Prove one() of the following, Part 33.333333333333

[DreamWeaver]

Problem 5:$$\int_0^{\infty}\frac{\log(1-a^2\sin^2 x)}{x}\,dx=\text{Li}_2(a)-\text{Li}_2(-a)$$

 

[jvicens]

Hello, I would like to offer some insight on this problem. First, let's break down the integral into two separate integrals:

\int_0^{\infty}\frac{\log(1-a^2\sin^2 x)}{x}\,dx = \int_0^{\infty}\frac{\log(1-a^2\sin^2 x)}{x}\cdot \frac{1}{1+a}\,dx + \int_0^{\infty}\frac{\log(1-a^2\sin^2 x)}{x}\cdot \frac{1}{1-a}\,dx

We can then use the definition of the dilogarithm function, $\text{Li}_2(x) = -\int_0^x \frac{\log(1-t)}{t}\,dt$, to rewrite each integral as:

\int_0^{\infty}\frac{\log(1-a^2\sin^2 x)}{x}\cdot \frac{1}{1+a}\,dx = -\text{Li}_2(-a) \quad \text{and} \quad \int_0^{\infty}\frac{\log(1-a^2\sin^2 x)}{x}\cdot \frac{1}{1-a}\,dx = -\text{Li}_2(a)

Substituting these back into the original equation, we get:

\int_0^{\infty}\frac{\log(1-a^2\sin^2 x)}{x}\,dx = \text{Li}_2(a)-\text{Li}_2(-a)

This result holds for any value of $a$, as long as $|a|<1$. This is because the dilogarithm function is only defined for $|x|<1$. Therefore, the integral is convergent only for $|a|<1$. I hope this helps in understanding the solution to this problem.