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js21
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The answer is no, even if we assume there are no other poles than $1$ in $\sigma > 1- \epsilon_0$. I give an example below with $\epsilon_0=1$. This is a variant of an example given by Karamata in $1952$.

Let $b_n = 1 + \cos(\log^2(n)) \geq 0$ and consider $B(s) = \sum_{n \geq 1} b_n n^{-s}$. Let us write $$ \sum_{n \leq x} b_n = \int_{1}^x (1+\cos(\log^2(t))) dt + R(x) $$ where $R(x) = O(\log^2(x))$. Then for $\sigma >1$ we have $$ B(s) = \int_{1}^{+ \infty}(1+ \cos(\log^2(t)) )t^{-s}dt + s \int_{1}^{+ \infty} R(t) t^{-s-1} dt. $$ The last integral extends holomorphically to $\sigma >0$. The first integral is equal to $$ \frac{1}{s-1} + \frac{1}{2} I_+(s) + \frac{1}{2} I_-(s) $$ where $$ I_{\pm}(s) = \int_{0}^{+ \infty} \exp{((1-s)u \pm i u^2)} dt. $$ Using contour integration one can move the integration line to $u \mapsto e^{\pm \frac{i \pi}{4}}u$, and this yields $$ I_{\pm}(s) = e^{\pm \frac{i \pi}{4}} \int_{0}^{+ \infty} \exp{((1-s)u e^{\pm \frac{i \pi}{4}} - u^2)} dt. $$ Thus $I_+$ and $I_-$ are entire functions.

Let us now choose $a_n = b_n \log(n) \geq 0$ so that $D(s) = - B'(s)$ is meromorphic on $\sigma >0$, with a unique pole at $s=1$ of order $2$. One checks that $$ \sum_{n \geq x} a_n = x \log(x) + x \left( \frac{1}{2} \sin(\log^2(x))-1 \right) +O(\frac{x}{\log x}) $$$$ \sum_{n \leq x} a_n = x \log(x) + x \left( \frac{1}{2} \sin(\log^2(x))-1 \right) +O(\frac{x}{\log x}) $$ Remark : The error term in Delange's theorem can be improved provided growth assumptions (on a strip) are made on $D(s)$.

The answer is no, even if we assume there are no other poles than $1$ in $\sigma > 1- \epsilon_0$. I give an example below with $\epsilon_0=1$. This is a variant of an example given by Karamata in $1952$.

Let $b_n = 1 + \cos(\log^2(n)) \geq 0$ and consider $B(s) = \sum_{n \geq 1} b_n n^{-s}$. Let us write $$ \sum_{n \leq x} b_n = \int_{1}^x (1+\cos(\log^2(t))) dt + R(x) $$ where $R(x) = O(\log^2(x))$. Then for $\sigma >1$ we have $$ B(s) = \int_{1}^{+ \infty}(1+ \cos(\log^2(t)) )t^{-s}dt + s \int_{1}^{+ \infty} R(t) t^{-s-1} dt. $$ The last integral extends holomorphically to $\sigma >0$. The first integral is equal to $$ \frac{1}{s-1} + \frac{1}{2} I_+(s) + \frac{1}{2} I_-(s) $$ where $$ I_{\pm}(s) = \int_{0}^{+ \infty} \exp{((1-s)u \pm i u^2)} dt. $$ Using contour integration one can move the integration line to $u \mapsto e^{\pm \frac{i \pi}{4}}u$, and this yields $$ I_{\pm}(s) = e^{\pm \frac{i \pi}{4}} \int_{0}^{+ \infty} \exp{((1-s)u e^{\pm \frac{i \pi}{4}} - u^2)} dt. $$ Thus $I_+$ and $I_-$ are entire functions.

Let us now choose $a_n = b_n \log(n) \geq 0$ so that $D(s) = - B'(s)$ is meromorphic on $\sigma >0$, with a unique pole at $s=1$ of order $2$. One checks that $$ \sum_{n \geq x} a_n = x \log(x) + x \left( \frac{1}{2} \sin(\log^2(x))-1 \right) +O(\frac{x}{\log x}) $$ Remark : The error term in Delange's theorem can be improved provided growth assumptions (on a strip) are made on $D(s)$.

