Let $f:\mathbb{N} \to \mathbb{C}$ be a function for which there exists a positive constant $A$ such that \begin{equation} \lim_{x\to \infty} \frac{1}{x}\sum_{n\leq x} f(n) = A. \end{equation} Prove that \begin{equation} \sum_{n\leq x} f(n)\log n = A(1+o(1))x\log x ~~~~ (x\to \infty) \end{equation}
What I did so far is:
We are given that \begin{equation} \sum_{n\leq x} f(n) = Ax + o(x) \end{equation}
Hence by Abel summation, if we put $F(x) = \sum_{n\leq x} f(n),$ we have \begin{align} \sum_{n\leq x}f(n)\log n &= F(x) \log x - \int_{1}^x \frac{F(t)}{t}dt \\ &= (Ax + o(x))\log x -\int_{1}^x \frac{At+o(t)}{t}dt \\ &=(Ax + o(x))\log x - A(x-1) -\int_{1}^x \frac{o(t)}{t}dt \end{align} I dont know what to do next, Also I dont know if I can write \begin{equation} (Ax + o(x))\log x = A(1+o(1))x\log x \end{equation}
Thanks in advance for the help.