $$\boxed{\operatorname{Li}_3 (e^{\frac{2 \pi i}{3}} ) = -\frac{4}{9} \zeta (3) + \frac{2 \pi^2}{81}i}$$

The third quantity

In order to evaluate this quantity we will make use of the following result (for a proof of this result, see here) $$\operatorname{Li}_{3}(z) + \operatorname{Li}_{3}(1-z)+ \operatorname{Li}_{3}\left(1 - \frac{1}{z}\right) = \zeta(3) + \frac{\ln^{3} (z)}{6}+ \frac{\pi^{2} \ln (z) }{6}- \frac{\ln^{2} (z) \ln(1-z)}{2}. \qquad (*)$$ If we set $$z = e^{\frac{2 \pi i}{3}}$$, note that $$1 - \frac{1}{z} = \overline{1 - z}.$$ Thus $$\operatorname{Li}_3 \left (1 - \frac{1}{z} \right ) = \operatorname{Li}_3 (\overline{1 - z}) = \overline{\operatorname{Li}_3 (1 - z)},$$ allowing us to write $$\operatorname{Li}_3 \left (1 - z \right ) + \operatorname{Li}_3 \left (1 - \frac{1}{z} \right ) = \operatorname{Li}_3 (1 - z) + \overline{\operatorname{Li}_3 (1 - z)} = 2 \operatorname{Re} \operatorname{Li}_3 (1 - z).$$ So ($$*$$) can be rewritten as $$\operatorname{Re} \operatorname{Li}_3 (1 - z) = \frac{1}{2} \zeta (3) + \frac{1}{12} \ln^3 (z) + \frac{\pi^2}{12} \ln (z) - \frac{1}{4} \ln^2 (z) \ln (1 - z) - \frac{1}{2} \operatorname{Li}_3 (z).$$ Setting $$z = e^{\frac{2 \pi i}{3}}$$, one has \begin{align} \operatorname{Re} \operatorname{Li}_3 (1 - e^{\frac{2 \pi i}{3}}) &= \frac{1}{2} \zeta (3) - \frac{1}{12} \cdot \frac{8 i\pi^3}{27} + \frac{\pi^2}{12} \cdot \frac{2 \pi i}{3} + \frac{1}{4} \cdot \frac{4 \pi^2}{9} \left (\frac{1}{2} \ln 3 - \frac{i \pi}{6} \right )\\ & \qquad - \frac{1}{2} \left (-\frac{4}{9} \zeta (3) + \frac{2 i \pi^3}{81} \right )\\ &= \frac{13}{18} \zeta (3) + \frac{\pi^2}{18} \ln 3 \end{align} Thus $$\boxed{\operatorname{Re} \operatorname{Li}_3 (1 - e^{\frac{2 \pi i}{3}}) = \frac{13}{18} \zeta (3) + \frac{\pi^2}{18} \ln 3}$$

The fourth quantity

To evaluate this quantity we will make use of Euler' reflexion formula, namely $$\operatorname{Li}_2 (z) + \operatorname{Li}_2 (1 - z) = \zeta (2) - \ln (z) \ln (1 - z).$$ Setting $$z = e^{\frac{2 \pi i}{3}}$$ we see that $$\operatorname{Li}_2 (1 - e^{\frac{2\pi i}{3}}) = \zeta (2) - \frac{2 \pi i}{3} \ln (1 - e^{\frac{2 \pi i}{3}}) - \operatorname{Li}_2 (e^{\frac{2 \pi i}{3}}) \qquad (**)$$

Finding $$\operatorname{Li}_2 (e^{\frac{2 \pi i}{3}})$$. For the real part: \begin{align} \operatorname{Re} \operatorname{Li}_2 (e^{\frac{2 \pi i}{3}}) &= \operatorname{Re} \sum_{n = 1}^\infty \frac{1}{n^2} e^{\frac{2 \pi i n}{3}}\\ &= \sum_{n = 1}^\infty \frac{1}{n^2} \cos \left (\frac{2 \pi n}{3} \right )\\ &= -\frac{1}{2} \cdot \frac{1}{1^2} - \frac{1}{2} \cdot \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{2} \cdot \frac{1}{4^2} - \frac{1}{2} \cdot \frac{1}{5^2} + \frac{1}{6^2} - \cdots\\ &=\frac{3}{2} \left (\frac{1}{3^2} + \frac{1}{6^2} + \frac{1}{9^2} + \cdots \right ) - \frac{1}{2} \left (\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots \right )\\ &= \frac{3}{2 \cdot 3^2} \left (\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots \right ) - \frac{1}{2} \left (\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots \right )\\ &= \frac{1}{6} \sum_{n = 1}^\infty \frac{1}{n^2} - \frac{1}{2} \sum_{n = 1}^\infty \frac{1}{n^2}\\ &= -\frac{1}{3} \sum_{n = 1}^\infty \frac{1}{n^2}\\ &= -\frac{1}{3} \zeta (2), \end{align} giving $$\boxed{\operatorname{Re} \operatorname{Li}_2 (e^{\frac{2 \pi i}{3}}) = -\frac{\pi^2}{18}}$$

