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There are few known closed form for values of the dilogarithm at specific points. Sometimes only the real part or only the imaginary part of the value is known, or a relation between several different values is known: [1][2][3][4][5][6]. Discovering a new identity of this sort is always of a great interest.

I numerically discovered the following conjectured closed form and now am looking how to prove it: $$\operatorname{Li}_2\!\left(\sqrt{2-\sqrt3}\cdot e^{i\pi/12}\right)=\operatorname{Li}_2\!\left(\tfrac12+i\left(1-\tfrac{\sqrt3}2\right)\right)\stackrel{\color{gray}?}=\\\frac{23\pi^2}{288}-\frac18\ln^2\!\left(2+\sqrt3\right)+i\left[\frac{2\;\!G}3+\frac{\pi^2}{12\sqrt3}-\frac\pi{24}\ln\left(2+\sqrt3\right)-\frac1{8\sqrt3}\psi^{\small(1)}\!\left(\tfrac13\right)\right]$$

If you have got any ideas please share them.

$G$ is the Catalan constant, and $\psi^{\small(1)}(z)$ is the trigamma function.


To respond questions in comments: my approach to find closed forms like this is to evaluate the expression with hundreds or sometimes thousands digits of precision and then use integer relation algorithms to find a matching linear combination with rational coefficients from a pool of candidate terms. The tricky part is that the pool cannot be too big, otherwise the algorithms begin to work extremely slowly or fail and produce false positives, so I need some intuition to select a restricted set of candidates that are likely to appear in a closed form I'm looking for. I usually select candidates from known closed forms of similar expressions, or from known closed forms of integrals that contain the expression in question along with other simpler terms (its "siblings"). Sometimes I also add some variations of existing terms, e.g. multiply them by a simple irrational coefficient like $\sqrt3$, or replace occurences of $\ln2$ with $\ln3$. When I find a matching linear combination, I verify it with much higher precision (usually tens or hundreds of thousand of decimal digits) to reduce chances that it is an accidental close but non-exact expression.

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  • $\begingroup$ To how many decimals have you numerically checked this? $\endgroup$ – Semiclassical Sep 10 '15 at 22:50
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    $\begingroup$ Sorry for my ignorance, what is $G$? $\endgroup$ – Nikolay Gromov Sep 10 '15 at 23:11
  • $\begingroup$ @Nikolay Gromov it is catalan constant: $G=\sum_{n=0}^{\infty}\frac{{(-1)}^{n}}{{(2n+1)}^{2}}$ $\endgroup$ – Oussama Boussif Sep 10 '15 at 23:16
  • $\begingroup$ @Matt Samuel that's not just guessing. He uses some techniques and algorithms to find the appropriate closed forms. $\endgroup$ – Oussama Boussif Sep 10 '15 at 23:24
  • $\begingroup$ @Semiclassical More than $40000$ digits and counting. $\endgroup$ – Vladimir Reshetnikov Sep 11 '15 at 0:01
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Reduction of dilogarithm. Known identities for the dilogarithm allow us the following simple transform:

$$ \operatorname{Li}_2\left(\tfrac{1}{2} + \tfrac{i}{2} \tan\theta\right) = - \operatorname{Li}_2(e^{i(\pi+2\theta)}) -\tfrac{1}{2}\log^2\left(\tfrac{1}{2} - \tfrac{i}{2} \tan\theta \right). $$

In this case, we have $\theta = \tfrac{1}{12}\pi$ and hence the dilogarithm in question is written as

$$ \operatorname{Li}_2\left(\tfrac{1}{2} + \tfrac{i}{2}\tan\tfrac{\pi}{12} \right) = - \operatorname{Li}_2(e^{7\pi i/6}) -\tfrac{1}{2}\log^2\left( \tfrac{1}{2} - \tfrac{i}{2}\tan\tfrac{\pi}{12} \right). \tag{1} $$

Now utilizing the Fourier series of the Bernoulli polynomial, we know that $\operatorname{Li}_2(e^{i\theta})$ reduces to

$$ \operatorname{Li}_2(e^{i\theta}) = \sum_{k=1}^{\infty} \tfrac{1}{k^2}(\cos (k\theta) + i\sin(k\theta)) = \tfrac{1}{6}\pi^2 - \tfrac{1}{2}\pi\theta + \tfrac{1}{4}\theta^2 + i \operatorname{Cl}_2(\theta) \tag{2} $$

for $0 \leq \theta \leq 2\pi$, where $\operatorname{Cl}_2(\theta) = \sum_{k=1}^{\infty} k^{-2}\sin(k\theta)$ is the Clausen function. So it remains to simplify $\operatorname{Cl}_2 \left( \tfrac{7}{6}\pi \right)$.

