This exercise is in the book by P.T. Bateman and H.G. Diamond which i'm not sure whether i perceived it correctly about the integrator involving in the convolution and integral, I was wondering if $$\sum_{n\le x} (T1*\mu)(n)=\int_{1-}^x dF*dG$$ Where $\displaystyle F(x)=\sum_{n\le x} T1(n),G(x)=\sum_{n\le x} \mu(n)$ is true. Note that $T1(n)=n$ and the eminent Mobius function $\mu(n)$.
The exercise wants the readers to prove that $$\sum_{n\le x}(T1*\mu)(n)=\dfrac{x^2}{2\zeta(2)}+O(x\log x)$$ And I want some hint to this problem because I'm not sure what I've stated above is true or not and even though it's true I have no idea to choose the bound for $F,G$ so that $\zeta(2)$ will stand out using Stability Theorem described in the chapter of the exercise.
I beg on the hint (or at least the clarification on the above) to the bound of maybe either $F,G$ . What gives me the confusions is that $G(x)=O(x)$ trivially
However, I don't think this one can have very sharp bound, and therefore, give not so big pace since even Riemann Hypothesis is true the bound should be $O(x^{1/2+\epsilon})$, and, I expect that it should be done with the term of $F(x)$.
Edited: Here I add the Stability Theorem that states in the chapter that they suggest readers to use the theorem
Theorem (Stability) Let $F,G\in$ the function class $\nu$ and satisfy for all $x\ge 1$ $F(x)=x^sP(\log x)+O(x^\theta\log^K x)$ and $G(x)=O(x^\tau\log^H x)$, where $s\in\mathbb{C}$; $P$ is a polynomial of degree $m$; $K,H\ge 0$;$0\le\tau<\Re(s)$ and $0\le\theta\le \Re(s)$. Then $$\int_{1-}^x dF*dG=x^sP^*(\log x)+R(x)$$ where $\displaystyle P^*(u)=\sum_{l=0}^m\dfrac{1}{l!}P^{(l)}(u)\int_{1-}^\infty (-\log t)^lt^{-s}dG(t)$, in which, $\deg(P)\le m$ and $$R(x) = \cases{O(x^\theta\log^K x) & , $\theta > \tau$ \cr\cr O(x^\theta(\log x)^{H+\max(m,K+1)}) & , $\theta = \tau$ \cr\cr O(x^\tau(\log x)^{H+m}) & , $\theta<\tau$} $$