Integrating Cyclotomic Polynomials

Another delightful problem to do at the breakfast table on a Saturday morning! Thanks, Comrade!

First note that

$\int_{0}^{1}\frac{\ln(1-x^n)}{x}dx=-\frac{Li_2(1)}{n}=-\frac{\pi^2}{6n}$

Now $1-x^n=\prod_{d|n}f_d(x)$ where $f_d(x)$ is the $d$ th cyclotomic polynomial except for $f_1(x)=1-x$ . Thus $\ln(1-x^n)=\sum_{d|n}\ln(f_d(x))$ and , by Möbius inversion, $\ln(f_n(x))=\sum_{d|n}\mu(d)\ln(1-x^{n/d})$ . Then $\int_{0}^{1}\frac{\ln(f_n(x))}{x}dx=-\frac{\pi^2}{6n}\sum_{d|n}\mu(d)d=-\frac{\pi^2}{6n}\prod_{p|n}(1-p)$ where $p$ is prime.

For $n=2015=5\times 13\times 31$ this gives $\frac{\phi(2015)}{2015}\times \frac{\pi^2}{6}=\frac{48\pi^2}{403}$ and the answer is $\boxed{451}$ .

Small discrepancy: My answer comes out positive while yours is negative; let's both double-check (the sign error may well be on my end). According to my work, the answer comes out positive iff $n$ has an odd number of distinct prime factors.