Is $\sigma(n)$ has explicit compositional inverse formula since it has series representation?

Question

let $\sigma_x(n)$ be a power of sum divisor function such that : $\sigma_x(n)=\sum_{d|n} d^x$ . My question is: Does $\sigma$ has explicit compositional inverse formula since it has series representation? and the same with Euler totient function ?

Answer 1

Yes. Since the Dirichlet series generating function for $\sigma_x$ is $\zeta(s)\zeta(s-x)$, its Dirichlet-convolution inverse is given by the coefficients of $$ \frac1{\zeta(s)\zeta(s-x)} = \prod_p \big( 1-p^{-s} \big) \big( 1-p^x p^{-s} \big) = \prod_p \big( 1 - (1+p^x) p^{-s} + p^x p^{-2s} \big). $$ In other words, the inverse of $\sigma_x$ is the multiplicative function $f$ with values on prime powers $f(p) = -1-p^x$, $f(p^2) = p^x$, and $f(p^k)=0$ for $k\ge3$.