What's about the Frullani extension for infinite series? I was thinking about the extension of Frullani that explains Jameson in the notes The Frullani integrals, (lectures notes from Lancaster University). I am saying FRU2 in page 3.
I tried write speculative examples of this extension in the infnite case. I am saying when one take series instead of finite sums. But I don't know if the LHS of the identity  in my examples make sense. For example, if I take weights as $m_n=\frac{\mu(n)}{n}$ for each $a_n=n\geq 1$ an integer, where $\mu(n)$ is the Möbius function, then I can write a RHS that converges with $f(x)=e^{-x}$ , but my LHS is
$$\int_0^\infty\frac{\sum_{n=1}^\infty\frac{\mu(n)}{n}e^{-nx}}{x}dx.$$

Question 1. Can you convice to me that this integral doesn't make sense? Or is it convergent? I try do reasonings with the exponential integral, but I don't know what say.

I would like to see some example, if make sense of an example of FRU2 but in the case infinite.

Question. Does make sense the same theorem FRU2, that is the extension of Frullani explained in the lecture notes, in the case which the series are infinite? If you know that it is a theorem (for suitable hypothesis), please refers where we can find the literature, and if it's possible create and provide us a simple example. If isn't possible, that is such theorem isn't feasible  explain us why. Many thanks and good nights.

 A: We may start noticing that $\sum_{n\geq 1}\frac{\mu(n)}{n}=0$ - this is essentially equivalent to the prime number theorem, and allows us to consider the better-behaved series $\sum_{n\geq 1}\frac{\mu(n)}{nx}\left(e^{-nx}-e^{-x}\right)$. For any $s>1$ we have $\sum_{n\geq 1}\frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)}$ and the $\zeta$ function has a simple pole at $s=1$ with residue $1$. By Frullani's theorem
$$ \int_{0}^{+\infty}\frac{e^{-nx}-e^{-x}}{x}\,dx = -\log(n) \tag{1}$$
hence by applying the dominated convergence theorem with extreme caution we have:
$$ \sum_{n\geq 1}\int_{0}^{+\infty}\frac{\mu(n)}{n}\cdot\frac{e^{-nx}-e^{-x}}{x}\,dx =-\sum_{n\geq 1}\frac{\mu(n)\log(n)}{n}.\tag{2}$$
On the other hand, for any $s>1$
$$ \frac{d}{ds}\left(\frac{1}{\zeta(s)}\right) = -\sum_{n\geq 1}\frac{\mu(n)\log(n)}{n^s} = -\frac{\zeta'(s)}{\zeta(s)^2}=-\frac{1}{\zeta(s)}\cdot\frac{d}{ds}\left(\log\zeta(s)\right)\tag{3}$$
and:
$$ \lim_{s\to 1^+}-\frac{\zeta'(s)}{\zeta(s)^2} =\lim_{s\to 1^+}-\frac{(s-1)^2 \zeta'(s)}{(s-1)^2\zeta^2(s)}=-\frac{-1}{1}=\color{red}{1}\tag{4}$$
by the Laurent expansion of $\zeta(s)$ and $\zeta'(s)$ in a right neighbourhood of $s=1$.
So $\color{red}{1}$ is the only reasonable value for the original integral. However, there are a lot of details to be dealt with care: for starters, that the original integral and the RHS of $(2)$ are convergent, but also that we are allowed to use $(3)$ to compute $(2)$ by considering $\lim_{s\to 1^+}-\frac{\zeta'(s)}{\zeta(s)^2}$.
