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.
Well, the integral is fairly easy to do: \begin{equation} \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) \end{equation}
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.