# Convergence of prime zeta function for $\mathfrak R(s)=1$?

By doing some estimates for the partial sums of the Prime zeta function $P(s)=\sum_p p^{-s}$ for $\mathfrak R(s)=1$ I got that $P(1+i\alpha)$ converges for every $\alpha\neq0$... Since I did not encounter a source mentioning that it converges for these values of $s$, I thought something went wrong.

Question. Is it true that $P(1+i\alpha)$ converges for $\alpha\neq0$? If not, where's the mistake in the proof below?

Let $S_\alpha(x)=\sum_{p\leq x}p^{-1-i\alpha}$. Using Abel's summation formula and the PNT in the form $\pi(x)=\frac x{\log x}+O(\frac x{\log^2 x})$ we have

\begin{aligned}S_\alpha(x)&=\pi(x)x^{-1-i\alpha} +(1+i\alpha)\int_2^x\pi(t)t^{-2-i\alpha}dt\\ &=O(1/\log x)+(1+i\alpha)\color{blue}{\int_2^xt^{-2-i\alpha}\left(\pi(t)-\frac t{\log t} \right )dt}+(1+i\alpha)\color{darkred}{\int_2^x\frac{t^{-1-i\alpha}}{\log t}dt}\end{aligned}

The first integral converges to $\color{blue}{\int_2^\color{red}\infty t^{-2-i\alpha}\left(\pi(t)-\frac t{\log t} \right )dt}$ with error term $\int_x^\infty t^{-2-i\alpha}\left(\pi(t)-\frac t{\log t} \right )dt=\int_x^\infty O(\frac1{t\log^2t})=O(1/\log x)$.

For the second integral we have (where $\rm Li$ denotes the logarithmic integral)

\begin{aligned}\color{darkred}{I(x):=\int_2^x\frac{t^{-1-i\alpha}}{\log t}dt}&=\left[{\rm Li}(t)t^{-1-i\alpha} \right ]_2^x+(1+i\alpha)\int_2^x{\rm Li}(t)t^{-2-i\alpha}dt\\ &=O(1/\log x)+(1+i\alpha)\color{green}{\int_2^x\left({\rm Li}(t)-\frac t{\log t}\right)t^{-2-i\alpha}dt}+(1+i\alpha)\color{darkred}{I(x)}\end{aligned} hence \begin{aligned}\frac{i\alpha}{1+i\alpha}\color{darkred}{I(x)}&=\color{green}{\int_2^x\left({\rm Li}(t)-\frac t{\log t}\right)t^{-2-i\alpha}dt}+O_\alpha(1/\log x)\\ &=\color{green}{\int_2^\color{red}\infty\left({\rm Li}(t)-\frac t{\log t}\right)t^{-2-i\alpha}dt}+O_\alpha(1/\log x)\end{aligned} (using ${\rm Li}(t)=\frac t{\log t}+O(\frac t{\log^2t})$ in the last step).

So I get $S_\alpha(x)=\color{blue}{\int_2^\infty t^{-2-i\alpha}\left(\pi(t)-\frac t{\log t} \right )dt}+\frac{(1+i\alpha)^2}{i\alpha}\color{green}{\int_2^\infty\left({\rm Li}(t)-\frac t{\log t}\right)t^{-2-i\alpha}dt}+O_\alpha(1/\log x)$,

which means $P(1+i\alpha)$ converges.

• What makes you think there must be something wrong? – Daniel Fischer Jan 17 '16 at 14:34
• See edit; neither Wikipedia nor Wolfram Mathworld mention that it converges for these values. – punctured dusk Jan 17 '16 at 14:37
• Okay. That might just mean they didn't consider it worthwhile to mention it. Or it might mean you made a mistake. Don't know yet which. – Daniel Fischer Jan 17 '16 at 14:39
• Perhaps it's just me, but I am absolutely repelled by coloured maths. – Klangen Nov 24 '18 at 8:25

## 2 Answers

Checking your calculations, I see a sign error, which however is inconsequential. You have

$$I(x) = O(1/\log x) + (1+i\alpha) B(x) + (1+i\alpha)I(x),$$

which gives you

$$\frac{-i\alpha}{1+i\alpha}I(x) = B(x) + O_{\alpha}(1/\log x),$$

where you wrote the factor $\dfrac{i\alpha}{1+i\alpha}$.

As mentioned before, that doesn't affect the conclusion that you have

$$S_{\alpha}(x) = C + O(1/\log x),$$

whence the convergence of the series for $s = 1+i\alpha$ when $\alpha \neq 0$.

