Question Define $f(z)=(s-1)\zeta(s)$ where $s=\frac{1}{1+z^2}$ and $\zeta$ denotes the Riemann zeta function. Prove that if $\rho$ denotes the non trivial zeros of $\zeta(s)$ then, $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{\Re(\rho)>1/2} \log\left|\frac{\rho}{1-\rho}\right|$$ I am reading a paper by Balazard et al. on the zeta function where both sums converge - https://www.semanticscholar.org/paper/Notes-sur-la-fonction%CE%B6de-Riemann%2C-2-Balazard-Saias/1e9f1a23efa0b569f4786944172d0f21b8863089

My try- $\rho=\frac{1}{1+\alpha^2}$ then $\alpha^2=\frac{1-\rho}{\rho}$ so that $\alpha=\pm \sqrt{\frac{1-\rho}{\rho}}$ $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi<\arg(\alpha)\leq \pi,|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2} $$ Since the sum on the right hand side is absolutely convergent so we can write the sum in any order.$$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi<\arg(\alpha)\leq 0}\log \frac{1}{|\alpha|^2}+ \sum_{0<\arg(\alpha)\leq \pi}\log \frac{1}{|\alpha|^2} $$ $\rho=\frac{1}{1+\alpha^2}$ is injective on $-\pi<\arg(\alpha)\leq 0$ and also it is injective on $0<\arg(\alpha)\leq \pi$. So using $\rho=\frac{1}{1+\alpha^2}$ we get, $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right|+ \sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right| $$

Question Define $f(z)=(s-1)\zeta(s)$ where $s=\frac{1}{1+z^2}$ and $\zeta$ denotes the Riemann zeta function. Prove that if $\rho$ denotes the non trivial zeros of $\zeta(s)$ then, $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{\Re(\rho)>1/2} \log\left|\frac{\rho}{1-\rho}\right|$$ I am reading a paper by Balazard et al. on the zeta function where both sums converge.

My try- $\rho=\frac{1}{1+\alpha^2}$ then $\alpha^2=\frac{1-\rho}{\rho}$ so that $\alpha=\pm \sqrt{\frac{1-\rho}{\rho}}$ $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi<\arg(\alpha)\leq \pi,|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2} $$ Since the sum on the right hand side is absolutely convergent so we can write the sum in any order.$$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi<\arg(\alpha)\leq 0}\log \frac{1}{|\alpha|^2}+ \sum_{0<\arg(\alpha)\leq \pi}\log \frac{1}{|\alpha|^2} $$ $\rho=\frac{1}{1+\alpha^2}$ is injective on $-\pi<\arg(\alpha)\leq 0$ and also it is injective on $0<\arg(\alpha)\leq \pi$. So using $\rho=\frac{1}{1+\alpha^2}$ we get, $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right|+ \sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right| $$

Question Define $f(z)=(s-1)\zeta(s)$ where $s=\frac{1}{1+z^2}$ and $\zeta$ denotes the Riemann zeta function. Prove that if $\rho$ denotes the non trivial zeros of $\zeta(s)$ then, $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{\Re(\rho)>1/2} \log\left|\frac{\rho}{1-\rho}\right|$$ I am reading a paper by Balazard et al. on the zeta function- https://www.semanticscholar.org/paper/Notes-sur-la-fonction%CE%B6de-Riemann%2C-2-Balazard-Saias/1e9f1a23efa0b569f4786944172d0f21b8863089

My try- $\rho=\frac{1}{1+\alpha^2}$ then $\alpha^2=\frac{1-\rho}{\rho}$ so that $\alpha=\pm \sqrt{\frac{1-\rho}{\rho}}$ $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi<\arg(\alpha)\leq \pi,|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2} $$ Since the sum on the right hand side is absolutely convergent so we can write the sum in any order.$$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi<\arg(\alpha)\leq 0}\log \frac{1}{|\alpha|^2}+ \sum_{0<\arg(\alpha)\leq \pi}\log \frac{1}{|\alpha|^2} $$ $\rho=\frac{1}{1+\alpha^2}$ is injective on $-\pi<\arg(\alpha)\leq 0$ and also it is injective on $0<\arg(\alpha)\leq \pi$. So using $\rho=\frac{1}{1+\alpha^2}$ we get, $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right|+ \sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right| $$

