5

I came across this polylogrithmic integral that evaluates to a close form of dilogarithmic value $$\int_0^1 \frac{\ln\frac{1+a}{1+x}}{a-\frac1a{x^2}}dx=\frac{\pi^2}{24}-\frac14\operatorname{Li}_2\bigg[\bigg(\frac{a-1}{a+1}\bigg)^2\bigg] $$ for $a>1$. So far, I am not able to derive it. Knowing the form of the result, I expect to utilize the identity $$\operatorname{Li}_2(z)+\operatorname{Li}_2(-z)= \frac12\operatorname{Li}_2(z^2)$$ But, I was not able to manipulate the integrand to fit it into the desired sum of dilogarithm integrals. It may not be as straightforward as I expected.

A-Level Student
  • 6,520
Quanto
  • 97,352
  • I would try to use series: for $0 \le x \le 1 < a$, $$\frac{1}{1-x^2/a^2} = \sum_{n=0}^{+\infty} \frac{x^{2n}}{a^{2n}},$$ use normal convergence to switch the sum and the integral, and integrate by parts each term $x^{2n}\ln \frac{1+x}{1+a}$, but I do not know whether it gives the result or not. – Christophe Leuridan Apr 28 '22 at 21:17

1 Answers1

6

$$I=\int_0^1 \frac{\ln\left(\frac{1+a}{1+ax}\right)}{a-\frac{1}{a}x^2}dx\overset{\large x\to\, a\frac{1-x}{1+x}}=\frac12\int_k^1\frac{\ln\left(\frac{1+x}{1-kx}\right)}{x}dx=I(k); \ k=\frac{a-1}{a+1}$$

$$\frac{\partial}{\partial k}I(k)=\frac{\ln(1-k)}{2k}+\frac12\int_k^1 \frac{1}{1-kx}dx=\frac{\ln(1-k^2)}{2k}$$

$$\Rightarrow I=\frac12\int_{-1}^k\frac{\ln(1-x^2)}{x}dx=-\frac14\operatorname{Li}_2(x^2)\bigg|_{-1}^k=\frac{\pi^2}{24}-\frac14\operatorname{Li}_2(k^2)$$

Zacky
  • 27,674
  • 2
    This is clever! – Quanto Apr 28 '22 at 23:44
  • Thank you! I don't know what's the story behind the integral, however I find it quite interesting that it relates to another integral (from here). I'm not sure about proving a direct relationship, but there is: $$\int_0^\infty\frac{\arctan(ax)\operatorname{arccot}(ax)}{1+x^2}dx=\frac{\pi}{2}\int_0^1 \frac{\ln\left(\frac{1+a}{1+ax}\right)}{a-\frac{1}{a}x^2}dx$$ – Zacky Apr 29 '22 at 00:05
  • 2
    Mine originates form this parity $\int_0^a\int_0^\frac1a \frac1{(1+x)(1+y)(1+xy)}dxdy$. They ought to be related. – Quanto Apr 29 '22 at 00:17
  • 2
    Also note $$\int_k^1\frac{\ln\left(\frac{1+x}{1-kx}\right)}{x}dx=\text{Li}2(kx)-\text{Li}_2(-x)|_k^1=\frac12\zeta(2)+\underbrace{\text{Li}_2(k)+\text{Li}_2(-k)}{\frac12\text{Li}_2(k^2)}-\text{Li}_2(k^2)$$ – Ali Shadhar Jun 15 '22 at 18:27