Continuous version of the Möbius inversion theorem

Question

Is there a continuous version of Möbius Inversion. Essentially, using integrals instead of sums.

Answer 1

The "divisor convolution" of two arithmetic functions $a_n$ and $b_n$ is the arithmetic function $(a\star b)(n) = \sum_{d \mid n} a_db_{n/d}$.

If $\sum a(n) n^{-s} = L(a, s)$ is the Dirichlet series of $a$, then we have the relation $$L(a, s)L(b, s) = L(a \star b, s).$$

In particular, the Möbius transform is $$L(\mu \star a, s) = L(a, s)/\zeta(s).$$

We can extract it in its usual form by comparing Dirichlet coefficients on both sides.

A possible continuous analogue of a Dirichlet series is the $L$-transform (not to be confused with the Laplace transform)

$$L(f,s) = \frac{1}{\Gamma(s)}\mathcal{M}_f(s) = \frac{1}{\Gamma(s)}\int_0^\infty f(x)\: x^{s} \frac{dx}{x}.$$

The analogue of the divisor convolution is

$$(f \star g)(x) := \int_0^\infty f(t)g(t/x) \frac{dt}{t}.$$

The analogue of the relation

$$L(a, s)L(b, s) = L(a \star b, s)$$

is the identity

$$L(f, s)L(g,s) = L(f\star g, s).$$

Now, what is a possible analogue of the zeta function?

Remark: In order to extract the "Dirichlet coefficients" of $L(f, s)$, we have to take an inverse Mellin transform, because we want the "$x$-th Dirichlet coefficient" to be $f(x)$...