Answer
$\infty$; Divergent
Work Step by Step
Simplify the improper integral $\int_{0}^{\infty} \dfrac{1}{x^2} dx$
Let us consider that $I=\int_{0}^{\infty} \dfrac{1}{x^2} dx$
Integrate the integral by using limits.
$I=\lim\limits_{z \to 0^{+}}\int_{z}^1 \dfrac{1}{x^2} da+\lim\limits_{z \to \infty}\int_{1}^z\\=\lim\limits_{z \to 0^{+}}[-1+\dfrac{1}{z}]+\lim\limits_{z \to \infty}[-\dfrac{1}{z}+1]\\=\infty$
So, the given integral is divergent.
