Approximate: $\sum_{\Im(\rho)>0}\frac{1}{|\rho-\frac{1}{2}|^2}$ where $\rho$ is non trivial zeros of the zeta function I am reading Kevin Broughan's Equivalents of Riemann Hypothesis Vol. 1 (p. 38). If $\rho$ denote the non trivial zeros of Riemann zeta function in the strip $0<\Re(\rho)<1$, consider,$$S=\sum_{\Im(\rho)>0}\frac{1}{|\rho-\frac{1}{2}|^2}.$$
Question: Find the approximate value of $S$.
Attempt:
Riemann $\xi$ function can be written as
$$\xi(s)=\xi(0)\prod_{\Im(\rho)>0}(1-\frac{s}{\rho})(1-\frac{s}{\bar{\rho}}),$$
where $\bar{\rho}$ denotes complex conjugate.
Taking logarithm on both sides,
$$\log(\xi(s))=\log(\xi(0))+\sum_{\Im(\rho)>0}\log(1-\frac{s}{\rho})+\sum_{\Im(\rho)>0}\log(1-\frac{s}{\bar{\rho}})$$
and differentiating both sides w.r.t. $s$,
$$\frac{\xi'(s)}{\xi(s)}=\sum_{\Im(\rho)>0}[\frac{1}{s-\rho}+\frac{1}{s-\bar{\rho}}].$$
After this I cannot think on it.
Please find an approximate value of $$S=\sum_{\Im(\rho)>0}\frac{1}{|\rho-\frac{1}{2}|^2}.$$
Edit
$$\xi(s)=\xi(0)\prod_{\rho}(1-\frac{s}{\rho})$$
Taking log we get
$$\log(\xi(s))=\log(\xi(0))+\sum_{\rho}\log(1-\frac{s}{\rho})$$
Differentiating w.r.t. s,
$$\frac{\xi'(s)}{\xi(s)} = \sum_{\rho}\frac{1}{s-\rho}$$
Differentiating w.r.t. s again,
$$\frac{d}{ds}\frac{\xi'(s)}{\xi(s)} =- \sum_{\rho}\frac{1}{(s-\rho)^2}$$
Putting $s=1/2$ and using $\xi'(1/2)=0$,
$$\frac{\xi''(\frac{1}{2})}{\xi(\frac{1}{2})}=-\sum_{\rho} \frac{1}{(\frac{1}{2}-\rho)^2}$$

$$\xi(s)=\xi(0)\prod_{\rho}(1-\frac{s}{\rho})$$
Taking log we get
$$\log(\xi(s))=\log(\xi(0))+\sum_{\rho}\log(1-\frac{s}{\rho})$$
Differentiating w.r.t. s,
$$\frac{\xi'(s)}{\xi(s)} = \sum_{\rho}\frac{1}{s-\rho}$$
Differentiating w.r.t. s again,
$$\frac{d}{ds}\frac{\xi'(s)}{\xi(s)} =- \sum_{\rho}\frac{1}{(s-\rho)^2}$$
Putting $s=1/2$ and using $\xi'(1/2)=0$,
$$\frac{\xi''(\frac{1}{2})}{\xi(\frac{1}{2})}=-\sum_{\rho} \frac{1}{(\frac{1}{2}-\rho)^2}$$