Reduction of Clausen function. To this end, we group the terms

$$ \operatorname{Cl}_2 \left(\tfrac{7}{6}\pi\right) = \sum_{k=1}^{\infty} \frac{1}{k^2} \sin \left(\tfrac{7}{6}k\pi\right) $$

according the value of sine and simplifying each group as in this proof, we find that

$$ \operatorname{Cl}_2 \left(\tfrac{7}{6}\pi\right) = \tfrac{1}{144} \sum_{j=1}^{6} \sin \left(\tfrac{7}{6}j\pi\right) \left( \psi^{(1)}\left(\tfrac{j}{12}\right) - \psi^{(1)}\left(\tfrac{1}{2}+\tfrac{j}{12}\right) \right). \tag{3} $$

Now utilizing the reflection formula and the duplication and triplication formula extensively, we can simplify the above sum as

$$ \operatorname{Cl}_2 \left(\tfrac{7}{6}\pi\right) = -\tfrac{2}{3}G - \tfrac{1}{12\sqrt{3}}\pi^2 + \tfrac{1}{8\sqrt{3}} \psi ^{(1)}\left(\tfrac{1}{3}\right). \tag{4} $$

Indeed, we expand the summation in (3) and utilize the triplication formula to the green-colored groups and the duplication formula to the blue-colored groups to obtain

\begin{align*} \operatorname{Cl}_2 \left(\tfrac{7}{6}\pi\right) & = - \tfrac{1}{288} \psi^{(1)}\left(\tfrac{1}{12}\right) + \tfrac{1}{96 \sqrt{3}} \psi^{(1)}\left(\tfrac{1}{6}\right) - \tfrac{1}{144} \psi^{(1)}\left(\tfrac{1}{4}\right) + \tfrac{1}{96 \sqrt{3}} \psi^{(1)}\left(\tfrac{1}{3}\right) - \tfrac{1}{288} \psi^{(1)}\left(\tfrac{5}{12}\right) \\ &\quad + \tfrac{1}{288} \psi^{(1)}\left(\tfrac{7}{12}\right) - \tfrac{1}{96\sqrt{3}} \psi^{(1)}\left(\tfrac{2}{3}\right) + \tfrac{1}{144} \psi^{(1)}\left(\tfrac{3}{4}\right) - \tfrac{1}{96 \sqrt{3}} \psi^{(1)}\left(\tfrac{5}{6}\right) + \tfrac{1}{288} \psi^{(1)}\left(\tfrac{11}{12}\right) \\ & = - \tfrac{1}{288} \color{green}{\left( \psi^{(1)}\left(\tfrac{1}{12}\right) + \psi^{(1)}\left(\tfrac{5}{12}\right) + \psi^{(1)}\left(\tfrac{9}{12}\right) \right)} - \tfrac{3}{288} \psi^{(1)}\left(\tfrac{1}{4}\right) \\ &\quad + \tfrac{1}{288} \color{green}{\left( \psi^{(1)}\left(\tfrac{3}{12}\right) + \psi^{(1)}\left(\tfrac{7}{12}\right) + \psi^{(1)}\left(\tfrac{11}{12}\right) \right)} + \tfrac{3}{288} \psi^{(1)}\left(\tfrac{3}{4}\right) \\ &\quad + \tfrac{1}{96 \sqrt{3}} \color{blue}{\left( \psi^{(1)}\left(\tfrac{1}{6}\right) + \psi^{(1)}\left(\tfrac{2}{3}\right) \right)} + \tfrac{1}{48 \sqrt{3}} \psi^{(1)}\left(\tfrac{1}{3}\right) \\ &\quad - \tfrac{1}{96 \sqrt{3}} \color{blue}{\left( \psi^{(1)}\left(\tfrac{1}{3}\right) + \psi^{(1)}\left(\tfrac{5}{6}\right) \right)} - \tfrac{1}{48\sqrt{3}} \psi^{(1)}\left(\tfrac{2}{3}\right) \\ & = \tfrac{1}{16 \sqrt{3}} \left( \psi^{(1)}\left(\tfrac{1}{3}\right) - \psi^{(1)}\left(\tfrac{2}{3}\right) \right) - \tfrac{1}{24} \left( \psi^{(1)}\left(\tfrac{1}{4}\right) - \psi^{(1)}\left(\tfrac{3}{4}\right) \right). \end{align*}

Now we focus on the last line. Applying the reflection formula to the first term and comparing the definition of the Catalan constant $G$ with the second term, we obtain (4) as claimed.

Finally, plugging this back gives the desired result.


