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

4$\begingroup$ Note that $\Lambda(d)$ restricts $d = p^k$ for some prime $p$ and positive integer $k$, and $\phi(p^k)/p^k = 1  \frac{1}{p}$. This should give you a good start on evaluating the sum. Intuitively, you should think of $\phi(n)/n$ as 1 on overage, so your sum is essentially $\sum_{n\leq x} \sum_{d\mid n} \Lambda(d) = \sum_{n\leq x} \log n = x\log x + O(x)$. $\endgroup$– Joshua StuckyMar 19 at 5:16

$\begingroup$ Please use a highlevel tag like "nt.numbertheory". I added this tag now. $\endgroup$– GH from MOMar 19 at 5:37
1 Answer
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 $$\sum_{p\leq x}(\log p)\left(1\frac{1}{p}\right)\frac{x}{p1}=x\sum_{p\leq x}\frac{\log p}{p}=x\log x+O(x),$$ 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 $$\sum_{n\leq x}\sum_{dn}\Lambda(d)\frac{\phi(d)}{d}=x\log x+O(x).$$