Does exist a variant of Frullani's theorem valid for $f(x)=\pi(x)/x$ or $f(x)=\psi(x)/x$, where the numerators are prime-counting functions?

Question

Frullani's theorem is a deep theorem in real analysis with applications, see the Wikipedia Frullani integral and other uses and contexts (see [2]). I wrote two imaginative examples of what can be deduced using this theorem, I add these as the first comment, as illustration of Frullani's integral.

I'm interested if it is possible to know if Frullani's theorem, or some variant of this theorem, can be used for the functions

$$f(x)=\frac{\pi(x)}{x}$$

and/or

$$f(x)=\frac{\psi(x)}{x},$$ where respectively $\pi(x)$ denotes the prime-counting function and $\psi(x)$ the second Chebyshev function $\sum_{1\leq x}\Lambda(n)$, being thus $\Lambda(n)$ the von Mangoldt function.

Fact. These functions are closely related and satisfy the prime number theorem (see, as reference, [3]).

The situation is clear, my belief is the these function don't safisfy all requirements of Frullani's theorem: these are discontinuous functions, and I don't know how to define $f(0)$. Thus I believe that these functions don't satisfy the theorem in [1].

On the other hand I know certain formulas that are in the literature, a representation due to Riemann see $(4)$ and $(7)$ of Riemann Prime Counting Function from the encyclopedia MathWorld and the identity $(10)$ from the article Prime Counting Function of MathWorld, and the article Explicit Formula also from MathWorld.

Question. Is it possible to get an application of Frullani theorem, I say similar than the showed in the cited Wikipedia Frullani integral for the functions $f(x)=\frac{\pi(x)}{x}$ or $f(x)=\frac{\psi(x)}{x}$ (maybe for a different lower limit of the integral $\int_a^\infty$, and $f(a)$ playing the role of $f(0)$)?

If it is impossible, does possible to create a variant of Frullani's theorem/integral with the purpose to get it (I say create a theorem from which I can invoke it, to get a similar statement than is showed in the cited Wikipedia Frullani integral, for a suitalbe choice of $f(0)$ or $f(a)$ the corresponding lower limit in the integral $\int_a^\infty$ )? Many thanks.

I know that certain set of discontinuities, for the Lebesgue integral, have Lebesgue measure $0$. Thus I don't know if there is in the literature a suitable integral from which one can write a suitable variant of Frullani theorem.

References:

[1] Juan Arias-de-Reyna, On the theorem of Frullani, Amer. Math. Soc. 109 (1990), 165-175.

[2] Frullani 's theorem in a complex context, from Mathematics Stack Exchange (May, 2016).

[3] Tom M. Apostol, Introduction to Analytic Number Theory, UTM, Springer (1978)

Answer 1

As you mentioned, there is an explicit formula for $\psi(x)$ of the form $$ \psi(x) = x - \sum_\rho \frac{x^\rho}\rho - \log2\pi - \frac12\log(1-x^{-2}) $$ (if we define the left-hand side suitably at its discontinuities). Assuming the Riemann hypothesis, this implies that $\psi(x)-x$ is equal to $\sqrt x$ times a function that has a limiting (logarithmic) distribution function; roughly speaking, this means that there is a random variable $Y$ out there such that a "randomly chosen" value of $\psi(x)-x$ will be equal to $Y\sqrt x$.

One can plug this formula into an integral and obtain $$ f(x) = \int_0^x \frac{\psi(ax)-\psi(bx)}x \,dx = (a-b)x - \sum_\rho (a^\rho-b^\rho) \frac{x^\rho}{\rho^2} - g(x) $$ where $g(x)$ is a Fruliani-type integral of $\frac12\log(1-x^{-2})$. In particular, assuming RH, one can again conclude that there is a random variable $Z$ out there such that a "randomly chosen" value of $f(x)-(a-b)x$ will be equal to $Z\sqrt x$.