# 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. – Will Jagy Mar 1 '13 at 4:57
• @WillJagy Yes, that was my question. It had not been getting much response. – Ryouka Hamasaki Mar 1 '13 at 5:05
• Why do you think there is anything to be said about this? And what are you doing? – Will Jagy Mar 1 '13 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? – Gerry Myerson Mar 1 '13 at 11:43
• @GerryMyerson I spoke with joriki about it; won't happen again. Sorry for not linking the questions. – Ryouka Hamasaki Mar 1 '13 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? – Ryouka Hamasaki Mar 1 '13 at 4:51
• – Will Jagy Mar 1 '13 at 4:56
• When $l=0$ you obtain the result at the displayed equation in my answer above. – Eric Naslund Mar 1 '13 at 6:31
• @Will Jagy: I didn't see the question on Math Stack Exchange until after answering this. – Eric Naslund Mar 1 '13 at 6:31
• Eric, please see math.stackexchange.com/questions/317635/… – Will Jagy Mar 1 '13 at 7:16