Proving Identity Involving Dilogarithms And Π / 9 \pi/9 Π /9

May 17, 2025 by ADMIN 61 views

**Proving Identity Involving Dilogarithms and π/9**

Introduction

The dilogarithm function, also known as the polylogarithm of order 2, is a special function that has been extensively studied in mathematics. It is defined as $\operatorname{Li}_2(x) = \sum_{n=1}^{\infty} \frac{x^n}{n^2}$ for $|x| \leq 1$ . In this article, we will focus on proving an identity involving dilogarithms and $\pi/9$ , which was mentioned in a textbook "The dilogarithm in algebraic fields" and on Mathworld, but no proof was provided.

The Identity

The identity we want to prove is:

\operatorname{Li}_2(-a) + \operatorname{Li}_2(a^2) - \frac{1}{3} \operatorname{Li}_2(-a^3) = \frac{\pi^2}{9} - \frac{1}{3} \log^2 \left( \frac{1 + a}{1 - a} \right)

where $a$ is a real number such that $|a| < 1$ .

Proof

To prove this identity, we will use the following properties of the dilogarithm function:

$\operatorname{Li}_2(x) = -\operatorname{Li}_2(1-x)$
$\operatorname{Li}_2(x) = \operatorname{Li}_2(x^2) + \log(1-x) \log(x)$

Using these properties, we can rewrite the left-hand side of the identity as:

\operatorname{Li}_2(-a) + \operatorname{Li}_2(a^2) - \frac{1}{3} \operatorname{Li}_2(-a^3)

= -\operatorname{Li}_2(1+a) + \operatorname{Li}_2(a^2) - \frac{1}{3} (-\operatorname{Li}_2(1-a^3))

= -\operatorname{Li}_2(1+a) + \operatorname{Li}_2(a^2) + \frac{1}{3} \operatorname{Li}_2(1-a^3)

Now, we can use the second property of the dilogarithm function to rewrite the expression as:

= -\operatorname{Li}_2(1+a) + \operatorname{Li}_2(a^2) + \frac{1}{3} \operatorname{Li}_2(1-a^3)

= -\operatorname{Li}_2(1+a) + \operatorname{Li}_2(a^2) + \frac{1}{3} \left( \operatorname{Li}_2(a^6) + \log(1-a^3) \log(a^3) \right)

= -\operatorname{Li}_2(1+a) + \operatorname{Li}_2(a^2) + \frac{1}{3} \operatorname{Li}_2(a^6) + \frac{1}{3} \log(1-a^3) \log(a^3)

Now, we can use the first property of the dilogarithm function to rewrite the expression as:

= -\operatorname{Li}_(1+a) + \operatorname{Li}_2(a^2) + \frac{1}{3} \operatorname{Li}_2(a^6) + \frac{1}{3} \log(1-a^3) \log(a^3)

= -\operatorname{Li}_2(1+a) + \operatorname{Li}_2(a^2) + \frac{1}{3} \operatorname{Li}_2(a^6) + \frac{1}{3} \left( \log(1-a) \log(a) + \log(1+a) \log(a) \right)

= -\operatorname{Li}_2(1+a) + \operatorname{Li}_2(a^2) + \frac{1}{3} \operatorname{Li}_2(a^6) + \frac{1}{3} \log(1-a) \log(a) + \frac{1}{3} \log(1+a) \log(a)

Now, we can use the fact that $\log(1-a) \log(a) = -\log(1+a) \log(a)$ to rewrite the expression as:

= -\operatorname{Li}_2(1+a) + \operatorname{Li}_2(a^2) + \frac{1}{3} \operatorname{Li}_2(a^6) - \frac{1}{3} \log(1+a) \log(a) + \frac{1}{3} \log(1+a) \log(a)

= -\operatorname{Li}_2(1+a) + \operatorname{Li}_2(a^2) + \frac{1}{3} \operatorname{Li}_2(a^6)

Now, we can use the fact that $\operatorname{Li}_2(a^2) = \operatorname{Li}_2(a^6) + \log(1-a^2) \log(a^2)$ to rewrite the expression as:

= -\operatorname{Li}_2(1+a) + \operatorname{Li}_2(a^6) + \log(1-a^2) \log(a^2) + \frac{1}{3} \operatorname{Li}_2(a^6)

= -\operatorname{Li}_2(1+a) + \frac{4}{3} \operatorname{Li}_2(a^6) + \log(1-a^2) \log(a^2)

Now, we can use the fact that $\operatorname{Li}_2(a^6) = \operatorname{Li}_2(a^2) + \log(1-a^6) \log(a^6)$ to rewrite the expression as:

= -\operatorname{Li}_2(1+a) + \frac{4}{3} \operatorname{Li}_2(a^2) + \frac{4}{3} \log(1-a^6) \log(a^6) + \log(1-a^2) \log(a^2)

= -\operatorname{Li}_2(1+a) + \frac{4}{3} \operatorname{Li}_2(a^2) + \frac{4}{3} \log(1-a^6) \log(a^6) + \log(1-a^2) \log(a^2)

Now, we can use the fact that $\log(1-a^2) \log(a^) = -\log(1+a^2) \log(a^2)$ to rewrite the expression as:

= -\operatorname{Li}_2(1+a) + \frac{4}{3} \operatorname{Li}_2(a^2) + \frac{4}{3} \log(1-a^6) \log(a^6) - \log(1+a^2) \log(a^2)

= -\operatorname{Li}_2(1+a) + \frac{4}{3} \operatorname{Li}_2(a^2) + \frac{4}{3} \log(1-a^6) \log(a^6) - \log(1+a^2) \log(a^2)

Now, we can use the fact that $\log(1-a^6) \log(a^6) = -\log(1+a^6) \log(a^6)$ to rewrite the expression as:

= -\operatorname{Li}_2(1+a) + \frac{4}{3} \operatorname{Li}_2(a^2) - \frac{4}{3} \log(1+a^6) \log(a^6) - \log(1+a^2) \log(a^2)

= -\operatorname{Li}_2(1+a) + \frac{4}{3} \operatorname{Li}_2(a^2) - \frac{4}{3} \log(1+a^6) \log(a^6) - \log(1+a^2) \log(a^2)

Now, we can use the fact that $\log(1+a^2) \log(a^2) = -\log(1-a^2) \log(a^2)$ to rewrite the expression as:

= -\operatorname{Li}_2(1+a) + \frac{4}{3} \operatorname{Li}_2(a^2) - \frac{4}{3} \log(1+a^6) \log(a^6) + \log(1-a^2) \log(a^2)

= -\operatorname{Li}_2(1+a) + \frac{4}{3} \operatorname{Li}_2(a^2) - \frac{4}{3} \log(1+a^6) \log(a^6) + \log<br/> **Q&A: Proving Identity Involving Dilogarithms and π/9** =====================================================

Q: What is the dilogarithm function?

A: The dilogarithm function, also known as the polylogarithm of order 2, is a special function that has been extensively studied in mathematics. It is defined as $\operatorname{Li}_2(x) = \sum_{n=1}^{\infty} \frac{x^n}{n^2}$ for $|x| \leq 1$ .

Q: What is the identity we want to prove?

A: The identity we want to prove is: