$\int_{1}^{x}\frac{M(t)}{t}dt=o(x)$ implies $M(x)=o(x)$

Question

Show that $$\int_{1-}^{x}\frac{M(t)}{t}dt=o(x)$$ implies $$M(x)=o(x)$$ where $M(x)=\sum_{n\leq x}\mu(n)$. My idea is complicated and goes like this : Note that $$\frac{M(t)}{t}dt=dM*t^{-1}dt$$ Now multiplying an 'approximate inverse' for $t^{-1}dt$ in the above equation and then integrating and using the hypothesis , we get $$\frac{M(x)}{x}+\frac{\int_{1}^{x}M(t)dt}{x^2}=o(1) $$ Using this equation recurrently one can get the desired result. The 'approximate inverse' i used was $d\left(\frac{1}{t}\right)$. I think this is not the intended solution. This a exercise from Bateman and Diamond's text on Analytic number thoery.

Answer 1

It is not true in general, you need to use that $|M(x+n)-M(x)| \le n$.

Assume that $|M(x_k)| > a x_k$ for infinitely many $k$, and find a contradiction from $$\int_{x_k-a x_k}^{x_k} \frac{M(t)}{t}dt = \int_1^{x_k} \frac{M(t)}{t}dt- \int_1^{x_k-a x_k} \frac{M(t)}{t}dt=o(x_k)$$