6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What makes you think there must be something wrong? $\endgroup$ – Daniel Fischer Jan 17 '16 at 14:34
  • $\begingroup$ See edit; neither Wikipedia nor Wolfram Mathworld mention that it converges for these values. $\endgroup$ – punctured dusk Jan 17 '16 at 14:37
  • $\begingroup$ 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. $\endgroup$ – Daniel Fischer Jan 17 '16 at 14:39
  • $\begingroup$ Perhaps it's just me, but I am absolutely repelled by coloured maths. $\endgroup$ – Klangen Nov 24 '18 at 8:25
4
$\begingroup$

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 $s \neq 1$.

$\endgroup$
  • $\begingroup$ Thanks for your effort! I'll await possible remarks from other users before accepting your answer, but probably I will. $\endgroup$ – punctured dusk Jan 17 '16 at 16:52
  • $\begingroup$ 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. $\endgroup$ – Daniel Fischer Jan 17 '16 at 19:28
1
$\begingroup$

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 $\Re(s) \ge 1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.