TITLE: taylor series involving Mobius function
QUESTION [1 upvotes]: is there a formulae for the function
$$ g(x)= \sum_{n=0}^{\infty} \mu (n) x^{n} $$
i presume that i must use the Lambert series
$$ x= \sum_ {n=0}^{\infty} \frac{\mu (n) x^{n}}{1-x^{n}} $$

REPLY [2 votes]: By Dirichlet convolution:
$$ \sum_{n\geq 1}\frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)} \tag{1}$$
for every $s\in\mathbb{C}$ such that $\text{Re}(s)>1$. It follows that:
$$ \int_{0}^{1}\frac{g(x)-1}{x}(-\log x)^{s-1}\,dx =\frac{\Gamma(s)}{\zeta(s)}\tag{2}$$
or, by setting $x=e^{-t}$,
$$ \frac{\Gamma(s)}{\zeta(s)} = \int_{0}^{+\infty}\left(g(e^{-t})-1\right)t^{s-1}\,dt = (\mathcal{M}\,f)(s)\tag{3} $$
with $f(x)=g(e^{-x})-1$ and $\mathcal{M}$ being the Mellin transform. By the inversion formula,
$$ g(e^{-x})-1 = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\Gamma(s)}{x^s \zeta(s)}\,ds\tag{4} $$
as soon as $c>1$.
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