# Term by term integration over a finite interval

I am reading The Riemann zeta function by H.M. Edwards , $$1974$$, pg. $$52-53$$. Here the author denotes by $$\rho$$ the non trivial zeros of Riemann zeta function

He proves that the following series in $$\rho$$ converges uniformly in any disc $$|s|\leq K$$ when the terms $$\rho$$ and $$1-\rho$$ are paired $$\sum_{\rho}\frac{1}{s-\rho}$$ For this he observes $$\left|\frac{1}{s-\rho}+\frac{1}{s-(1-\rho)}\right|=\left|\left[\frac{1}{\left(s-\frac{1}{2}\right)-\left(\rho-\frac{1}{2}\right)}\right]+\left[\frac{1}{\left(s-\frac{1}{2}\right)+\left(\rho-\frac{1}{2}\right)}\right]\right|$$ which gives $$\left|\frac{1}{s-\rho}+\frac{1}{s-(1-\rho)}\right|=\left|\frac{2\left(s-\frac{1}{2}\right)}{\left(s-\frac{1}{2}\right)^2-\left(\rho-\frac{1}{2}\right)^2}\right|$$ Now we have $$\left|\frac{2\left(s-\frac{1}{2}\right)}{\left(s-\frac{1}{2}\right)^2-\left(\rho-\frac{1}{2}\right)^2}\right|\leq \frac{C}{\left|\rho-\frac{1}{2}\right|^2}$$ for all sufficiently large $$\rho$$ once $$K$$ is fixed and because $$\sum_{\rho}\frac{1}{|\rho-\frac{1}{2}|^2}$$ converges.

Question: How do we have the last inequality for all sufficiently large $$\rho$$? How does it ensure the uniform convergence?

Can someone please clear this doubt. Thank you.

• @JyrkiLahtonen Sorry I could not follow. So we have an upper bound for the tail of the sum. How does it ensure uniform convergence?
– user1073119
Jul 14, 2022 at 15:58
• Wait a second. I only now realize that $\rho$ was not intended to be an integer variable. Scratch all that. Jul 14, 2022 at 16:22
• @JyrkiLahtonen Yes $\rho$ is a complex number with real part between $0$ and $1$.
– user1073119
Jul 14, 2022 at 16:23
• Is it known that $\sum_\rho|\rho-\dfrac12|^{-2}$ converges? When I didn't read carefully, and mistakenly assumed $\rho$ to be an integer variable, the convergence was obvious. If the convergence of that sum is known for some other reason, then an argument like Weierstrass M-test should still work. Jul 15, 2022 at 12:41
• @JyrkiLahtonen Yes that's okay. But I need an estimate valid for all $\rho$. Please see the answer below. But I am not able to understand why the second maximum exist or why $C_K$ should exist.
– user1073119
Jul 15, 2022 at 12:44

When $$|s|\le K$$, we have $$|s-\frac12|\le K+\frac12$$. By the properties of $$\rho$$, we know that when $$\Im(\rho)>\sqrt2(K+\frac12)$$, there is $$|\rho-\frac12|>\sqrt2(K+\frac12)$$, so $$|\rho-\frac12|>\sqrt2|s-\frac12|$$. Consequently, we have $$\left|{2\left(s-\frac12\right)\over\left(s-\frac12\right)^2-\left(\rho-\frac12\right)^2}\right|\le{2K+1\over|\rho-\frac12|^2-|s-\frac12|^2}<{4K+2\over|\rho-\frac12|^2}.$$

• Thanks. But how do we know that $\Im (\rho)$ is large enough? We have the first $\Im (\rho)$ at $\Im(\rho)\approx 14.1$. Then do we have similar inequality to the above inequality for small $\rho$?
– user1073119
Jul 15, 2022 at 9:09
• Please prove an inequality valid for all non trivial zeros of Riemann zeta, $\rho$
– user1073119
Jul 15, 2022 at 10:15
• As long as you can find a constant valid for large $\Im(\rho)$, you can automatically deduce that replacing $4K+2$ by a larger number will make the inequality work for all $\rho$. Jul 15, 2022 at 11:55
• By large, I mean that there exists $T>0$ such that when $\Im(\rho)>T$, we have $|\rho-\frac12|>\sqrt2(K+\frac12)$, which is certainly possible because there are $\zeta(s)$ has infinitely many nontrivial zeros. Jul 15, 2022 at 11:56
• I thank you for your guidance. Please show that how can we automatically deduce that replacing $4K+2$ by a larger number which will make the inequality work for all $\rho$. Please include this in your answer.
– user1073119
Jul 15, 2022 at 12:03