= Möbius harmonic logarithmic moments
{c}
{title2=$A_j(x)$}
Put $A_j(x)=\sum_{d\leq x}\mu(d)\log^j(x/d)/d$. The elementary estimates are
$$
A_0(x)=O(1),\qquad A_1(x)=O(1),\qquad A_2(x)=2\log x+O(1).
$$
The identities $\sum_d\mu(d)\lfloor x/d\rfloor=1$ and $\sum_d\mu(d)H_{\lfloor x/d\rfloor}/d=1$ give the first two bounds. Convolve the <harmonic divisor sum> with $\mu(d)/d$: because $\mu*\tau=\mathbf1$, its left side is $H_{\lfloor x\rfloor}$. Thus $\tfrac12A_2+2\gamma A_1+cA_0=\log x+O(1)$. The summed error is $O(x^{-1/2}\sum_{d\leq x}d^{-1/2}\log(2x/d))=O(1)$.
This elementary smoothing argument is used in https://www.math.lsu.edu/~mahlburg/teaching/handouts/2014-7230/Selberg-ElemPNT1949.pdf[Selberg's original proof] and https://ramare-olivier.github.io/Maths/LecturesEasyChennai.pdf[Ramaré's sieve lectures, Lemma 4.2].
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