Showing that $Q(x)-\frac{6x}{\pi^2}=\Omega_{\pm}(x^{1/4})$ Let $Q(x)$ denote the number of square-free numbers not exceeding $x$.
How can we show that $Q(x)-\frac{6x}{\pi^2}=\Omega_{\pm}(x^{1/4})$, i.e.
$$\liminf_{x\to +\infty} \frac{Q(x) - \frac{6x}{\pi^2}}{x^{1/4}} < 0 \qquad\text{and}\qquad \limsup_{x \to +\infty} \frac{Q(x) - \frac{6x}{\pi^2}}{x^{1/4}} > 0\,?$$
The computation
\begin{align}
Q(x) &= \sum_{k \leqslant \sqrt{x}} \mu(k)\biggl\lfloor \frac{x}{k^2}\biggr\rfloor \\
&= \sum_{k \leqslant \sqrt{x}} \mu(k)\,\frac{x}{k^2} - \sum_{k \leqslant \sqrt{x}} \mu(k)\biggl\lbrace \frac{x}{k^2}\biggr\rbrace \\
&= \frac{6x}{\pi} - x\sum_{k > \sqrt{x}} \frac{\mu(k)}{k^2} - \sum_{k \leqslant \sqrt{x}} \mu(k)\biggl\lbrace \frac{x}{k^2}\biggr\rbrace
\end{align}
yields $Q(x) - \frac{6x}{\pi^2} = O(\sqrt{x})$ easily, but it is not at all obvious if we can obtain a lower bound on the magnitude of $Q(x) - \frac{6x}{\pi^2}$ from this. How does one go about proving these?
 A: I will consider $R(x) := Q(x) - \frac{6}{\pi^2} \,\lfloor x\rfloor$ instead of $R_0(x) := Q(x) - \frac{6}{\pi^2}\,x$ because the former has a slightly more convenient Mellin transform. Since $R(x) - R_0(x) = \frac{6}{\pi^2} \,\lbrace x\rbrace$ is bounded, these two have the same $\Omega_{\pm}(x^{\alpha})$ characteristics for all $\alpha > 0$, so showing $R(x) = \Omega_{\pm}(x^{1/4})$ also solves the original question.
First we note that for $\operatorname{Re} s > 1$
$$s\int_1^{+\infty} \frac{Q(x)}{x^{s+1}}\,dx = \sum_{n = 1}^{+\infty} \frac{\lvert\mu(n)\rvert}{n^s} = \prod_{p}\biggl(1 + \frac{1}{p^s}\biggr) = \prod_{p}\frac{1 - \frac{1}{p^{2s}}}{1 - \frac{1}{p^s}} = \frac{\zeta(s)}{\zeta(2s)}\,.$$
Then, since
$$\zeta(s) = \sum_{n = 1}^{+\infty} \frac{1}{n^s} = s\int_1^{+\infty} \frac{\lfloor x\rfloor}{x^{s+1}}\,dx\,,$$
we obtain
$$H(s) := s\int_1^{+\infty} \frac{R(x)}{x^{s+1}}\,dx = \frac{\zeta(s)}{\zeta(2s)} - \frac{6}{\pi^2}\zeta(s) = \biggl(\frac{1}{\zeta(2s)} - \frac{1}{\zeta(2)}\biggr)\,\zeta(s)\tag{1}$$
first for $\operatorname{Re} s > 1$, then by the identity theorem on the half-plane where $R(x)\cdot x^{-s-1}$ is integrable. By the estimate $R(x) \in O(\sqrt{x})$, this holds at least for $\operatorname{Re s} > \frac{1}{2}$. The right hand side of $(1)$ gives a meromorphic continuation of $H$ to the entire plane. The pole of $\zeta(s)$ at $1$ is cancelled by the zero of $\frac{1}{\zeta(2s)} - \frac{1}{\zeta(2)}$ there, so the poles of $H$ are precisely the zeros (trivial and nontrivial) of $\zeta(2s)$. From
$$\zeta(s) = s\int_1^{+\infty} \frac{x-\lbrace x\rbrace}{x^{s+1}}\,dx = s\int_1^{+\infty} \frac{dx}{x^s} - s\int_1^{+\infty} \frac{\lbrace x\rbrace}{x^{s+1}}\,dx = \frac{s}{s-1} - s\int_1^{+\infty} \frac{\lbrace x\rbrace}{x^{s+1}}\,dx$$
we can read off that $\zeta(s) < 0$ on the real interval $(0,1)$. Since $\zeta(s) > 0$ on $(1,+\infty)$, it follows that $H$ has no pole on the positive real axis. We know that $\zeta$ has zeros on the line $\operatorname{Re} s = \frac{1}{2}$, hence $H$ has poles on the line $\operatorname{Re} s = \frac{1}{4}$. This implies that we cannot have a bound $\lvert R(x)\rvert \leqslant C\cdot x^{1/4-\varepsilon}$ for any $\varepsilon > 0$, since such a bound would imply that $H$ is holomorphic on the half-plane $\operatorname{Re} s > \frac{1}{4} - \varepsilon$ as the integral defining $H$ would converge on that half-plane.
