1
$\begingroup$

Consider the line integral

$$I=\int_{1/2 -i\infty}^{1/2 + i\infty} \frac{\log((s-1)\zeta(s))}{s} \mathrm{d}s-\int_{1/2 -i \infty}^{1/2 + i\infty} \frac{i\arg \zeta (s)}{s}\mathrm{d}s$$

where $\zeta$ denotes the Riemann zeta function, using complex analysis (or otherwise). Note that $I$ exists (is well-defined) since $|\zeta(s)|=o(|s|)$ for $\Re(s)=1/2$.

Is the argument of https://arxiv.org/abs/1306.0856 (proofs of Theorem 1.2-1.3) applicable in evaluating $I$ ?

$\endgroup$
1
$\begingroup$
  • Do you know the Cauchy integral theorem ? $$\int_{1/2 -i\infty}^{1/2 + i\infty} \frac{\log(s-1)}{s^2} ds= \lim_{a\to 0}\int_{C_a} \frac{\log(s-1)}{s^2} ds - \int_{L_a} \frac{\log(s-1)}{s^2} ds$$ where $C_a$ is a closed-contour $1/2-i\infty \to 1/2-ia \to 1+a-ia \to 1+a+ia\to 1/2+ia \to 1/2+i\infty \to \infty +i\infty \to \infty-i\infty \to 1/2-i\infty$ and $L_a$ is the portion $1/2-ia \to 1+a-ia \to 1+a+ia\to 1/2+ia$.

Since $\frac{\log(s-1)}{s^2}$ is analytic inside $S_a$ and decreases in $O(s^{-1-\epsilon})$ then $\int_{C_a} \frac{\log(s-1)}{s^2} ds = 0$ and since $\log(s-1) \mapsto \log(s-1)+2i\pi$ when rotating counterclockwise around $s-1$ then $$\lim_{a\to 0}\int_{L_a} \frac{\log(s-1)}{s^2} ds= \int_1^{1/2} \frac{2i\pi}{s^2}ds$$

  • For $F$ analytic decreasing in $O(s^{-1-\epsilon})$, assuming the implied branch of $\log (\zeta(s)(s-1))$ is $O(s^{c}), c < \epsilon$ the same argument (with a contour excluding each non-trivial zero) gives $$\lim_{T \to \infty}\int_{1/2 -iT}^{1/2 + iT} \log(\zeta(s)(s-1))F(s) ds = \lim_{T \to \infty}\sum_{\rho = \sigma+it, \sigma > 1/2, |t| < T} \int_{1/2+it}^{\sigma+it} 2i \pi F(s)ds$$

  • In your question, what branch of $\log (\zeta(s)(s-1))$ are you considering ?

    Look at $$\lim_{T \to \infty}\int_{1/2 -iT}^{1/2 + iT} \frac{\log(\zeta(s)(s-1))}{s^{1+1/T}} ds$$

$\endgroup$

Your Answer

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