For completeness, let's include a different proof of convergence:

By bounds obtained by Pierre Dusart, we have

$$\log n + \log \log n - 1 < \frac{p_n}{n} < \log n + \log \log n$$

for $n \geqslant 6$. Setting $t_n = n\log n$, we then have $0 < p_n - t_n < n\log \log n$ for $n \geqslant 6$, and hence

$$\lvert p_n^{-s} - t_n^{-s}\rvert = \biggl\lvert s\int_{t_n}^{p_n} \frac{du}{u^{s+1}}\biggr\rvert \leqslant \lvert s\rvert \int_{t_n}^{p_n} \frac{du}{u^{1 + \operatorname{Re} s}} < \lvert s\rvert \frac{p_n-t_n}{t_n^{1 + \operatorname{Re} s}} < \lvert s\rvert \frac{\log \log n}{n^{\operatorname{Re} s}(\log n)^{1 + \operatorname{Re} s}}.$$

We see that for $\operatorname{Re} s \geqslant 1$

$$\sum_{n = 2}^{\infty} \biggl\lvert \frac{1}{p_n^{s}} - \frac{1}{t_n^{s}}\biggr\rvert < +\infty,$$

and hence $\sum p^{-s}$ converges if and only if $\sum t_n^{-s}$ converges.

The convergence of the latter series for $s = 1 + i\alpha,\; \alpha \in \mathbb{R}\setminus \{0\}$ is relatively easy to show.

First we note that via integration by parts

$$T(m) := \sum_{n = 2}^m \frac{1}{n^s} = \frac{1}{2}\bigl(2^{-s} + m^{-s}\bigr) + \int_2^m \frac{dt}{t^s} - s\int_2^m \frac{\{t\} -\tfrac{1}{2}}{t^{s+1}}\,dt,$$

where $\{t\}$ denotes the fractional part of $t$. The last integrand is Lebesgue-integrable on $[2,+\infty)$, and the first integral is $\frac{m^{1-s} - 2^{1-s}}{1-s}$, which, since $1-s$ is purely imaginary, is bounded by $\frac{2}{\lvert \operatorname{Im} s\rvert}$. So there is a $C\in \mathbb{R}$ with $\lvert T(m)\rvert \leqslant C\tag{1}$ for all $m$.

Next, for $0 < a < b$ we have

$$\biggl\lvert \frac{1}{a^s} - \frac{1}{b^s}\biggr\rvert = \biggl\lvert s\int_a^b \frac{dt}{t^{s+1}}\biggr\rvert \leqslant \lvert s\rvert \int_a^b \frac{dt}{\lvert t^{s+1}\rvert} = \lvert s\rvert \int_a^b \frac{dt}{t^2} = \lvert s\rvert\biggl(\frac{1}{a} - \frac{1}{b}\biggr). \tag{2}$$

Now a summation by parts shows

\begin{align} \sum_{n = k}^m \frac{1}{n^s}\cdot \frac{1}{(\log n)^s} &= \sum_{n = k}^m \bigl(T(n) - T(n-1)\bigr)\frac{1}{(\log n)^s}\\ &= \sum_{n = k}^m \frac{T(n)}{(\log n)^s} - \sum_{n = k-1}^{m-1} \frac{T(n)}{\bigl(\log (n+1)\bigr)^s}\\ &= \frac{T(m)}{\bigl(\log (m+1)\bigr)^s} - \frac{T(k-1)}{(\log k)^s} + \sum_{n = k}^m T(n)\biggl(\frac{1}{(\log n)^s} - \frac{1}{\bigl(\log (n+1)\bigr)^s}\biggr). \end{align}

The triangle inequality, $(1)$ and $(2)$ now yield

\begin{align} \Biggl\lvert \sum_{n = k}^m \frac{1}{n^s}\cdot \frac{1}{(\log n)^s}\Biggr\rvert &\leqslant \frac{C}{\log (m+1)} + \frac{C}{\log k} + C\lvert s\rvert \sum_{n = k}^m \biggl(\frac{1}{\log n} - \frac{1}{\log (n+1)}\biggr)\\ &= \frac{C(1+\lvert s\rvert)}{\log k} - \frac{C(\lvert s\rvert - 1)}{\log (m+1)}\\ &< \frac{C(1+\lvert s\rvert)}{\log k}, \end{align}

so the series

$$\sum_{n = 2}^{\infty} \frac{1}{t_n^s} = \sum_{n = 2}^{\infty} \frac{1}{n^s(\log n)^s}$$

is convergent. By the introductory remark, it follows that

$$\sum_p \frac{1}{p^s}$$

is convergent for $\operatorname{Re} s = 1$ and $s \neq 1$.

• Thanks for your effort! I'll await possible remarks from other users before accepting your answer, but probably I will. – punctured dusk Jan 17 '16 at 16:52
• I have a proof via approximation ($p_n \sim n\log n$ is good enough that one series converges iff the other converges for $\operatorname{Re} s \geqslant 1$) and summation by parts. You might like it. – Daniel Fischer Jan 17 '16 at 19:28

the prime number theorem says that there exists $A$ such that $\ln \zeta(s) + \ln(s-1)$ doesn't have any singularity for $\sigma > 1-\frac{A}{1+|\log t|}$,

also, for $\Re(s) \ge 1/2+\epsilon$ : $\ln \zeta(s) = \mathcal{O}(1) + \sum_p p^{-s}$ you'll find that all this implies that $\sum_{p \le x} p^{-s} - \sum_{n \le x} \frac{n^{-s}}{\ln n}$ converges when $x \to \infty$ and $\sigma > 1-\frac{A}{1+|\log t|}$, thus for $\Re(s) \ge 1$