Prove that von Mangoldt function satisfies $\sum_{n \le x} \Lambda(n) \lfloor{\frac{x}{n}}\rfloor= x \ln(x)-x+O(\ln x)$ The picture above is what I have, which has an error that is too big.
 A: Indeed, changing the $\lfloor \frac xn \rfloor$ to $\frac xn + O(1)$ will immediately incur an error that is $O\big(\sum_{n\le x} \Lambda(n)\big) = O(x)$.
One can write $\lfloor \frac xn \rfloor$ as the number of integers up to $\frac xn$, which leads to the following manipulation:
$$
\sum_{n\le x} \Lambda(n) \frac xn = \sum_{n\le x} \Lambda(n) \sum_{m\le x/n} 1 = \sum_{m\le x} \sum_{n\le x/m} \Lambda(n) = \sum_{m\le x} \psi\bigg(\frac xm\bigg).
$$
Unfortunately, each summand has a reasonably large error term, so this doesn't seem like the way to go either. Indeed, to get an error term as small as $O(\log x)$, we don't want to use anything about primes at all—they simply aren't that well distributed.
However, $\Lambda$ is related to the smooth function $\log$ by $\log m = \sum_{n\mid m} \Lambda(n)$; so we should aim to create such a sum. And indeed we can do so, by writing $\lfloor \frac xn \rfloor$ not as the number of integers up to $\frac xn$ but rather as the number of multiples of $n$ up to $x$:
$$
\sum_{n\le x} \Lambda(n) \frac xn = \sum_{n\le x} \Lambda(n) \sum_{\substack{m\le x \\ n\mid m}} 1 = \sum_{m\le x} \sum_{\substack{n\le x \\ n\mid m}}\Lambda(n) = \sum_{m\le x} \sum_{n\mid m}\Lambda(n) = \sum_{m\le x} \log m.
$$
Now the result follows by comparing this sum with the integral $\int_0^x \log t\,dt$.