While for the imaginary part: \begin{align} \operatorname{Im} \operatorname{Li}_2 (e^{\frac{2 \pi i}{3}}) &= \operatorname{Im} \sum_{n = 1} \frac{1}{n^2} e^{\frac{2 \pi i n}{3}}\\ &= \sum_{n = 1}^\infty \frac{1}{n^2} \sin \left (\frac{2 \pi n}{3} \right )\\ &= \frac{\sqrt{3}}{2} \underbrace{\sum_{\substack{n = 1\\n \in 1,4,7,\ldots}}^\infty \frac{1}{n^2}}_{n \, \mapsto \, 3n + 1} - \frac{\sqrt{3}}{2} \underbrace{\sum_{\substack{n = 1\\n \in 2,5,8,\ldots}}^\infty \frac{1}{n^2}}_{n \, \mapsto \, 3n + 2}\\ &= \frac{\sqrt{3}}{2} \sum_{n = 1}^\infty \frac{1}{(3n + 1)^2} - \frac{\sqrt{3}}{2} \sum_{n = 1}^\infty \frac{1}{(3n + 2)^2}\\ &= \frac{\sqrt{3}}{18} \sum_{n = 1}^\infty \frac{1}{(n + \frac{1}{3})^2} - \frac{\sqrt{3}}{18} \sum_{n = 1}^\infty \frac{1}{(n + \frac{2}{3})^2}\\ &= \frac{\sqrt{3}}{18} \left [\psi^{(1)} \left (\frac{1}{3} \right ) -\psi^{(1)} \left (1 - \frac{1}{3} \right ) \right ]\\ &= \frac{\sqrt{3}}{18} \left [\psi^{(1)} \left (\frac{1}{3} \right ) - \left \{\frac{4 \pi^2}{3} - \psi^{(1)} \left (\frac{1}{3} \right ) \right \} \right ]\\ &= \frac{1}{3 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right ) - \frac{2}{9 \sqrt{3}} \pi^2 \end{align} Thus we see that $$\boxed{\operatorname{Li}_2 (e^{\frac{2 \pi i}{3}}) = -\frac{\pi^2}{18} + \frac{i}{3 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right ) - \frac{2i \pi^2}{9 \sqrt{3}}}$$

So on making use of Euler's reflexion formula, we see that $$\operatorname{Li}_2 (1 - e^{\frac{2 \pi i}{3}}) = \frac{\pi^2}{9} - \frac{\pi i}{3} \ln 3 + \frac{2i}{9\sqrt{3}} - \frac{i}{3 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right ).$$ So at last \begin{align} \operatorname{Re} \left [\ln (1 - e^{\frac{2 \pi i}{3}}) \operatorname{Li}_2 (1 - e^{\frac{2 \pi i}{3}}) \right ] &= \operatorname{Re} \left [\left \{\frac{1}{2} \ln 3 - \frac{\pi i}{6} \right \} \right. \times \\ & \qquad \left. \left \{\frac{\pi^2}{9} - \frac{\pi i}{3} \ln 3 + \frac{2i}{9\sqrt{3}} - \frac{i}{3 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right ) \right \} \right ]\\ &= \frac{\pi^3}{27 \sqrt{3}} - \frac{\pi}{18 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right ). \end{align} Thus $$\boxed{\operatorname{Re} \left [\ln (1 - e^{\frac{2 \pi i}{3}}) \operatorname{Li}_2 (1 - e^{\frac{2 \pi i}{3}}) \right ] = \frac{\pi^3}{27 \sqrt{3}} - \frac{\pi}{18 \sqrt{3}} \psi^{(1)} \left (\frac{1}{3} \right )}$$