Addendum: Fourier series of the Bernoulli polynomial. Taking imaginary part of

$$ \log(1-e^{2\pi i x}) = - \sum_{k=1}^{\infty} \frac{e^{2\pi i k x}}{k} $$

for $x \in (0, 1)$, we find that the Bernoulli polynomial $B_1(x)$ of degree 1 is written as

$$ B_1( x ) = x - \tfrac{1}{2} = - \frac{1}{2\pi i} \sum_{k \neq 0} \frac{e^{2\pi i k x}}{k}, \quad 0 < x < 1.$$

Integrating both sides repeatedly and using the relation $B_n'(x) = nB_{n-1}(x)$, we find that for any $n \geq 1$,

$$ B_n( x ) = - \frac{n!}{(2\pi i)^n} \sum_{k \neq 0} \frac{e^{2\pi i k x}}{k^n}, \quad 0 < x < 1.$$

Notice that depending on whether $n$ is even or odd, this reduces to either cosine series or sine series. For example, when $n = 2$ we have

$$ x^2 - x + \tfrac{1}{6} = B_2(x) = \frac{1}{\pi^2} \sum_{k=1}^{\infty} \frac{\cos(2\pi k x)}{k^2} $$

and hence we obtain the formula which was used in our solution:

$$ \sum_{k=1}^{\infty} \frac{\cos(k x)}{k^2} = \pi^2 B_2\left(\tfrac{1}{2\pi} x\right) = \tfrac{1}{4} x^2 - \tfrac{1}{2}\pi x + \tfrac{1}{6}\pi^2. $$

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  • $\begingroup$ Could you please elaborate on the Fourier series of the Bernoulli polynomial? $\endgroup$ – Vladimir Reshetnikov Sep 11 '15 at 15:44
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We use the main three steps from my answer to one of your previous questions.

$$\operatorname{Li}_2(z) = -\operatorname{Li}_2\left(\frac{z}{z-1}\right)-\frac{1}{2}\ln^2(1-z), \quad z \notin (1,\infty).\tag{$\diamondsuit$}$$ $$\operatorname{Li}_2\left(e^{i\theta}\right) = \operatorname{Sl}_2(\theta)+i\operatorname{Cl}_2(\theta), \quad \theta \in [0,2\pi).\tag{$\heartsuit$} $$ $$ \operatorname{Sl}_2(\theta) = \frac{\pi^2}{6}-\frac{\pi\theta}{2}+\frac{\theta^2}{4}, \quad \theta \in [0,2\pi).\tag{$\spadesuit$} $$ For definitions and notations see my previous answer.


Let $z:=\tfrac{1}{2}+i\left(1-\tfrac{\sqrt3}{2}\right).$ Note that $$ \frac{z}{z-1} = -\frac{\sqrt 3}{2} - \frac{i}{2}, $$ and therefore $\left|\frac{z}{z-1}\right|=1$.

The equation $$ e^{i\theta} = \frac{z}{z-1} = -\frac{\sqrt 3}{2} - \frac{i}{2} $$ has the only solution $\theta=\tfrac76 \pi$ 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 $z=\tfrac{1}{2}+i\left(1-\tfrac{\sqrt3}{2}\right)$ and $\theta =\tfrac76 \pi$.

For the logarithm term we get $$ \Re{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = \frac{1}{8}\ln^2\left(2-\sqrt{3}\right)-\frac{\pi^2}{288} $$ and $$ \Im{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = -\frac{\pi}{24}\ln\left(2-\sqrt{3}\right). $$ 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 $$ \color{red}{\operatorname{Sl}_2(\theta)} =-\frac{11\pi^2}{144}. $$ Now we could obtain your conjectured closed-form for the real part: $$\Re\left[\operatorname{Li}_2(z)\right] = -\color{red}{\operatorname{Sl}_2(\theta)} - \Re{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = \frac{23\pi^2}{288}-\frac{1}{8}\ln^2\left(2-\sqrt{3}\right).$$

For the imaginary part we have $$\begin{align}\Im\left[\operatorname{Li}_2(z)\right] &= -\color{green}{\operatorname{Cl}_2(\theta)} - \Im{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} \\ &= -\operatorname{Cl}_2\left(\tfrac{7}{6}\pi\right)+\frac{\pi}{24}\ln\left(2-\sqrt{3}\right).\end{align}$$ Now by using the relationship between Clausen function and polygamma function and by using some polygamma tricks, we can get that $$\operatorname{Cl}_2 \left(\frac{7}{6}\pi\right) = -\frac{2}{3}G - \frac{\pi^2}{12\sqrt{3}} + \frac{1}{8\sqrt{3}} \psi ^{(1)}\left(\frac{1}{3}\right).$$ This completes the proof.

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