I'm trying to prove this identity involving the $\psi$ Chebyshev function: $\sum_{n\leq x}\log(n)=\sum_{n\leq x}\psi\left(\frac{x}{n}\right)$. Here's my attempt:
The TFA states:
$$\log(n)=\sum_{j\mid n}\Lambda(j)$$
Hence,
$$\sum_{n\leq x}\log(n)=\sum_{n\leq x}\sum_{j\mid n}\Lambda(j)=\sum_{n\leq x}\sum_{i\mid n}\Lambda\left(\frac{n}{i}\right)$$
I know i need to use the following:
$$\sum_{n\leq x}\psi\left(\frac{x}{n}\right)=\boxed{\sum_{n\leq x}\sum_{i\leq \frac{x}{n}}\Lambda(i)=\sum_{n\leq x}\sum_{i\mid n}\Lambda\left(\frac{n}{i}\right)}$$
However, I'm not sure about how to provide a formal proof for the boxed identity. I appreciate any ideas.
The step $\displaystyle \sum_{n\leq x}\sum_{j\mid n}\Lambda(j)=\sum_{n\leq x}\sum_{i\mid n}\Lambda\bigl(\tfrac{n}{i}\bigr)$ is mathematically correct but, in my opinion, not the most helpful way to proceed. Instead, I recommend changing the order of summation of the left-hand side: $$ \sum_{n\leq x}\sum_{j\mid n}\Lambda(j) = \sum_{j\le x} \Lambda(j) \sum_{\substack{n\le x \\ j\mid n}} 1. $$ This particular type of double sum happens a lot in analytic number theory and it's worthwhile practicing how to exchange the summations. (It's also an example of the general procedure for interchanging a double sum: on the left, $n$ ranges over all possible values and then $j$ ranges over the values relevant to each particular $n$; on the right, we need $j$ to range over all possible values and then $n$ to range over the values relevant to each particular $j$.)
In this particular case, I think the best way to proceed further is the slightly sneaky $$ \sum_{\substack{n\le x \\ j\mid n}} 1 = \biggl\lfloor \frac xj \biggr\rfloor = \sum_{m\le x/j} 1. $$ Can you finish from there?