Continuous version of the Möbius inversion theorem Is there a continuous version of Möbius Inversion. Essentially, using integrals instead of sums.
 A: 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 $f(x)$...
