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How could I apply the Mobius inversion formulafor the equality

$$ f(x)\ln(x)+f(x/2)\ln(x/2)+f(x/3)\ln(x/3)+\dots.=g(x) \tag1$$

to get $ f(x)$ from the value of $ g(x) $ ??

The sum inside $(1)$ is infinite and $\ln$ is the natural logarithm. For the case without logarithm, I know the answer but with the logarithm I am lost.

We assume also the asymptotic $ f(x) \sim x $.

vitamin d
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Jose Garcia
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  • Could you state your question as a sum over the divisors involving the Möbius function or is that your question, how to sum over the divisors? – Mats Granvik Aug 22 '16 at 12:18
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    the sum is just a function sum over n in the form $ \sum_{n=1}^{\infty}f(\frac{x}{n})ln(\frac{x}{n}) $ – Jose Garcia Aug 22 '16 at 18:47
  • I believe that this is using Möbius inversion formula (a version of such theorem). See [1], then I believe (and I saying I believe since I don't know how discuss the convergence that I now understand as an equivalence) that the specialisation of (12) $f(x)\log (x)$ instead of $f(x)$ provide you the answer of your question : how get $f(x)$ from $g(x)$? References: [1] Benito, Navas and Varona, Möbius inversion formulas for flows of arithmetic semigroups, Journal of Number Theory Vol. 128.2, (2008). –  Nov 03 '16 at 10:50
  • There is a remark in the article for the convergence in the following paragraph after (12). –  Nov 03 '16 at 10:56

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