I'd like to know the estimate of the following sum
$$\sum\_{n\leq x}\sum\_{d|n}\Lambda(d)\frac{\phi(d)}{d} $$
where $\Lambda(d)$ is the von mangoldt function and $\phi(d)$ is the Euler totient function. Do you know how to compute this?



Estimating a sum involving the von Mangoldt function
Following Joshua Stucky's remark, the sum can be rewritten as the following sum over prime numbers:
$$\sum\_{p\leq x}(\log p)\left(1-\frac{1}{p}\right)\sum\_{k=1}^\infty\left\lfloor\frac{x}{p^k}\right\rfloor.$$
Hence the sum is upper bounded by
$$\sum\_{p\leq x}(\log p)\left(1-\frac{1}{p}\right)\frac{x}{p-1}=x\sum\_{p\leq x}\frac{\log p}{p}=x\log x+O(x),$$
and it is lower bounded by
$$\sum\_{p\leq x}(\log p)\left(1-\frac{1}{p}\right)\left(\frac{x}{p}-1\right)>x\sum\_{p\leq x}\frac{\log p}{p}-x\sum\_{p\leq x}\frac{\log p}{p^2}-\sum\_{p\leq x}\log p=x\log x+O(x).$$
Therefore, the sum is
$$\sum\_{n\leq x}\sum\_{d|n}\Lambda(d)\frac{\phi(d)}{d}=x\log x+O(x).$$


