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 $\xi'(1/2)=0$,
$$\frac{\xi''(\frac{1}{2})}{\xi(\frac{1}{2})}=-\sum_{\rho} \frac{1}{(\frac{1}{2}-\rho)^2}$$
 A: A comment that got too long:
Note that $S=\sum_{\Im(\rho)>0}\frac{1}{|\rho-\frac{1}{2}|^2}=\frac{1}{2}\sum_{\rho}\frac{1}{|\rho-\frac{1}{2}|^2}$ since for conjugate roots $|\rho-1/2|=|\bar \rho -1/2|$ and the results of Broughan below involve the full sum
$2S$ is not that different from $\sum_{\Im(\rho)>0}\frac{1}{|\rho|^2}$ which is estimated in Broughan by $0.04619..$, so an estimate of $0.0463$, for example, is probably very easy (leave details to OP as I am not that fond of pointless numerical calculations - note that )
Up to some height, $H$ (see Broughan who takes $H=600,269.6771$) where all the zeroes are on the critical line the difference is just the sum of $\frac{1}{4t^2(t^2+1/4)} \le \frac{1}{4t^4}$ where those roots are $1/2+it$ and we know that $t$ increases quite fast starting at $14+$ so $1/(4t^4)$ starts at something lower than $0.00001$ and decreasing fast.
So if one needs better numerics take the table of zeroes of $\zeta$ up to the Broughan range $H$ above and compute $\sum 1/(4t^4), \rho=1/2+it, 14 < t \le H$)
Beyond that range, $S$ and the Broughan sum part differ by at most a factor of $1-\frac{1}{H}$ and that is fairly negligible not to speak that the part of either sum beyond that range is really small.
A: $$\xi(s)=\xi(0)\prod_{\rho}(1-\frac{s}{\rho})$$
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 $\xi'(1/2)=0$,
$$\frac{\xi''(\frac{1}{2})}{\xi(\frac{1}{2})}=-\sum_{\rho} \frac{1}{(\frac{1}{2}-\rho)^2}$$
