For any non-negative $ t\in \mathbb{R}\setminus \mathbb{Z}$ applying Dirichlet’s hyperbola method, Abels summation formula, and several Fourier series expansions gives:

$ $ \sum_{n\leq x}\frac{\Lambda(n)}{n}\{nt\}=\frac{\log(x)}{2}-\frac{\gamma}{2}-\log\left(\Gamma(\{t\})\sqrt{\frac{|\sin(\pi t)|}{\pi}}\right)+[\gamma+\log(2\pi)]\left(\frac{1}{2}-\{t\}\right)+\mathcal{O}\left(\frac{1}{x}\frac{\log(x)}{\sin(\pi t)}+\int_{x}^\infty\frac{|\psi(t)-t|}{t^2} dt+\frac{1}{x}\sum_{n\leq x}\frac{\Lambda(n)}{|\sin\left(\pi n t\right)|}\right)$ $

To what extent can I simplify/reduce/bound that error estimate at the end? I’m pretty sure it should vanish as $ x\to \infty$ though I can’t prove it. I mean if one assumes the Riemann hypothesis and lets $ \mu_{t}(x)=\min(|\sin\left(\pi n t\right)|:1\leq n\leq x)$ then I could replace the error term with $ \mathcal{O}\left(\frac{\log(x)^2}{\mu_{t}(x)\sqrt{x}}\right)$ . So I’m not really sure what to do here and would appreciate any help. At the least I want to show:

$ $ \sum_{n\leq x}\frac{\Lambda(n)}{n}\{nt\}=\frac{1}{2}\log(x)+C_{t}+o(1)$ $

Where $ C_{t}$ is the previous expression involving the logarithms, gamma function etc.