Answer
Diverges
Work Step by Step
$$\eqalign{
& \int_0^\infty {\frac{1}{{\sqrt x }}} dx \cr
& {\text{Expressing the integral as a sum of improper integrals type 1}} \cr
& {\text{and type 2 as follows:}} \cr
& \int_0^\infty {\frac{1}{{\sqrt x }}} dx = \int_0^1 {\frac{1}{{\sqrt x }}} dx + \int_1^\infty {\frac{1}{{\sqrt x }}} dx,{\text{ }}0 > 1 \cr
& \cr
& {\text{*Calculating }}\int_0^1 {\frac{1}{{\sqrt x }}} dx \cr
& {\text{Using the definition of improper integrals }} \cr
& \int_0^1 {\frac{1}{{\sqrt x }}} dx = \mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\frac{1}{{\sqrt x }}} dx \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left[ {2\sqrt x } \right]_a^1 \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left[ {2\sqrt 1 - 2\sqrt a } \right] \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left[ {2 - 2\sqrt a } \right] \cr
& = \underbrace {\mathop {\lim }\limits_{a \to {0^ + }} \left[ 2 \right]}_{{\text{Tends to 2}}} - \underbrace {\mathop {\lim }\limits_{a \to {0^ + }} \left[ {2\sqrt a } \right]}_{{\text{Tends to 0}}} \cr
& = 2 + 0 \cr
& = 2 \cr
& \cr
& {\text{*Calculating }}\int_1^\infty {\frac{1}{{\sqrt x }}} dx \cr
& {\text{Using the definition of improper integrals }} \cr
& \int_1^\infty {\frac{1}{{\sqrt x }}} dx = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{1}{{\sqrt x }}} dx \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {2\sqrt x } \right]_1^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {2\sqrt b - 2\sqrt 1 } \right] \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {2\sqrt b - 2} \right] \cr
& = \underbrace {\mathop {\lim }\limits_{a \to {0^ + }} \left[ {2\sqrt b } \right]}_{{\text{Tends to }}\infty } - \underbrace {\mathop {\lim }\limits_{a \to {0^ + }} \left[ 2 \right]}_{{\text{Tends to 2}}} \cr
& {\text{Diverges}} \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \int_0^\infty {\frac{1}{{\sqrt x }}} dx = \underbrace {\int_0^1 {\frac{1}{{\sqrt x }}} dx}_{{\text{tends to 2}}} + \underbrace {\int_1^\infty {\frac{1}{{\sqrt x }}} dx}_{{\text{diverges}}},{\text{ }}0 > 1 \cr
& \int_0^\infty {\frac{1}{{\sqrt x }}} dx:{\text{ Diverges}} \cr} $$
