# A line integral involving $\zeta(s)$

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 $$I$$ ?

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