Conjecture $\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)=\frac{7\pi^2}{48}-\frac13\arctan^22-\frac16\arctan^23-\frac18\ln^2(\tfrac{18}5)$

I numerically discovered the following conjecture: $$\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)\stackrel{\color{gray}?}=\frac{7\pi^2}{48}-\frac{\arctan^22}3-\frac{\arctan^23}6-\frac18\ln^2\!\left(\frac{18}5\right).$$ It holds numerically with a precision of more than $30000$ decimal digits.

Could you suggest any ideas how to prove it?

Can we find a closed form for $\Im\,\operatorname{Li}_2\left(\frac12+\frac i6\right)$?

Is there a general method to find closed forms of expressions of the form $\Re\,\operatorname{Li}_2(p+iq)$, $\Im\,\operatorname{Li}_2(p+iq)$ for $p,q\in\mathbb Q$?

Solution 1:

I don't know how you got your conjecture but I checked and on further simplifications I found it to be correct.

We know that :

$$ {Li}_{2}(\bar{z})=\bar{{Li}_{2}(z)} $$

So :

$$ \Re{{Li}_{2}(z)}=\frac{\bar{{Li}_{2}(z)}+{Li}_{2}(z)}{2} $$

So:

$$ \Re{{Li}_{2}(\frac{1}{2}+\frac{i}{6})}=\frac{{Li}_{2}(\frac{1}{2}+\frac{i}{6})+{Li}_{2}(\frac{1}{2}-\frac{i}{6})}{2}\\ =\frac{{Li}_{2}(\frac{1}{2}+\frac{i}{6})+{Li}_{2}(1-(\frac{1}{2}+\frac{i}{6}))}{2} $$

Now let's use:

$$ {Li}_{2}(z)+{Li}_{2}(1-z)=\frac{{\pi}^{2}}{6}-\ln{z}\ln{(1-z)} $$

So we get:

$$ \Re{{Li}_{2}(\frac{1}{2}+\frac{i}{6})}=\frac{\frac{{\pi}^{2}}{6}-\ln{(\frac{1}{2}+\frac{i}{6})}\ln{(1-(\frac{1}{2}+\frac{i}{6}))}}{2}\\ =\frac{\frac{{\pi}^{2}}{6}-\ln{(\frac{1}{2}+\frac{i}{6})}\ln{(\frac{1}{2}-\frac{i}{6})}}{2} $$

Let's compute those logarithms:

$$ \ln(\frac{1}{2}\pm \frac{i}{6})=\ln{(\sqrt{\frac{1}{2^2}+\frac{1}{6^2}}{e}^{\pm i\arctan{\frac{1}{3}}})}\\ =\ln{(\sqrt{\frac{5}{18}}{e}^{\pm i\arctan{\frac{1}{3}}})}\\ =\frac{1}{2}\ln{(\frac{5}{18})}\pm i\arctan{\frac{1}{3}} $$

Taking their product: $$ \ln{(\frac{1}{2}+\frac{i}{6})}\ln{(\frac{1}{2}-\frac{i}{6})}=\frac{1}{4}{\ln{(\frac{5}{18}})}^{2}+{(\arctan{\frac{1}{3}})}^{2} $$

Finally:

$$ \Re{{Li}_{2}(\frac{1}{2}+\frac{i}{6})}=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln{(\frac{18}{5}})}^{2}-\frac{1}{2}{(\arctan{1/3})}^{2}\\ =\frac{7(\pi)^2}{48}-\frac{{(\arctan{2})}^{2}}{3}-\frac{{(\arctan{3})}^{2}}{6}-\frac{1}{8} {\ln{\frac{18}{5}}}^{2} $$

There's a general formula using the same method. But the imaginary part doesn't have a known closed form:

$$ \Re{{Li}_{2}(\frac{1}{2}+iq)}=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln{(\frac{1+4q^2}{4})}}^{2}-\frac{{\arctan{(2q)}}^{2}}{2} $$

Solution 2:

This is an answer finding $\Re\operatorname{Li}_2\left(\frac{1+ti}2\right)$ via integration method. (Personally I don't like the reflection formula)
First, we restrict $t\in\mathbb R$. One can prove $\Re\operatorname{Li}_2\left(\frac{1+ti}2\right)$ is differentiable and the following changing the positions of signs is correct. $$\begin{aligned} \Re\operatorname{Li}_2\left(\frac{1+ti}2\right)&=-\frac12\int_0^1\ln\left(1-x+\frac{1+t^2}4x^2\right)\frac{d x}x\\ &=-\frac12\int\int_0^1\frac{\partial}{\partial t}\ln\left(1-x+\frac{1+t^2}4x^2\right)\frac{d x}xd t\\ &=-\int\int_0^1\frac{tx}{4-4x+(1+t^2)x^2}d xd t\\ &=-\int\left(\frac1{1+t^2}\arctan t+\frac12\cdot\frac t{1+t^2}\ln\frac{1+t^2}4\right)d t\\ &=-\frac12\arctan^2t-\frac18\ln^2\frac{1+t^2}4+C \end{aligned}$$ Substitute $t=0$ into the equation, we get $\frac1{12}\pi^2-\frac12\ln^22=0-\frac18\ln^24+C$, or $C=\frac1{12}\pi^2$. Hence $$\Re\operatorname{Li}_2\left(\frac{1+ti}2\right)=\frac1{12}\pi^2-\frac12\arctan^2t-\frac18\ln^2\frac{1+t^2}4$$

