Does $\int_{2}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x$ converge on the real axis for $s>1/2$? [closed]

Question

Consider the prime zeta function, defined for $\Re(s)>1$, by the infinite series

$$\sum_{p} p^{-s} = \sum_{m=1}^{\infty}\frac{\mu(m)}{m}\log \zeta(ms)$$ where $p$ denotes a prime, $\mu$ the Mobius function and $\zeta$ the Riemann zeta function. By partial summation, one finds that

$$s\int_{2}^{\infty} \pi(x)x^{-s-1} \mathrm{d}x =\log \zeta(s) + \sum_{m=2}^{\infty}\frac{\mu(m)}{m}\log \zeta(ms)$$ for $\Re(s)>1$, where $\pi$ is the prime counting function. Let $Li(x)=\int_{2}^{x} \frac{dt}{\log t}$. We know that $s\int_{2}^{x} Li(x)x^{-s-1} \mathrm{d}x=-\log(s-1)+r(s)$, where $r(s)$ is entire. Putting this into the above gives

\begin{equation} s\int_{2}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x -\log ((s-1)\zeta(s)) -r(s)= \sum_{m=2}^{\infty}\frac{\mu(m)}{m}\log \zeta(ms) \\ \end{equation} for $\Re(s)>1$. Note that $\log \zeta(s)=s\int_{2}^{\infty}\Pi(x)x^{-s-1} dx$ hence $\log((s-1)\zeta(s))=s\int_{2}^{\infty}(\Pi(x)-Li(x))x^{-s-1} dx$ for $\Re(s)>1$ where $\Pi$ is the prime power counting function. Putting this into the above identity, we find that the identity can be written as

$$s\int_{2}^{\infty}(\pi(x)-\Pi(x))x^{-s-1} \mathrm{d}x-r(s)=\sum_{m=2}^{\infty}\frac{\mu(m)}{m}\log \zeta(ms)$$

for $\Re(s)>1$. Notice that both sides of the preceding identity are analytic and convergent for $\Re(s)>1/2$ since $\Pi(x)-\pi(x)=O(x^{1/2}), r(s)$ is entire and $\mu(m)\log \zeta(ms) \ll 2^{-m\Re(s)}$ as $m\rightarrow \infty$ for $\Re(s)>1/2$. By analytic continuation, this implies that the preceding identity also holds for $\Re(s)>1/2$.

Because $(s-1)\zeta(s)>0$ for every real $s>0$, it follows from this idenity that $\int_{2}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x$ converges on the real axis for $s>1/2$ ?

Answer 1

the answer to the original question is obviously we do not know (as noted the question is equivalent to RH), while the reasoning above doesn't work for the same reason that, for example, the fact that $\zeta(i+s)-\frac{1}{s+i-1}$ extends to an entire function from the original definition on $\Re s >1$ and obviously $\frac{1}{s+i-1}$ doesn't have any singularity for $s$ real, doesn't imply that the Dirichlet series (which can be expressed as an integral if one wishes) of $\zeta(i+s)$ converges beyond $\Re s >1$.

The reason in both cases that we cannot conclude anything definite from the extension to a bigger half-plane plus lack of singularities on the real line (while we have singularities somewhere else on the original abscissa of convergence) is that the coefficients of the series (or the integrand $\pi(x)-Li(x)$) are not eventually positive or negative - in my contrived example they are complex, in the example in the post the integrand is real but change sign indefinitely.

Edit later - just to note that we have for $\Re s >1$ the identity:

$\sum{\frac{1}{n^{i+s}}}-\frac{1}{i+s-1}=(s+i)\int_{1}^{\infty}\frac{[x]-x+.5}{x^{s+i+1}}dx + .5$ where RHS converges uniformly for $\Re s > \sigma >0$ so extends to the half plane $\Re s >0$, and also $\frac{1}{i+s-1}$ has no singularity for $s$ real, so we have the precise setting of the post but here we do know that $\sum{\frac{1}{n^{i+s}}}$ actually doesn't converge for any $\Re s \le 1$