We denote with $\arctan(x)$ the inverse tangent function, and for integers $n\geq 1$ the Möbius function $\mu(n)$, see its definition from this MathWorld.
We know that the limit of the antiderivative $\int \arctan(x)dx$ as $x\to0$ is $0$, and also (or from the graph of the inverse tangent function it is obvious) that doesn't exist $\lim_{x\to\infty}\int_0^x \arctan(t)dt$.
Question. Prove or refute that $$ \left| \int_0^x \mu \left( \lceil t\rceil \right)\arctan(t)\,dt \right|,\tag{1}$$ where $ \lceil y\rceil$ denotes the ceil function, is bounded as $x\to\infty$. Thanks in advance.
I know that, with the purpose to solve this exercise, should be interesting to know the Taylor series for the inverse tangent function, and split the interval of integration according to the distribution of squarefree integers with an odd/even number of distinct prime factors.
Example of notation and computational fact. One has $\mu \left( \lceil 0.1\rceil \right)=\mu(1)=1$ and $\mu \left( \lceil 1.1\rceil \right)=\mu(2)=-1$. See the graph that provided me Wolfram Alpha online calculator with this my code
int mu(ceil(x))arctan(x)dx, from x=0 to 50
It is unbounded.
$$\begin{aligned} \int_0^x {\mu (\left\lceil t \right\rceil )\arctan tdt} &= \int_1^x {\mu (\left\lceil t \right\rceil )\left( {\arctan t - \frac{\pi }{2}} \right)dt} + \frac{\pi }{2}\int_1^x {\mu (\left\lceil t \right\rceil )dt} + O(1) \\ &= \int_1^x {\mu (\left\lceil t \right\rceil )\left( { - \frac{1}{t} + O(\frac{1}{{{t^3}}})} \right)dt} + \frac{\pi }{2}\sum\limits_{n \le x} {\mu (n)} + O(1) \\ &= - \int_1^x {\frac{{\mu (\left\lceil t \right\rceil )}}{t}dt} + \frac{\pi }{2}M(x) + O(1) \\ &= - \frac{{M(x)}}{x} - \int_1^x {\frac{{M(t)}}{{{t^2}}}dt} + \frac{\pi }{2}M(x) + O(1) \end{aligned}$$
where we used summation by parts in the last line, with $M(x)$ being the Mertens function. The first term tends to $0$ as $x\to \infty$ (a consequence of prime number theorem). Now we claim that, the 2nd term is bounded, while 3rd is unbounded, both claims involve some deep results.
For 2nd term, we invoke the famous growth estimate (the best currently available): \begin{equation}\tag{1}M(x) = O(x\exp(-c\ln^{3/5}x (\ln \ln x)^{-1/5})\end{equation} where $c>0$ is a constant. Thus it suffices to show that $$\int_{}^\infty \frac{\exp(-c\ln^{3/5}t (\ln \ln t)^{-1/5})}{t} dt$$ converges. This in turn follows easily from integral test and the convergence of $$\int_{}^\infty e^{-cu^{3/5} (\ln u)^{-1/5}} du$$
For the 3rd term, it is easily obtained from the fact that \begin{equation}\tag{2}\limsup_{x\to \infty} \frac{M(x)}{\sqrt{x}} > 1 \end{equation} although this is again non-trivial. Perhaps there is a simpler way to prove that Merten's function is unbounded.
You can find the references regarding $(1),(2)$ here:http://www.cs.uleth.ca/~nathanng/RESEARCH/mobius2b.pdf.