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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 left-hand side (LHS) of the above identity reveals that the LHS is identical to $s\int_{2}^{\infty}(\pi(x)-\Pi(x)x^{-s-1} \mathrm{d}x-r(s)$. Thus it follows 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 $s>1/2$ ?

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 $s>1/2$ ?

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 left-hand side (LHS) of the above identity reveals that the LHS is identical to $s\int_{2}^{\infty}(\pi(x)-\Pi(x)x^{-s-1} \mathrm{d}x-r(s)$. Thus it follows that both sides of the preceding equation 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 $s>1/2$ ?

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 left-hand side (LHS) of the above identity reveals that the LHS is identical to $s\int_{2}^{\infty}(\pi(x)-\Pi(x)x^{-s-1} \mathrm{d}x-r(s)$. Thus it follows 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 $s>1/2$ ?

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 left-hand side (LHS) of the above identity reveals that the LHS is identical to $s\int_{2}^{\infty}(\pi(x)-\Pi(x)x^{-s-1} \mathrm{d}x-r(s)$. Thus it follows that both sides of the preceding equation 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 this identity also holds for $\Re(s)>1/2$.

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

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 left-hand side (LHS) of the above identity reveals that the LHS is identical to $s\int_{2}^{\infty}(\pi(x)-\Pi(x)x^{-s-1} \mathrm{d}x-r(s)$. Thus it follows that both sides of the preceding equation 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 $s>1/2$ ?