To finish the proof, we need a lemma similar to Landau's lemma for Dirichlet series with non-negative coefficients:
Lemma: Let $K \geqslant 1$ and $f \colon [K,+\infty) \to [0,+\infty)$ a measurable function. If
$$\inf\: \biggl\{ t\in \mathbb{R} : \int_K^{+\infty} \frac{f(x)}{x^t}\,dx < +\infty\biggr\} = 0\,,\tag{$\ast$}$$
then the holomorphic function defined on the right half-plane by
$$F(s) = \int_K^{+\infty} \frac{f(x)}{x^s}\,dx$$
has a singularity at $0$. (That is, there is no holomorphic function $G$ defined on a neighbourhood $U$ of $0$ such that $F(s) = G(s)$ for all $s\in U$ with $\operatorname{Re} s > 0$.)
Proof: Suppose $f(x)\cdot x^{-t}$ is integrable for all $t > 0$ and $F$ has an analytic continuation (also denoted by $F$) to a disk $D = \{ z : \lvert z\rvert < 2\varepsilon\}$. Then the Taylor series of $F$ about $\varepsilon$ converges (to $F$) at least on the disk with radius $\sqrt{5}\,\varepsilon$ and centre $\varepsilon$. Since
$$F^{(n)}(s) = \int_K^{+\infty} \frac{f(x)(-\log x)^n}{x^s}\,dx$$
for $\operatorname{Re} s > 0$, it follows that for $0 < y < \sqrt{5}\,\varepsilon$ we have
\begin{align}
F(\varepsilon - y)
&= \sum_{n = 0}^{+\infty} \frac{F^{(n)}(\varepsilon)}{n!}(-y)^n \\
&= \sum_{n = 0}^{+\infty} \frac{1}{n!}\int_K^{+\infty} \frac{f(x)(y\log x)^n}{x^{\varepsilon}}\,dx \\
&= \int_K^{+\infty} \frac{f(x)}{x^{\varepsilon}} \sum_{n = 0}^{+\infty} \frac{(y\log x)^n}{n!}\,dx \\
&= \int_K^{+\infty} \frac{f(x)}{x^{\varepsilon}} x^y\,dx \\
&= \int_K^{+\infty} \frac{f(x)}{x^{\varepsilon - y}}\,dx\,,
\end{align}
where the interchange of summation and integration is justified by the monotone convergence theorem (everything is non-negative). In particular, for $y = 2\varepsilon$ we find
$$\int_K^{+\infty} \frac{f(x)}{x^{-\varepsilon}}\,dx < +\infty\,,$$
so $(\ast)$ cannot hold if $F$ is regular at $0$.
