# How to rewrite this totient summation in terms of Mertens?

The totient sum function has an identity:

$\displaystyle\sum_{k=1}^{N}\varphi(k) = \sum_{k=1}^N {\rm M}(\lfloor\frac{N}{k}\rfloor)k$

$\varphi(k)$ is the Euler totient function, and $M$ is the Mertens function $\displaystyle M(N)=\sum_{k=1}^N \mu (k)$ where $\mu$ is the Moebius function.

My question: What is the Mertens function summation equivalent of $\displaystyle\sum_{k=1}^{N}k\varphi(k)$ and $\displaystyle\sum_{k=1}^{N}k^2\varphi(k)$?

• See math.stackexchange.com/questions/316376/… and math.stackexchange.com/questions/317482/… the latter deleted a few minutes ago. Mar 1, 2013 at 4:57
• @WillJagy Yes, that was my question. It had not been getting much response. Mar 1, 2013 at 5:05
• Why do you think there is anything to be said about this? And what are you doing? Mar 1, 2013 at 5:19
• Why why why why do you refuse, despite repeated urging, to link your questions to your earlier, related questions? Why does someone else always have to do this for you? Mar 1, 2013 at 11:43
• @GerryMyerson I spoke with joriki about it; won't happen again. Sorry for not linking the questions. Mar 1, 2013 at 15:09

Since $\varphi(n)=n\sum_{d|n}\frac{\mu(d)}{d},$ by switching the order of summation we have that for fixed $l$ $$\sum_{n\leq x}\varphi(n)n^{l}=\sum_{kd\leq x}\mu(d)k^{l}d^{l}k$$

$$=\sum_{k\leq x}k^{l+1}\sum_{d\leq\frac{x}{d}}\mu(d)d^{l}.$$ Now, $$\sum_{d\leq y}\mu(d)d^{l}=\int_{0}^{y}t^{l}d\left(M(t)\right)=M(y)y^{l}-l\int_{0}^{y}M(t)t^{l-1}dt,$$ so you can write $$\sum_{n\leq x}\varphi(n)n^{l}=\sum_{k\leq x}k^{l+1}\left(\frac{x}{k}\right)^{l}M\left(\frac{x}{k}\right)-l\int_{0}^{\frac{x}{k}}t^{l-1}M(t)dt.$$ This may be rearranged as

$$x^{l}\sum_{k\leq x}kM\left(\frac{x}{k}\right)-l\sum_{k\leq x}k^{l+1}\int_{0}^{\frac{x}{k}}t^{l-1}M(t)dt,$$ which is equal to

$$x^{l}\sum_{k\leq x}kM\left(\frac{x}{k}\right)-l\int_{0}^{x}t^{l-1}\sum_{k\leq x}k^{2}M\left(\frac{t}{k}\right)dt.$$

• Does this not have arithmetic closed forms like the l=0 case? Mar 1, 2013 at 4:51
• Mar 1, 2013 at 4:56
• When $l=0$ you obtain the result at the displayed equation in my answer above. Mar 1, 2013 at 6:31
• @Will Jagy: I didn't see the question on Math Stack Exchange until after answering this. Mar 1, 2013 at 6:31
• Eric, please see math.stackexchange.com/questions/317635/… Mar 1, 2013 at 7:16