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
$$ x= \sum_ {n=0}^{\infty} \frac{\mu (n) x^{n}}{1-x^{n}} $$
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What is the definition of $\mu$? – Fly by Night Dec 06 '16 at 19:53
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2This might interest you: http://mathoverflow.net/questions/126442/ordinary-generating-function-for-mobius – Olivier Oloa Dec 06 '16 at 20:02
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@FlybyNight apparently the https://en.wikipedia.org/wiki/M%C3%B6bius_function – reuns Dec 07 '16 at 16:55
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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 $c>1$.
Jack D'Aurizio
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What is $\mu(0) $ ? I think the OP meant $g(x) = \sum_{n=1}^\infty \mu(n) x^n$ (for $|x| < 1$) – reuns Dec 07 '16 at 16:55
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@user1952009: I assumed $\mu(0)=1$ in the above lines, but if the series defining $g(x)$ starts at $n=1$ there is very little to fix. – Jack D'Aurizio Dec 07 '16 at 17:07