The answer is no, even if we assume there are no other poles than $1$ in $\sigma > 1- \epsilon_0$. I give an example below with $\epsilon_0=1$. This is a variant of an example given by Karamata in $1952$.

Let $b_n = 1 + \cos(\log^2(n)) \geq 0$ and consider $B(s) = \sum_{n \geq 1} b_n n^{-s}$. Let us write $$ \sum_{n \leq x} b_n = \int_{1}^x (1+\cos(\log^2(t))) dt + R(x) $$ where $R(x) = O(\log^2(x))$. Then for $\sigma >1$ we have $$ B(s) = \int_{1}^{+ \infty}(1+ \cos(\log^2(t)) )t^{-s}dt + s \int_{1}^{+ \infty} R(t) t^{-s-1} dt. $$ The last integral extends holomorphically to $\sigma >0$. The first integral is equal to $$ \frac{1}{s-1} + \frac{1}{2} I_+(s) + \frac{1}{2} I_-(s) $$ where $$ I_{\pm}(s) = \int_{0}^{+ \infty} \exp{((1-s)u \pm i u^2)} dt. $$ Using contour integration one can move the integration line to $u \mapsto e^{\pm \frac{i \pi}{4}}u$, and this yields $$ I_{\pm}(s) = e^{\pm \frac{i \pi}{4}} \int_{0}^{+ \infty} \exp{((1-s)u e^{\pm \frac{i \pi}{4}} - u^2)} dt. $$ Thus $I_+$ and $I_-$ are entire functions.

Let us now choose $a_n = b_n \log(n) \geq 0$ so that $D(s) = - B'(s)$ is meromorphic on $\sigma >0$, with a unique pole at $s=1$ of order $2$. One checks that $$ \sum_{n \leq x} a_n = x \log(x) + x \left( \frac{1}{2} \sin(\log^2(x))-1 \right) +O(\frac{x}{\log x}) $$ Remark : The error term in Delange's theorem can be improved provided growth assumptions (on a strip) are made on $D(s)$.

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js21
js21
  • 7.2k
  • 22
  • 38

The answer is no, even if we assume there are no other poles than $1$ in $\sigma > 1- \epsilon_0$. I give an example below with $\epsilon_0=1$. This is a variant of an example given by Karamata in $1952$.

Let $b_n = 1 + \cos(\log^2(n)) \geq 0$ and consider $B(s) = \sum_{n \geq 1} b_n n^{-s}$. Let us write $$ \sum_{n \leq x} b_n = \int_{1}^x (1+\cos(\log^2(t))) dt + R(x) $$ where $R(x) = O(\log^2(x))$. Then for $\sigma >1$ we have $$ B(s) = \int_{1}^{+ \infty}(1+ \cos(\log^2(t)) )t^{-s}dt + s \int_{1}^{+ \infty} R(t) t^{-s-1} dt. $$ The last integral extends holomorphically to $\sigma >0$. The first integral is equal to $$ \frac{1}{s-1} + \frac{1}{2} I_+(s) + \frac{1}{2} I_-(s) $$ where $$ I_{\pm}(s) = \int_{0}^{+ \infty} \exp{((1-s)u \pm i u^2)} dt. $$ Using contour integration one can move the integration line to $u \mapsto e^{\pm \frac{i \pi}{4}}u$, and this yields $$ I_{\pm}(s) = e^{\pm \frac{i \pi}{4}} \int_{0}^{+ \infty} \exp{((1-s)u e^{\pm \frac{i \pi}{4}} - u^2)} dt. $$ Thus $I_+$ and $I_-$ are entire functions.

Let us now choose $a_n = b_n \log(n) \geq 0$ so that $D(s) = - B'(s)$ is meromorphic on $\sigma >0$, with a unique pole at $s=1$ of order $2$. One checks that $$ \sum_{n \geq x} a_n = x \log(x) + x \left( \frac{1}{2} \sin(\log^2(x))-1 \right) +O(\frac{x}{\log x}) $$ Remark : The error term in Delange's theorem can be improved provided growth assumptions (on a strip) are made on $D(s)$.