# 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

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).

• As was said the closed-form of series can involve the Möbius function and particular values of special function, aren't required that you provide me the sum (because get closed-form of series involving the Möbius function is very difficutl/impossible). Thus is required express as a series the result $(1)$, or well if it is your approach study an approximation. Always in any approach, focusing on or adding the justification/reasoning of the convergence issues. Many thanks. – user243301 Jan 9 '18 at 9:20
• Everyone, I think that $(1)$ is convergent. – user243301 Jan 13 '18 at 13:12
• As you think (1) is convergent, could you please write out your proof as an answer? – Hans Jan 22 '18 at 19:38
• @Hans many thanks for your attention, you think that am I wong? – user243301 Jan 22 '18 at 19:42
• No, I do not think you are wrong. But that does not mean I think you are right, either. You can always numerical compute the series to give you an idea of the likelihood of convergence. But that does not tell us the answer for sure. We all need a proof to back up one's claim. Do you not agree? – Hans Jan 22 '18 at 19:47

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. $$a_{10^8} = -0.922276586957362$$ $$a_{10^9} = -0.922303796034064$$

• Many thanks I am going to study your answer. Many thanks thus for your effort, details and calculations. – user243301 Jan 19 '18 at 16:24
• @user243301 You are welcome. Let me know if you manage to show the convergence. – mucciolo Jan 19 '18 at 16:41
• I think that it is enough for me, of course many thanks for your great answer. If you want feel free to study questions of the same kind. – user243301 Jan 20 '18 at 19:34