18
$\begingroup$

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 $p,q\in\mathbb Q$?

$\endgroup$
  • 2
    $\begingroup$ I love your conjecture. Maybe better than Ramanujan ! :) $\endgroup$ – ParaH2 Sep 6 '15 at 21:49
8
$\begingroup$

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:

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

$\endgroup$
  • 1
    $\begingroup$ (+1) I would say it is pretty finished, your answer proves that the dilogarithm reflection formula gives a way for computing the closed form of $\text{Re}\,\text{Li}_2\left(\frac{1}{2}+iq\right)$ for any $q\in\mathbb{R}$. $\endgroup$ – Jack D'Aurizio Sep 6 '15 at 21:16
  • 1
    $\begingroup$ Yes I was going to say that once I'm finished but there's no closed form for the imaginary part tough. $\endgroup$ – Oussama Boussif Sep 6 '15 at 21:17
  • 1
    $\begingroup$ @OussamaBoussif: You can say that again. $\endgroup$ – Lucian Sep 6 '15 at 23:49
  • 2
    $\begingroup$ @OussamaBoussif. Thanks for the great answer! I suppose Lucian meant that this is not the first time when we know a closed form for the real part of a polylogarithm, but not for its imaginary part. $\endgroup$ – Vladimir Reshetnikov Sep 7 '15 at 17:42
  • 3
    $\begingroup$ There are some know results: $$\Im\,\operatorname{Li}_2(i)=G,$$ $$\Im\,\operatorname{Li}_2(1+i)=G+\frac\pi4\ln2,$$ $$\Im\,\operatorname{Li}_2\!\left(\frac{1+i}2\right)=G-\frac\pi8\ln2,$$ $$\Im\,\operatorname{Li}_2\left(i\left(2+\sqrt3\right)\right)=\frac23G-\frac{5 \pi}{12}\ln\left(2-\sqrt3\right),$$ $$\Im\,\operatorname{Li}_2\!\left(\frac{1+i}{\sqrt2}\right)=\frac{1-\sqrt8}4G- \frac{1+\sqrt2}{16} \pi^2+ \frac{\sqrt2}{32}\psi^{(1)}\!\left(\tfrac18\right).$$ $\endgroup$ – Vladimir Reshetnikov Sep 7 '15 at 18:42
3
$\begingroup$

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 $\left|\frac{z}{z-1}\right|=1$.

$\endgroup$
3
$\begingroup$

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 $$\Re\operatorname{Li}_2\left(\frac{1+ti}2\right)=\frac1{12}\pi^2-\frac12\arctan^2t-\frac18\ln^2\frac{1+t^2}4$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.