First we calculate ${\displaystyle {\int }_2^T\frac{1}{x^2} dx=(-x^{-1}){|}_2^T=(-\frac{1}{T}+\frac{1}{2}).}$ Now we observe that ${\displaystyle \frac{1}{T}\to 0}$ as ${\displaystyle T\to \infty }$, so the integral converges, and its value is ${\displaystyle 1 \text{/} 2}$.