Real analysis and convergence versus $\int_0^1\sum_{n=1}^\infty x^2\mu(n)\frac{1-\cos(nx)}{n(1-\cos(x))}dx$, where $\mu(n)$ is the Möbius function

Question

This morning I was playing (using Wolfram Alpha online calculator) with series involving the Möbius function $\mu(n)$ and the so-called Fejér kernel, see if you need it this Wikipedia.

My conclusion is that my example, next Question, has mathematical meaning (notice that the form of my integrand is due that I believe that such example has mathematical meaning, that is defined on $[0,1]$).

I did some unfinished calculations and without justifications (from my calculations I think that some of the resulting identities hold, but also that maybe is very difficult to justify those).

Question. Is it possible to calculate a good approximation (justifying it), or can you express as a closed-form as a series involving particular values of special functions (justifying it) the integral $$\int_0^1\sum_{n=1}^\infty x^2\mu(n)\left(\frac{1-\cos(nx)}{n(1-\cos(x))}\right)dx\,?\tag{1}$$ Many thanks.

Thus I am asking about what manipulations from real analysis and convergence one can to perform to express $(1)$ with the sign of summation outside of the integral, and after simplify the result in terms of an approximation or as closed-forms of series of particular values of special functions (if it is feasible).

Answer 1

Not sure if it enough but I hope it helps somehow.

Firstly write $ F(n, x) = \dfrac{1}{n}\left(\dfrac{1-\cos(nx)}{1-\cos(x)}\right) $ and $ f(n, x) = x^2 \mu(n) F(n, x) $ so $(1)$ becomes $ \int_0^1\sum_{n=1}^\infty f(n, x)\ dx $. Thus if $ \int_0^1\sum_{n=1}^\infty |f(n, x)|\ dx$ or $ \sum_{n=1}^\infty \int_0^1|f(n, x)|\ dx $ converges then the integration and the summation commutes by the Fubini-Tonelli theorem.

To see that suppose that one of them converges and let $X = [0,1]$, $\mathcal{L}(X)$ the collection of Lebesgue measurable sets of $X$, $\lambda$ the Lebesgue measure and $\mu$ the counting measure. Then both $(X, \mathcal{L}(X), \lambda)$ and $(\mathbb{N}, \mathscr{P}(\mathbb{N}), \mu)$ are $\sigma$-finite measure spaces.

So all Fubini-Tonelli theorem hypotheses are satisfied. Therefore

$$ \int_0^1\sum_{n=1}^\infty f(n, x)\ dx = \int_X \left( \int_\mathbb{N} f(n,x)\ d\mu \right) d\lambda = \int_\mathbb{N} \left( \int_X f(n,x)\ d\lambda \right) d\mu = \sum_{n=1}^\infty \int_0^1 f(n, x)\ dx $$

Furthermore, by the definition of the Fejér kernel we can obtain an integral-free form of $(1)$ $$ \begin{align} \sum_{n=1}^\infty \int_0^1 f(n, x)\ dx &= \sum_{n=1}^\infty \int_0^1 x^2 \mu(n) \frac{1}{n} \sum_{k=0}^{n-1} \sum_{s=-k}^{k} \exp(isx) \ dx \\ &= \sum_{n=1}^\infty \frac{\mu(n)}{n} \sum_{k=0}^{n-1} \sum_{s=-k}^{k} \int_0^1 x^2 \exp(isx) \ dx \\ &= \sum_{n=1}^\infty \frac{\mu(n)}{n} \left(\sum_{k=0}^{n-1} \left( \frac{1}{3} + \sum_{s=-k\\ s \neq 0}^{k} \frac{e^{i s} ((2-i s) s+2 i)-2 i}{s^3} \right) \right) \\ &= \sum_{n=1}^\infty \frac{\mu(n)}{3} \left(\sum_{k=0}^{n-1} \sum_{s=-k\\ s \neq 0}^{k} \frac{e^{i s} ((2-i s) s+2 i)-2 i}{s^3} \right) \\ \end{align} $$

Although also believing $(1)$ converges I could not manage to show it. However if it is really convergent, I guess that a good way to go is writing $(1)$ as $$ \int_0^1 \frac{x^2}{1-\cos(x)} \sum_{n=1}^\infty \mu(n) \frac{1-\cos(n x)}{n}\ dx $$ and showing that the sequence $$ a_k = \sum_{n=1}^{k} \mu(n) \frac{1-\cos(n x)}{n} $$ is Cauchy, because I have computed an approximation for $\lim_\limits{k \to \infty} a_k$ and it seems to have three fixed decimal places. $$ a_{10^8} = -0.922276586957362 $$ $$ a_{10^9} = -0.922303796034064 $$