On functions that satisfy $\int_0^1\frac{f(x)}{1-x}dx<\infty$ as a version of the Prime Number Theorem for the Möbius function

Question

Using the integral representation for harmonic numbers $H_k$ showed by this Wikipedia I can prove that (I presume that there are no problems to justify the issues about convergence)

Fact. Let $\mu(k)$ the Möbius funtion then the Prime Number Theorem implies $$\int_0^1\left(\sum_{k=1}^\infty\frac{\mu(k)}{k}x^k\right)\frac{dx}{1-x}=-\sum_{k=1}^\infty\frac{\mu(k)}{k}H_k\equiv\text{ a constant}.$$

The recipe is easy: multiply the integral representation by $\frac{\mu(k)}{k}$, take the summation from $k=1$ to $k=n$, and after take the limit as $n\to\infty$ inside the integration combinig with the Prime Number Theorem. Here of course we presume that we can conclude that $-\sum_{k=1}^\infty\frac{\mu(k)}{k}H_k$ is convergent (notice that the absolute convergence says nothing to me).

Computational evidence. See these codes with Wolfram Alpha online calculator

int (sum mu(k)/k x^k/(1-x), from k=1 to 300) dx, from x=0 to 0.99

sum -mu(k)/k H_k, from k=1 to 2000

I hope that next question is clear (a characterization of all functions, well defined, that satisfy a similar statement in terms of the coefficients $a_k$ from $\sum_{k=1}^\infty a_k x^k$ following if it is possible a version of the Prime Number Theorem).

Question. Let $f(x)$ a function defined as $$\sum_{k=1}^\infty a_k x^k$$ for $0<x<1$, I presume that $f(0)=0$ and $f(1)=\text{something}$.

Can you find a condition in terms of the coefficients $a_k$ guaranteeing that $$\int_0^1\frac{f(x)}{1-x}dx$$ is convergent? Many thanks.

Explanation of the question. The Prime Number Theorem is the condition that seems that guarantees the convergence of the integral in my Fact., and such condition is well known that was written in terms of the coefficients of the function of our example, that was $\sum_{k=1}^\infty\frac{\mu(k)}{k}x^k$. Thus I am asking for a similar condition in terms of our coefficients $a_k$ and the details of such theorem (if the theorem that I am evoking is feasible via a good defition of our functions $f(x)$). Notice that the Liouville function also seems that satisfy our intended theorem, because the Liouville's function also$\dagger$ satisfies a Prime Number Theorem and thus we can follow my recipe to get a similar statement is my Fact.

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$\dagger$ If we are agree to say that our intended theorem is a Prime Number Theorem version, then maybe it is possible to write a new version of our intended theorem as the Riemann Hypothesis version, because it is well known that the Möbius function and Liouville function satisfy a Riemann Hypothesis and I believe that we could try to modify my recipe with the purpose to rewrite the Question and the intended theorem as a new version, the Riemann Hypothesis version.

Answer 1

• Elementary answer :

The prime number theorem is that $\sum_{n=1}^\infty \mu(n)n^{-s}$ as well as all its derivatives converges for $\Re(s) = 1$. Thus $\sum_{n=1}^\infty \frac{\mu(n)}{n}H_n = \sum_{n=1}^\infty \frac{\mu(n)}{n}\log n+\gamma\sum_{n=1}^\infty \frac{\mu(n)}{n}+\sum_{n=1}^\infty \frac{\mu(n)}{n}\mathcal{O}(n^{-1})$ $=\int_0^1 \frac{\sum_{n=1}^\infty \frac{\mu(n)}{n} (1-x^n)}{1-x} dx=\int_0^1 \frac{-\sum_{n=1}^\infty \frac{\mu(n)}{n} x^n}{1-x} dx $ converges

• More about the theory of $\zeta(s)$ :

Since $\int_0^\infty e^{-nx} x^{s-1}dx=n^{-s} \Gamma(s)$, for $\Re(s) > 0$ $$\frac{\Gamma(s)}{\zeta(s+1)} = \int_0^\infty (\sum_{n=1}^\infty \frac{\mu(n)}{n} e^{-nx}) x^{s-1}dx$$

By inverse Fourier/Laplace/Mellin transform, for $\sigma > 0$ $$\sum_{n=1}^\infty \frac{\mu(n)}{n} e^{-nx} =\frac{1}{2i\pi} \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{\Gamma(s)}{\zeta(s+1)}x^{-s}ds$$ The zero-free region for $\zeta(s)$ obtained when proving the prime number theorem implies we can shift the contour $\int_{\sigma-i\infty}^{\sigma+i\infty}$ to $t \in (-\infty,\infty), \sigma = -\frac{A}{1+\log |t|}$ (for some $A > 0$) so that $$\sum_{n=1}^\infty \frac{\mu(n)}{n} e^{-nx} = o(\log^{-m} x) \qquad (x \to 0^+)$$ $\qquad$ (note the Riemann hypothesis is $o(x^{1/2-\epsilon})$)

And hence the following converges absolutely $$\int_0^\infty \frac{1}{1-e^{-x}} \sum_{n=1}^\infty \frac{\mu(n)}{n} e^{-nx}e^{-x}dx= \int_0^1 \frac{1}{1-x} \sum_{n=1}^\infty \frac{\mu(n)}{n} x^n dx$$

• $f(x)=\frac{1}{1-x}\sum_{n=1}^\infty a_n x^n$ is analytic (for $|x| < 1$), with $F$ its primitive, there is obviously no general criterion for if $F$ is continuous as $x \to 1^-$.