Möbius inversion formula for two functions f(x) and g(x) Given the 2 functions $$ g(x)= \sum_{n=1}^{\infty}f\left(\frac{x}n\right)\log(n)\;, $$ how can I use Möbius inversion to recover $f$ from $g$?? I believe that
$$ f(x)=  \sum_{n=1}^{\infty}\mu (n)g\left(\frac{x}n\right)\log(n)\;. $$ Here 'mu' is the Möbius function.
 A: Suppose $f(x)=x^{10}$. Then $$g(x)=\sum_1^{\infty}x^{10}n^{-10}\log n=Cx^{10}$$ where $C=\sum_1^{\infty}n^{-10}\log n$ is a very small positive constant. Then $$\sum_1^{\infty}\mu(n)g(x/n)\log n=Cx^{10}\sum_1^{\infty}\mu(n)n^{-10}\log n=CDx^{10}$$ where $D=\sum_1^{\infty}\mu(n)n^{-10}\log n$ is a very small constant. We can't have $CD=1$, so we can't have $$f(x)=\sum_1^{\infty}\mu(n)g(x/n)\log n$$  
A: There is a well known inversion formula stated in Section 27.5 of the NIST Handbook of Mathematical Functions (DLMF online):
$$G(x) = \sum_{n \leq x} F(x/n) \iff F(x) = \sum_{n \leq x} \mu(n) G(x/n).$$ So there are two options:


*

*We can write your $g(x)$ with a floor function, as 
$$g(x) = \sum_{n \leq x} f\left(\left\lfloor \frac{x}{n} \right\rfloor\right) \log n,$$ and then apply the DLMF manual's formula; or 

*Essentially do this, but bound $f(x/n) \log n$ for large $n > x$, and then apply a variant of the stated formula.

A: I was re-reading through some old posts on this site. I want to respond to Gerry's concern given in the comment from above in more complete detail. For the series Gerry cites using $g(x) = x^{10}$, we have (by uniformly convergent series analysis, and differentiation of this formal series) that
$$g(x) = -\zeta^{\prime}(10) \cdot x^{10}.$$
Fair enough, here the exact $|\zeta^{\prime}(10)| \approx 0.000697033$, indeed a small constant factor. Let's take $C_0 := -\zeta^{\prime}(10)$. By similar reasoning on the second series expansion that results for $f(x)$, we see that
$$f(x) = \sum_{n \geq 1} \mu(n) g(x/n) \log n = C_0 \cdot x^{10} \times \frac{d}{ds}\left[\frac{1}{\zeta(s)}\right]\Biggr\rvert_{s=10}.$$
By formally computing the last derivative of the reciprocal of the Riemann zeta function with respect to $s$, we have that $f(x) = x^{10}$, e.g., the former notation does in fact yield $CD = 1$ in this case.
Note that we have used the property that the Dirichlet series over the Moebius function satisfies $$\sum_{n \geq 1} \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)}, \forall \Re(s) > 1.$$
