In the book Equivalents of Riemann Hypothesis Kevin Broughan Volume 1 , pg 38, Riemann Xi function is defined as $\xi(s)=\xi(0)\prod_{\Im(\rho)>0}(1-\frac{s(1-s)}{\rho(1-\rho)})$. Then he says that this product above is convergent if $\sum_{\Im(\rho)>0}\frac{1}{\mid{\rho(1-\rho)}\mid}<\infty$. Then on completing the square
$$\sum_{\Im(\rho)>0}\frac{1}{\mid{\rho(1-\rho)}\mid}< \sum_{\Im(\rho)>0}\frac{1}{\mid{\rho-\frac{1}{2}}\mid^2}\ll \int_{1}^{\infty} \frac{d(t\log t)}{t^2}.$$
I am not able to understand the last inequality involving integration
Let $N(t)$ be the number of zeros in the critical strip whose imaginary part is positive and is at most $t$. It is known that $N(t) \ll t\log t$. Thus $$ \sum\limits_{0 < \Im (\rho ) \le T} {\frac{1}{{\left| {\rho - \frac{1}{2}} \right|^2 }}} = \int_1^T {\frac{{dN(t)}}{{\left| {t - \frac{1}{2}} \right|^2 }}} \ll \int_1^T {\frac{{d(t\log t)}}{{t^2 }}} $$ ($N(t)=0$ for $0<t<1$). Now let $T\to +\infty$.