# Convergence of $\sum_{n=1}^\infty\mu(n)\int_0^1\cos\left(\sqrt{\pi n z}\right)dz$, where $\mu(n)$ is the Möbius function

I was interested from the MathWorld's article about the so-called Meijer G-Function to compute integrals of similar functions $$\int_0^1\cos\left(\sqrt{\pi n z}\right)dz.\tag{1}$$ And from this $(1)$ to compute using Wolfram Alpha online calculator partial sums of $$\sum_{n=1}^\infty\mu(n)\int_0^1\cos\left(\sqrt{\pi n z}\right)dz=\frac{2}{\pi}\sum_{n=1}^\infty\frac{\mu(n)}{n}\left(\sqrt{\pi n}\sin\left(\sqrt{\pi n }\right)+\cos\left(\sqrt{\pi n }\right)-1\right),\tag{2}$$ where $\mu(n)$ denotes the Möbius function, see this MathWorld.

Question. Is it possible to prove that $$\sum_{n=1}^\infty\mu(n)\int_0^1\cos\left(\sqrt{\pi n z}\right)dz$$ does converge? Thanks in advance.

If this question or similar was in the literature feel free to refer it, answering this question as a reference request. Then I can to search and read those statements.

• LOL. Do you want to prove Riemann Hypothesis with a little help from MSE? – user436658 Jan 4 '18 at 12:57
• Good afternoon @ProfessorVector I don't know how is related my Question with the Riemann Hypothesis. If you know it and want to explain such relation feel free to add a comment. – user243301 Jan 4 '18 at 13:22
• I'll comment on that as soon as I've seen your own thoughts about this problem, its significance, motivation for asking this... in a word, context. – user436658 Jan 4 '18 at 13:25
• @barto You see context, here?! Wow. I see links to thoughts of other people, but not any own research, not any own thought. And according to the rules of this site, that's not asking too much. – user436658 Jan 4 '18 at 20:06
• @ProfessorVector: I seem to recall that proving the convergence of $\sum_{n\geq 1}\frac{\mu(n)}{n^{1/2+\varepsilon}}$ for any $\varepsilon>0$ is more or less equivalent to RH; but I also seem to recall that $$\sum_{n\geq 1}\frac{\mu(n)}{\sqrt{n}}$$ can be proved to be non-convergent: mathoverflow.net/q/164874/30521 – Jack D'Aurizio Jan 4 '18 at 23:08

Well, the integral is fairly easy to do: $$\int_0^1 \cos (\sqrt{\pi n z}) dz = 2\int_0^1 y \cos(\sqrt{\pi n} y) dy = 2 [\frac{1}{\sqrt{\pi n}} y \sin(\sqrt{\pi n} y) + \frac{1}{\pi n} \cos(\sqrt{\pi n} y)] |_0^1 = \frac{2}{\sqrt{\pi n}} \sin(\sqrt{\pi n}) + \frac{2}{\pi n} (\cos(\sqrt{\pi n}) -1)$$

Now, the sum involving $\mu(n)/n$ might(?) converge, and the sum involving $\mu(n)/\sqrt{n}$ -- well, that may converge conditional on the Riemann Hypothesis. But I can't see any easy way of proving either statement. And I also don't see any reason there should be a cancellation between the two.

• Many thanks for your remarks and calculations. – user243301 Jan 8 '18 at 21:05
• I don't know if can be interesting do the same experiments with different erratic functions instead of the Möbius function, I say the Liouville function $\lambda(n)$, or their summatory functions of Möbius, the Mertens function, or $\sum_{n=1}^N \lambda(n)$. I don't know if it was in the literature or as I am saying is interesting (I didn't experiments), but feel free to study/explore it in your home. – user243301 Jan 12 '18 at 22:46

We want to calculate $S=\sum_{n=1}^\infty\mu(n)\int_0^1\cos\left(\sqrt{\pi n z}\right)dz$. First we try to calculate $\int_0^1\cos\left(\sqrt{\pi n z}\right)dz$ somehow. To do that we do the followings: $$\int_0^1\cos\left(\sqrt{\pi n z}\right)dz=\int_0^1 2u\cos\left(\sqrt{\pi n }u\right)du=2(\frac{\sin \sqrt{\pi n}}{\sqrt{\pi n}}+\frac{\cos \sqrt{\pi n}-1}{\pi n})$$ Therefore the summation can be written as follows: $$S=2\sum_{n=1}^\infty\mu(n)(\frac{\sin \sqrt{\pi n}}{\sqrt{\pi n}}+\frac{\cos \sqrt{\pi n}-1}{\pi n})$$ Now we prove that $X=\sum_{n=1}^\infty\frac{\sin \sqrt{\pi n}}{\sqrt{\pi n}}+\frac{\cos \sqrt{\pi n}-1}{\pi n}$ is convergent and then we conclude the same for $S$. By Cauchy root test we have: $$L=lim_{n\to\infty}|\frac{\sin \sqrt{\pi n}}{\sqrt{\pi n}}+\frac{\cos \sqrt{\pi n}-1}{\pi n}|^{\frac{1}{n}}$$$$=lim_{u\to 0}|{u\sin {1\over u}}+u^2({\cos {1\over u}-1)}|^{\pi u^2}$$ By substituting using Taylor series we get: $$L=lim_{u\to 0}|1-{1\over{6u^2}}+O(\frac{1}{u^4})-\frac{1}{2}+\frac{1}{24u^2}+O(\frac{1}{u^4})|^{\pi u^2}$$$$=lim_{u\to 0}|\frac{1}{2}-{1\over{8u^2}}+O(\frac{1}{u^4})|^{\pi u^2}$$$$=lim_{u\to 0}(\frac{1}{2})^{\pi u^2}|1-{1\over{4u^2}}+O(\frac{1}{u^4})|^{\pi u^2}$$$$=lim_{u\to 0}|1-{1\over{4u^2}}|^{\pi u^2}=lim_{u\to 0}|1-{1\over{4u^2}}|^{{4u^2}{\pi\over 4}}=1$$ which implies on $X$ being convergent. Since $\mu(n)$ is a function taking on values $-1,0,1$ therefore it keeps the limit of the general sequence $<1$ and also implies in convergence of S.

• Note that $\lim_{u\to \color{red}{0}} |1-(2u)^{-2}|^{4u^2}=1\neq e^{-1}$ – Shashi Jan 14 '18 at 1:01
• Assume $w=4u^2$ so we are seeking for finding $L=lim_{w\to 0} |1-\frac{1}{w}|^w$. Taking ln from both sides you have $\ln L=lim_{w\to 0}\frac{ln(1-{1\over u})}{{1\over u}}=lim_{u\to 0} \frac{u}{1-u}=-1$ by L'Hopital rule. therefore $ln L=-1\to L=\frac{1}{e}$ – Mostafa Ayaz Jan 14 '18 at 1:14
• please look carefully at every step of the last comment you have sent to me. First of all $w\to 0^+$. Secondly you drop the absolute bars without justifying. Thirdly, you have a negative argument in the logarithm. – Shashi Jan 14 '18 at 1:23
• You are right the limit is 1 i fixed it.... of course it doesn't change the conclusion of convergence – Mostafa Ayaz Jan 14 '18 at 8:52
• I thought that getting $1$ does not say anything about the convergence. The root test is then inconclusive.. – Shashi Jan 14 '18 at 9:55