Solution 3:

First of all we know that:

$$ \operatorname{Li}_2(z) = -\operatorname{Li}_2\left(\frac{z}{z-1}\right)-\frac{1}{2}\ln^2(1-z), \quad z \notin (1,\infty).\tag{$\diamondsuit$} $$

Furthermore, we have the following relationship between the dilogarithm and the Clausen functions:

$$\operatorname{Li}_2\left(e^{i\theta}\right) = \operatorname{Sl}_2(\theta)+i\operatorname{Cl}_2(\theta), \quad \theta \in [0,2\pi).\tag{$\heartsuit$}$$

where $\operatorname{Cl}_2$ and $\operatorname{Sl}_2$ are the standard Clausen functions, defined as: $$\begin{align} \operatorname{Cl}_2(\theta) &= \sum_{k=1}^{\infty}\frac{\sin(k\theta)}{k^2}, \\ \operatorname{Sl}_2(\theta) &= \sum_{k=1}^{\infty}\frac{\cos(k\theta)}{k^2}. \end{align}$$ Using the relationship between SL-type Clausen functions and Bernoulli polynomials, we have that

$$ \operatorname{Sl}_2(\theta) = \frac{\pi^2}{6}-\frac{\pi\theta}{2}+\frac{\theta^2}{4}, \quad \theta \in [0,2\pi).\tag{$\spadesuit$} $$

Now let $z:=\tfrac{1}{2}+\tfrac{i}{6}$. Because $\left|\tfrac{z}{z-1}\right| = 1$, the equation $$ e^{i\theta} = \frac{z}{z-1} = -\frac{4}{5}-\frac{3}{5}i, $$ has the only solution $\theta = \arctan\left(\tfrac{3}{4}\right) + \pi$ in $[0,2\pi)$.

Because of $(\diamondsuit)$ and $(\heartsuit)$ we have $$ \operatorname{Li}_2(z) = -\color{red}{\operatorname{Sl}_2(\theta)} - i \color{green}{\operatorname{Cl}_2(\theta)} - \color{blue}{\frac{1}{2}\ln^2(1-z)}, $$ for $z=\tfrac{1}{2}+\tfrac{i}{6}$ and $\theta = \arctan\left(\tfrac{3}{4}\right) + \pi$.

For the logarithm term we get $$ \Re{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = \frac{1}{8}\left(\ln^2\left(\frac{18}{5}\right)-(\pi-2\arctan 3)^2\right) $$ and $$ \Im{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = \frac{1}{4}\ln\left(\frac{18}{5}\right)(\pi-2\arctan 3). $$

We know that $\color{red}{\operatorname{Sl}_2(\theta)}$ and $\color{green}{\operatorname{Cl}_2(\theta)}$ are real quantities. By using $(\spadesuit)$ for the SL-type Clausen term we get $$ \color{red}{\operatorname{Sl}_2(\theta)} = \frac{\pi^2}{12}-\frac{1}{4}\arctan^2\left(\frac{3}{4}\right). $$

Now we could obtain your conjectured closed-form: $$ \Re\left[\operatorname{Li}_2(z)\right] = -\color{red}{\operatorname{Sl}_2(\theta)} - \Re{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = \frac{7\pi^2}{48}-\frac{\arctan^22}3-\frac{\arctan^23}6-\frac18\ln^2\!\left(\frac{18}5\right). $$

For the imaginary part we have $$\begin{align} \Im\left[\operatorname{Li}_2(z)\right] &= -\color{green}{\operatorname{Cl}_2(\theta)} - \Im{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} \\ &= -\operatorname{Cl}_2\left(\arctan\left(\frac{3}{4}\right)+\pi\right)-\frac{1}{4}\ln\left(\frac{18}{5}\right)(\pi-2\arctan 3). \end{align}$$

By using $(\diamondsuit), (\heartsuit)$ and $(\spadesuit)$, you could generalize this process for all $z \in \mathbb{C}$ such that $\left|\frac{z}{z-1}\right|=1$.