# A rigorous proof of $\Re\int_0^{2\pi}\left(\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\log\left(\frac{1}{1-ne^{i\theta}}\right)\right)d\theta=2\pi$

We denote the real part of a complex number $z$, as $\Re z$, then I was thinking an exercise. Just I've combined a formula from a course notes and did some manipulations, now I would like to know if we can prove it rigorously. Thus I don't know if there is some special issue about convergence that we need to justify.

For integers $n\geq 1$, wedenote the Möbius function as $\mu(n)$; see the definition of this arithmetic function from this MathWorld.

Question. Prove justifying rigorously each step that (I presume that my identity is right) $$\Re\int_0^{2\pi}\left(\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\log\left(\frac{1}{1-ne^{i\theta}}\right)\right)d\theta=2\pi.\tag{1}$$ Thanks you in advance.

Thus I omit my symbolic manipulations and combinations, waiting for your rigorous proof.

Sketch of my calculations. First I've take the specialization $n\geq 1$ of a formula in [1], page 55, a formlula related to the Poisson kernel, and from this I wrote $$\Re\int_0^{2\pi}\left(\frac{\mu(n)}{n}\log\left(1-ne^{i\theta}\right)\right)d\theta=2\pi\frac{\mu(n)}{n}\log n,$$ after I've combined with the formula $(11)$ of previous MathWorld's article that I've cited. As I've said I don't know if there is some issue of convergence and how justify it rigorousliy.$\square$

From $$\int_0^{2\pi}\left(\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\log\left(\frac{1}{1-ne^{i\theta}}\right)\right)d\theta=-\int_0^{2\pi}\left(\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\left(\log n\log+\left(\frac{1}{n}-e^{i\theta}\right)\right)\right)d\theta,$$ I can to write $$2\pi-\Re\int_0^{2\pi}\left(\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\log\left(\frac{1}{n}-e^{i\theta}\right)\right)d\theta=2\pi.$$
• While it is true that $\int_0^{2\pi} \log(1- ne^{i\theta}) = 2\pi \log n$, I do not see how the Möbius function enters the game? – Fabian Feb 5 '18 at 20:28