We note that the singularity of $F$ at $0$ need not be a pole, like in Landau's lemma. An $f$ satisfying the hypotheses of the lemma can be integrable. Now we come to the
Proof that $R(x) = \Omega_{\pm}(x^{1/4})$: Suppose to the contrary that
$$\liminf_{x\to +\infty} \frac{R(x)}{x^{1/4}} \geqslant 0\,.$$
Then for every $\varepsilon > 0$ there is a $K = K(\varepsilon) \geqslant 1$ such that $R(x) + \varepsilon \,x^{1/4} \geqslant 0$ for $x \geqslant K$. Writing
$$\frac{H(s)}{s} + \frac{\varepsilon}{s-\frac{1}{4}} = \int_1^{+\infty} \frac{R(x) + \varepsilon\, x^{1/4}}{x^{s+1}}\,dx = \int_1^K \frac{R(x) + \varepsilon\, x^{1/4}}{x^{s+1}}\,dx + \int_K^{+\infty} \frac{R(x) + \varepsilon\, x^{1/4}}{x^{s+1}}\,dx\,,$$
we note that the first term on the right hand side defines an entire function. Since the left hand side has no singularities on $\bigl(\frac{1}{4},+\infty\bigr)$, the lemma tells us that
$$\int_K^{+\infty} \frac{R(x) + \varepsilon\, x^{1/4}}{x^{s+1}}\,dx$$
exists for $\operatorname{Re} s > \frac{1}{4}$ (and consequently our assumption implies the Riemann hypothesis - in fact, already the assumption that the $\liminf$ is greater than $-\infty$ does). Now let $\rho = \frac{1}{4} + i\gamma$ a simple zero of $\zeta(2s)$ and $a = \operatorname{Res}(H;\rho)$. Then on the one hand
$$\lim_{\sigma \downarrow \frac{1}{4}} \biggl(\sigma - \frac{1}{4}\biggr)\biggl(H(\sigma + i\gamma) + \varepsilon\,\frac{\sigma + i\gamma}{\sigma - \frac{1}{4} + i\gamma}\biggr) = a\,,$$
and on the other hand
\begin{align}
\Biggl\lvert H(s) + \varepsilon\,\frac{s}{s-\frac{1}{4}}\Biggr\rvert
&\leqslant \lvert s\rvert \int_1^K \frac{\lvert R(x) + \varepsilon\, x^{1/4}\rvert}{x^{\sigma + 1}}\,dx + \lvert s\rvert \int_K^{+\infty} \frac{R(x) + \varepsilon\, x^{1/4}}{x^{\sigma + 1}}\,dx \\
&= \lvert s\rvert \int_1^K \frac{\lvert R(x) + \varepsilon\, x^{1/4}\rvert}{x^{\sigma + 1}}\,dx - \lvert s\rvert \int_1^K \frac{R(x)}{x^{\sigma + 1}}\,dx + \lvert s\rvert \int_1^{+\infty} \frac{R(x)}{x^{\sigma+1}}\,dx + \lvert s\rvert \int_K^{+\infty} \frac{\varepsilon\,dx}{x^{\sigma - \frac{1}{4} + 1}} \\
&= \lvert s\rvert (A(\sigma) - B(\sigma)) + \frac{\lvert s\rvert}{\sigma} H(\sigma) + \frac{\lvert s\rvert \varepsilon}{\sigma - \frac{1}{4}}\, K^{\frac{1}{4}-\sigma}\,,
\end{align}
where $A$ and $B$ are entire functions given by the integrals over $[1,K]$, shows
$$\lvert a\rvert = \lim_{\sigma \downarrow \frac{1}{4}} \biggl(\sigma - \frac{1}{4}\biggr)\biggl\lvert H(\sigma + i\gamma) + \varepsilon\,\frac{\sigma + i\gamma}{\sigma - \frac{1}{4} + i\gamma}\biggr\rvert \leqslant \biggl\lvert \frac{1}{4} + i\gamma\biggr\rvert\, \varepsilon.$$
This is a contradiction if we choose $0 < \varepsilon < \bigl\lvert\frac{a}{\frac{1}{4} + i\gamma}\bigr\rvert$, whence
$$\liminf_{x\to +\infty} \frac{R(x)}{x^{1/4}} < 0.$$
The same argument applied to $-\bigl(R(x) - \varepsilon\, x^{1/4}\bigr)$ shows that the assumption
$$\limsup_{x\to +\infty} \frac{R(x)}{x^{1/4}} \leqslant 0$$
also leads to a contradiction. Thus our proof is complete.
