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Consider this integral

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

$M$ is the Mertens function.

Assume that $l$ and $x$ are given. How does one evaluate this integral?

I am trying to understand the last piece to Eric's answer to Totient-summation.

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What do you mean by "evaluate?" To find an asymptotic, one can undo the computation I did in my previous answer, and then use the hyperbola method. This allows us to prove that $$\sum_{n\leq x} \varphi(n)n^l = \frac{6}{\pi^2 (l+2)}x^{l+2}+O\left(x^{l+1}\log x\right).$$

To calculate the above asymptotic, we note that since $\varphi(n)=n\sum_{d|n}\mu(d)/d$ we have $$\sum_{n\leq x}\varphi(n)n^{l}=\sum_{n\leq x}n^{l+2}\sum_{d|n}\frac{\mu(d)}{d}=\sum_{d\leq x}\mu(d)d^{l}\sum_{n\leq\frac{x}{d}}n^{l+1}. $$ It is not hard to prove that $$\sum_{n\leq\frac{x}{d}}n^{l+1}=\frac{1}{l+2}\left(\frac{x}{d}\right)^{l+2}+O\left(\frac{x^{l+1}}{d^{l+1}}\right),$$ so that $$\sum_{n\leq x}\varphi(n)n^{l}=\frac{x^{l+2}}{l+2}\sum_{n\leq x} \frac{\mu(d)}{d^2}+O\left(x^{l+1}\log x\right)$$ which implies the asymptotic in the first line.

Your previous question asked how to put this sum in terms of $M(x)$. At a glance, I am not sure why one would want to do this, but I assumed you had a reason. If your goal is to find an asymptotic, this is not the correct way to proceed.

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