Question Define $f(z)=(s-1)\zeta(s)$ where $s=\frac{1}{1+z^2}$ and $\zeta$ denotes the Riemann zeta function. Prove that if $\rho$ denotes the non trivial zeros of $\zeta(s)$ then, $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{\Re(\rho)>1/2} \log\left|\frac{\rho}{1-\rho}\right|$$ I am reading a paper by Balazard et al. on the zeta function where both sums converge - https://www.semanticscholar.org/paper/Notes-sur-la-fonction%CE%B6de-Riemann%2C-2-Balazard-Saias/1e9f1a23efa0b569f4786944172d0f21b8863089

My try- $\rho=\frac{1}{1+\alpha^2}$ then $\alpha^2=\frac{1-\rho}{\rho}$ so that $\alpha=\pm \sqrt{\frac{1-\rho}{\rho}}$ $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi<\arg(\alpha)\leq \pi,|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2} $$ Since the sum on the right hand side is absolutely convergent so we can write the sum in any order.$$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi<\arg(\alpha)\leq 0}\log \frac{1}{|\alpha|^2}+ \sum_{0<\arg(\alpha)\leq \pi}\log \frac{1}{|\alpha|^2} $$ $\rho=\frac{1}{1+\alpha^2}$ is injective on $-\pi<\arg(\alpha)\leq 0$ and also it is injective on $0<\arg(\alpha)\leq \pi$. So using $\rho=\frac{1}{1+\alpha^2}$ we get, $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right|+ \sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right| $$

Question Define $f(z)=(s-1)\zeta(s)$ where $s=\frac{1}{1+z^2}$ and $\zeta$ denotes the Riemann zeta function. Prove that if $\rho$ denotes the non trivial zeros of $\zeta(s)$ then, $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{\Re(\rho)>1/2} \log\left|\frac{\rho}{1-\rho}\right|$$

My try- $\rho=\frac{1}{1+\alpha^2}$ then $\alpha^2=\frac{1-\rho}{\rho}$ so that $\alpha=\pm \sqrt{\frac{1-\rho}{\rho}}$ $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi<\arg(\alpha)\leq \pi,|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2} $$ Since the sum on the right hand side is absolutely convergent so we can write the sum in any order.$$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi<\arg(\alpha)\leq 0}\log \frac{1}{|\alpha|^2}+ \sum_{0<\arg(\alpha)\leq \pi}\log \frac{1}{|\alpha|^2} $$ $\rho=\frac{1}{1+\alpha^2}$ is injective on $-\pi<\arg(\alpha)\leq 0$ and also it is injective on $0<\arg(\alpha)\leq \pi$. So using $\rho=\frac{1}{1+\alpha^2}$ we get, $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right|+ \sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right| $$

Question Define $f(z)=(s-1)\zeta(s)$ where $s=\frac{1}{1+z^2}$ and $\zeta$ denotes the Riemann zeta function. Prove that if $\rho$ denotes the non trivial zeros of $\zeta(s)$ then, $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{\Re(\rho)>1/2} \log\left|\frac{\rho}{1-\rho}\right|$$ I am reading a paper by Balazard et al. on the zeta function- https://www.semanticscholar.org/paper/Notes-sur-la-fonction%CE%B6de-Riemann%2C-2-Balazard-Saias/1e9f1a23efa0b569f4786944172d0f21b8863089

My try- $\rho=\frac{1}{1+\alpha^2}$ then $\alpha^2=\frac{1-\rho}{\rho}$ so that $\alpha=\pm \sqrt{\frac{1-\rho}{\rho}}$ $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi<\arg(\alpha)\leq \pi,|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2} $$ Since the sum on the right hand side is absolutely convergent so we can write the sum in any order.$$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi<\arg(\alpha)\leq 0}\log \frac{1}{|\alpha|^2}+ \sum_{0<\arg(\alpha)\leq \pi}\log \frac{1}{|\alpha|^2} $$ $\rho=\frac{1}{1+\alpha^2}$ is injective on $-\pi<\arg(\alpha)\leq 0$ and also it is injective on $0<\arg(\alpha)\leq \pi$. So using $\rho=\frac{1}{1+\alpha^2}$ we get, $$\sum_{|\alpha|<1,f(\alpha)=0}\log \frac{1}{|\alpha|^2}=\sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right|+ \sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right| $$