Let $Q(x)$ denote the number of squarefree integers up to $x$. I want to show using Perron's formula that $$ Q(x) = \frac{x}{\zeta(2)} + O(x^{1/2} \exp(-c \sqrt{\log x})), $$ where $c$ is a positive constant. Using elementary argument (in particular, the identity $\mu(n)^2 = \sum_{d^2 | n} \mu(d)$) one obtains slightly weaker estimate with error term being $O(x^{1/2})$. I know that one obtains the above estimate using the estimate $\sum_{n \leq x} \mu(n) \ll x\exp(-c\sqrt{\log x})$ but I want to drive it directly from Perron's formula. By Perron's formula we have $$ \sum_{n \leq x}' \mu(n)^2 = \frac{1}{2 \pi i} \int_{\sigma_0 - i T}^{\sigma_0 + iT} \frac{\zeta(s)}{\zeta(2s)} \frac{x^s}{s} \, ds + O \left( \frac{x}{T} \log x \right), $$ where $\sigma_0 = 1 + 1/\log x$. If $\sigma \geq 1 - \frac{c}{\log |t|}$, $|t| \geq 1$ is the classical zero-free region of $\zeta(s)$, $\sigma_1 = \frac{1}{2} -\frac{c}{2 \log T}$ and $R_T$ is the rectangle with vertices at $\sigma_0 - iT, \sigma_0 + iT, \sigma_1 + iT, \sigma_1 - iT$, then $$ \frac{1}{2 \pi i}\int_{R_T} \frac{\zeta(s)}{\zeta(2s)} \frac{x^s}{s} \, ds = x. $$ The integral over the horizontal segments can be estimated as $$ \int_{\sigma_1 \pm iT}^{\sigma_0 \pm iT} \frac{\zeta(s)}{\zeta(2s)} \frac{x^s}{s} \, ds \ll (\log T)^2 \int_{\sigma_1}^{\sigma_0} (x/T)^{\sigma} \, d\sigma \ll (\log T)^2 \left( \frac{x}{T} \right)^{\sigma_0} $$ since $\zeta(s) \ll (1 + \tau^{1 - \sigma}) \min(|\sigma - 1|^{-1}, \log \tau)$ for $\sigma \geq \delta$ and $|t| \geq 1$. As for the integral over the vertical segment to the left we have $$ \int_{\sigma_1 - iT}^{\sigma_1 + iT} \frac{\zeta(s)}{\zeta(2s)} \frac{x^s}{s} \, ds \ll x^{\sigma_1} (\log T) T^{1 - \sigma_1}, $$ where I use the estimate $1/\zeta(s) \ll \log \tau$ for $\sigma \geq 1 - \frac{c}{\log \tau}$. But these estimates combined do not lead to the error term of the order $x^{1/2} \exp(-c \sqrt{\log x})$.
Any help will be highly appreciated.