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edited May 12, 2019 at 10:05

As suggested by @Peter we can prove the claim as follows. We will use of the following results from the book "Multiplicative Number Theory, I" by Montgomery-Vaughan.

Lemma 12.2 For each real number $T\geq 2$ there is a $T_1$ with $T\leq T_1\leq T+1$ such that $$ \frac{\zeta^{\prime}}{\zeta}(\sigma+iT_1)\ll(\log T)^2 $$ uniformly for $-1\leq \sigma\leq 2$.

and

Lemma 12.4 Let $\mathcal{A}$ denote the set of points of $s\in\mathbb{C}$ such that $\sigma\leq -1$ and $|s+2k|\geq 1/4$ for every positive integer $k$. Then $$ \frac{\zeta^{\prime}}{\zeta}(s)\ll\log(|s|+1) $$ uniformly for $s\in\mathcal{A}$.

Using these two results we proceed as follows (basically following the proof of Theorem 12.5 in the same book): let $K$ be a positive integer and consider the contour of integration consisting of the line segments connecting $$ c-iT_1,\quad -K-iT_1,\quad -K+iT_1,\quad c+iT_1 $$ Moreover we split the horizontal segments $$ [-K\pm iT_1,\,c\pm iT_1] $$ in $$ [-K\pm iT_1,\,-1-\Re(s)\pm iT_1]\cup [-1-\Re(s)\pm iT_1,\,c\pm iT_1] $$ Since $|\sigma+iT_1|\geq T$, by Lemma 12.2 if follows that $$ \int_{-1-\Re(s)\pm iT_1}^{c\pm iT_1}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw\ll\frac{(\log T)^2}{T^2}\int_{-1-\Re(s)}^{c}x^{\sigma}d\sigma\ll\frac{x(\log T)^2}{T^2 \log x} $$ which goes to 0 as $T\to \infty$. Similarly, by Lemma 12.4 we have $$ \int_{-K\pm iT_1}^{-1-\Re(s)\pm iT_1}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw\ll \frac{\log T}{T^2}\int_{-K}^{-1-\Re(s)}x^{\sigma}d\sigma\ll\frac{\log T}{T^2}\int_{-\infty}^{-1}x^{\sigma}d\sigma\\ \ll\frac{\log T}{xT^2 \log x} $$ which again tends to 0 as $T\to \infty$. It remain roremains to bound the vertical segment. Using again Lemma 12.4 and subsadditivity of the logarithm, we get $$ \int_{-K-iT_1}^{-K+iT_1}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw\ll\frac{\log (K+T)}{K^2}x^{-K}\int_{-T_1}^{T_1}1\,dt\ll\frac{T\log KT}{K^2x^K} $$ which tends to 0 as $K\to \infty$, proving our claim.

As suggested by @Peter we can prove the claim as follows. We will use of the following results from the book "Multiplicative Number Theory, I" by Montgomery-Vaughan.

Lemma 12.2 For each real number $T\geq 2$ there is a $T_1$ with $T\leq T_1\leq T+1$ such that $$ \frac{\zeta^{\prime}}{\zeta}(\sigma+iT_1)\ll(\log T)^2 $$ uniformly for $-1\leq \sigma\leq 2$.

and

Lemma 12.4 Let $\mathcal{A}$ denote the set of points of $s\in\mathbb{C}$ such that $\sigma\leq -1$ and $|s+2k|\geq 1/4$ for every positive integer $k$. Then $$ \frac{\zeta^{\prime}}{\zeta}(s)\ll\log(|s|+1) $$ uniformly for $s\in\mathcal{A}$.

Using these two results we proceed as follows (basically following the proof of Theorem 12.5 in the same book): let $K$ be a positive integer and consider the contour of integration consisting of the line segments connecting $$ c-iT_1,\quad -K-iT_1,\quad -K+iT_1,\quad c+iT_1 $$ Moreover we split the horizontal segments $$ [-K\pm iT_1,\,c\pm iT_1] $$ in $$ [-K\pm iT_1,\,-1-\Re(s)\pm iT_1]\cup [-1-\Re(s)\pm iT_1,\,c\pm iT_1] $$ Since $|\sigma+iT_1|\geq T$, by Lemma 12.2 if follows that $$ \int_{-1-\Re(s)\pm iT_1}^{c\pm iT_1}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw\ll\frac{(\log T)^2}{T^2}\int_{-1-\Re(s)}^{c}x^{\sigma}d\sigma\ll\frac{x(\log T)^2}{T^2 \log x} $$ which goes to 0 as $T\to \infty$. Similarly, by Lemma 12.4 we have $$ \int_{-K\pm iT_1}^{-1-\Re(s)\pm iT_1}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw\ll \frac{\log T}{T^2}\int_{-K}^{-1-\Re(s)}x^{\sigma}d\sigma\ll\frac{\log T}{T^2}\int_{-\infty}^{-1}x^{\sigma}d\sigma\\ \ll\frac{\log T}{xT^2 \log x} $$ which again tends to 0 as $T\to \infty$. It remain ro bound the vertical segment. Using again Lemma 12.4 and subsadditivity of logarithm, we get $$ \int_{-K-iT_1}^{-K+iT_1}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw\ll\frac{\log (K+T)}{K^2}x^{-K}\int_{-T_1}^{T_1}1\,dt\ll\frac{T\log KT}{K^2x^K} $$ which tends to 0 as $K\to \infty$, proving our claim.

