Chapter 7 - Section 7.8 - Improper Integrals - 7.8 Exercises - Page 550: 67

$\pi $

$$\eqalign{ & \int_0^\infty {\frac{1}{{\sqrt x \left( {1 + x} \right)}}} 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 \left( {1 + x} \right)}}} dx = \int_0^1 {\frac{1}{{\sqrt x \left( {1 + x} \right)}}} dx + \int_1^\infty {\frac{1}{{\sqrt x \left( {1 + x} \right)}}} dx,{\text{ }} \cr & \cr & {\text{*Calculating }}\int_0^1 {\frac{1}{{\sqrt x \left( {1 + x} \right)}}} dx \cr & {\text{Using the definition of improper integrals }} \cr & \int_0^1 {\frac{1}{{\sqrt x \left( {1 + x} \right)}}} dx = \mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\frac{1}{{\sqrt x \left( {1 + x} \right)}}} dx \cr & {\text{Rewrite the integrand}} \cr & = \mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\frac{2}{{2\sqrt x \left( {1 + {{\left( {\sqrt x } \right)}^2}} \right)}}} dx \cr & = 2\mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\frac{1}{{\underbrace {\left( {1 + {{\left( {\sqrt x } \right)}^2}} \right)}_{1 + {u^2}}}}} \overbrace {\left( {\frac{1}{{2\sqrt x }}} \right)dx}^{du} \cr & {\text{Integrating}} \cr & = 2\mathop {\lim }\limits_{a \to {0^ + }} \left[ {\arctan \left( {\sqrt x } \right)} \right]_a^1 \cr & = 2\mathop {\lim }\limits_{a \to {0^ + }} \left[ {\arctan \left( {\sqrt 1 } \right) - \arctan \left( {\sqrt a } \right)} \right] \cr & = 2\mathop {\lim }\limits_{a \to {0^ + }} \left[ {\frac{\pi }{4} - \arctan \left( {\sqrt a } \right)} \right] \cr & = 2\underbrace {\mathop {\lim }\limits_{a \to {0^ + }} \left[ {\frac{\pi }{4}} \right]}_{{\text{Tends to }}\frac{\pi }{4}} - 2\underbrace {\mathop {\lim }\limits_{a \to {0^ + }} \left[ {\arctan \left( {\sqrt a } \right)} \right]}_{{\text{Tends to 0}}} \cr & = 2\left( {\frac{\pi }{4}} \right) + 0 \cr & = \frac{\pi }{2} \cr & \cr & {\text{*Calculating }}\int_1^\infty {\frac{1}{{\sqrt x \left( {1 + x} \right)}}} dx \cr & {\text{Using the definition of improper integrals }} \cr & \int_1^\infty {\frac{1}{{\sqrt x \left( {1 + x} \right)}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{1}{{\sqrt x \left( {1 + x} \right)}}} dx \cr & = 2\mathop {\lim }\limits_{b \to \infty } \left[ {\arctan \left( {\sqrt x } \right)} \right]_1^b \cr & = 2\mathop {\lim }\limits_{b \to \infty } \left[ {\arctan \left( {\sqrt b } \right) - \arctan \left( {\sqrt 1 } \right)} \right] \cr & = 2\mathop {\lim }\limits_{b \to \infty } \left[ {\arctan \left( {\sqrt b } \right) - \frac{\pi }{4}} \right] \cr & = 2\underbrace {\mathop {\lim }\limits_{b \to \infty } \left[ {\arctan \left( {\sqrt b } \right)} \right]}_{{\text{Tends to }}\infty } - 2\underbrace {\mathop {\lim }\limits_{a \to {0^ + }} \left[ {\frac{\pi }{4}} \right]}_{{\text{Tends to }}\frac{\pi }{4}} \cr & = 2\underbrace {\mathop {\lim }\limits_{b \to \infty } \left[ {\arctan \left( {\sqrt b } \right)} \right]}_{{\text{Tends to }}\frac{\pi }{2}} - 2\underbrace {\mathop {\lim }\limits_{a \to {0^ + }} \left[ {\frac{\pi }{4}} \right]}_{{\text{Tends to }}\frac{\pi }{4}} \cr & = 2\left( {\frac{\pi }{2}} \right) - 2\left( {\frac{\pi }{4}} \right) \cr & = \frac{\pi }{2} \cr & \cr & {\text{Therefore}}{\text{,}} \cr & \int_0^\infty {\frac{1}{{\sqrt x \left( {1 + x} \right)}}} dx = \underbrace {\int_0^1 {\frac{1}{{\sqrt x \left( {1 + x} \right)}}} dx}_{{\text{tends to }}\frac{\pi }{2}} + \underbrace {\int_1^\infty {\frac{1}{{\sqrt x \left( {1 + x} \right)}}} dx}_{{\text{tends to }}\frac{\pi }{2}},{\text{ }} \cr & \int_0^\infty {\frac{1}{{\sqrt x \left( {1 + x} \right)}}} dx = \frac{\pi }{2} + \frac{\pi }{2} \cr & \int_0^\infty {\frac{1}{{\sqrt x \left( {1 + x} \right)}}} dx = \pi \cr} $$