Average order of an arithmetic function: Difference between revisions

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Precision
The error bound for the case \alpha = 1 is not O(x), but O(x log x) as proved in several of the references of the article.
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\sum_{n\le x}\sigma_{\alpha}(n)=
\begin{cases}
\;\;\sum_{n\le x}\sigma_\alpha(n)=\frac{\zeta(\alpha+1)}{\alpha+1}x^{\alpha+1}+O(x^\beta) & \text{if } \alpha>0,\alpha \ne 1, \\
\;\;\sum_{n\le x}\sigma_{1}(n)=\frac{\zeta(2)}{2}x^2+O(x \log x) & \text{if } \alpha=1, \\
\;\;\sum_{n\le x}\sigma_{-1}(n)=\zeta(2)x+O(\log x) & \text{if } \alpha=-1, \\
\;\;\sum_{n\le x}\sigma_\alpha(n)=\zeta(-\alpha+1)x+O(x^{\max(0,1+\alpha)}) & \text{otherwise.}