As suggested by @Peter we can prove the claim as follows. We will use of the following results from the book "Multiplicative Number Theory, I" by Montgomery-Vaughan.

Lemma 12.2 For each real number $T\geq 2$ there is a $T_1$ with $T\leq T_1\leq T+1$ such that $$ \frac{\zeta^{\prime}}{\zeta}(\sigma+iT_1)\ll(\log T)^2 $$ uniformly for $-1\leq \sigma\leq 2$.

and

Lemma 12.4 Let $\mathcal{A}$ denote the set of points of $s\in\mathbb{C}$ such that $\sigma\leq -1$ and $|s+2k|\geq 1/4$ for every positive integer $k$. Then $$ \frac{\zeta^{\prime}}{\zeta}(s)\ll\log(|s|+1) $$ uniformly for $s\in\mathcal{A}$.

Using these two results we proceed as follows (basically following the proof of Theorem 12.5 in the same book): let $K$ be a positive integer and consider the contour of integration consisting of the line segments connecting $$ c-iT_1,\quad -K-iT_1,\quad -K+iT_1,\quad c+iT_1 $$ Moreover we split the horizontal segments $$ [-K\pm iT_1,\,c\pm iT_1] $$ in $$ [-K\pm iT_1,\,-1-\Re(s)\pm iT_1]\cup [-1-\Re(s)\pm iT_1,\,c\pm iT_1] $$ Since $|\sigma+iT_1|\geq T$, by Lemma 12.2 if follows that $$ \int_{-1-\Re(s)\pm iT_1}^{c\pm iT_1}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw\ll\frac{(\log T)^2}{T^2}\int_{-1-\Re(s)}^{c}x^{\sigma}d\sigma\ll\frac{x(\log T)^2}{T^2 \log x} $$ which goes to 0 as $T\to \infty$. Similarly, by Lemma 12.4 we have $$ \int_{-K\pm iT_1}^{-1-\Re(s)\pm iT_1}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw\ll \frac{\log T}{T^2}\int_{-K}^{-1-\Re(s)}x^{\sigma}d\sigma\ll\frac{\log T}{T^2}\int_{-\infty}^{-1}x^{\sigma}d\sigma\\ \ll\frac{\log T}{xT^2 \log x} $$ which again tends to 0 as $T\to \infty$. It remains to bound the vertical segment. Using again Lemma 12.4 and subsadditivity of the logarithm, we get $$ \int_{-K-iT_1}^{-K+iT_1}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw\ll\frac{\log (K+T)}{K^2}x^{-K}\int_{-T_1}^{T_1}1\,dt\ll\frac{T\log KT}{K^2x^K} $$ which tends to 0 as $K\to \infty$, proving our claim.

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created May 12, 2019 at 9:07

asd

As suggested by @Peter we can prove the claim as follows. We will use of the following results from the book "Multiplicative Number Theory, I" by Montgomery-Vaughan.

Lemma 12.2 For each real number $T\geq 2$ there is a $T_1$ with $T\leq T_1\leq T+1$ such that $$ \frac{\zeta^{\prime}}{\zeta}(\sigma+iT_1)\ll(\log T)^2 $$ uniformly for $-1\leq \sigma\leq 2$.

and

Lemma 12.4 Let $\mathcal{A}$ denote the set of points of $s\in\mathbb{C}$ such that $\sigma\leq -1$ and $|s+2k|\geq 1/4$ for every positive integer $k$. Then $$ \frac{\zeta^{\prime}}{\zeta}(s)\ll\log(|s|+1) $$ uniformly for $s\in\mathcal{A}$.

Using these two results we proceed as follows (basically following the proof of Theorem 12.5 in the same book): let $K$ be a positive integer and consider the contour of integration consisting of the line segments connecting $$ c-iT_1,\quad -K-iT_1,\quad -K+iT_1,\quad c+iT_1 $$ Moreover we split the horizontal segments $$ [-K\pm iT_1,\,c\pm iT_1] $$ in $$ [-K\pm iT_1,\,-1-\Re(s)\pm iT_1]\cup [-1-\Re(s)\pm iT_1,\,c\pm iT_1] $$ Since $|\sigma+iT_1|\geq T$, by Lemma 12.2 if follows that $$ \int_{-1-\Re(s)\pm iT_1}^{c\pm iT_1}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw\ll\frac{(\log T)^2}{T^2}\int_{-1-\Re(s)}^{c}x^{\sigma}d\sigma\ll\frac{x(\log T)^2}{T^2 \log x} $$ which goes to 0 as $T\to \infty$. Similarly, by Lemma 12.4 we have $$ \int_{-K\pm iT_1}^{-1-\Re(s)\pm iT_1}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw\ll \frac{\log T}{T^2}\int_{-K}^{-1-\Re(s)}x^{\sigma}d\sigma\ll\frac{\log T}{T^2}\int_{-\infty}^{-1}x^{\sigma}d\sigma\\ \ll\frac{\log T}{xT^2 \log x} $$ which again tends to 0 as $T\to \infty$. It remain ro bound the vertical segment. Using again Lemma 12.4 and subsadditivity of logarithm, we get $$ \int_{-K-iT_1}^{-K+iT_1}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw\ll\frac{\log (K+T)}{K^2}x^{-K}\int_{-T_1}^{T_1}1\,dt\ll\frac{T\log KT}{K^2x^K} $$ which tends to 0 as $K\to \infty$, proving our claim.