Chapter 8: Techniques of Integration - Practice Exercises - Page 519: 62

$$\pi $$

$$\eqalign{ & \int_{ - \infty }^\infty {\frac{{4dx}}{{{x^2} + 16}}} \cr & {\text{By the definition of the improper integrals, we have }} \cr & \int_{ - \infty }^\infty {\frac{{4dx}}{{{x^2} + 16}}} = \mathop {\lim }\limits_{a \to - \infty } \int_a^c {\frac{{4dx}}{{{x^2} + 16}}} + \mathop {\lim }\limits_{b \to \infty } \int_c^b {\frac{{4dx}}{{{x^2} + 16}}} \cr & {\text{taking }}c = 0 \cr & \int_{ - \infty }^\infty {\frac{{4dx}}{{{x^2} + 16}}} = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{{4dx}}{{{x^2} + 16}}} + \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{{4dx}}{{{x^2} + 16}}} \cr & \cr & {\text{Integrating }}\int {\frac{{4dx}}{{{x^2} + 16}}{\text{ by tables}}} \cr & \int {\frac{{4dx}}{{{x^2} + 16}} = 4\int {\frac{1}{{{x^2} + {{\left( 4 \right)}^2}}}} dx} \cr & {\text{Use the formula }}\int {\frac{{dx}}{{{x^2} + {a^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + C \cr & \,\,4\int {\frac{1}{{{x^2} + {{\left( 4 \right)}^2}}}} dx = 4\left( {\frac{1}{4}{{\tan }^{ - 1}}\left( {\frac{x}{4}} \right)} \right) + C \cr & \,\,\,\,\,\int {\frac{1}{{{x^2} + {{\left( 4 \right)}^2}}}} dx = {\tan ^{ - 1}}\left( {\frac{x}{4}} \right) + C \cr & \cr & {\text{Then}}{\text{,}} \cr & \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{{4dx}}{{{x^2} + 16}}} = \mathop {\lim }\limits_{a \to - \infty } \left( {{{\tan }^{ - 1}}\left( {\frac{x}{4}} \right)} \right)_a^0 \cr & = \mathop {\lim }\limits_{a \to - \infty } \left( {{{\tan }^{ - 1}}\left( {\frac{{\left( 0 \right)}}{4}} \right) - {{\tan }^{ - 1}}\left( {\frac{a}{4}} \right)} \right) \cr & = - {\tan ^{ - 1}}\left( {\frac{a}{4}} \right) \cr & {\text{Evaluate the limit when }}a \to - \infty \cr & = - {\tan ^{ - 1}}\left( {\frac{{\left( { - \infty } \right)}}{4}} \right) \cr & = - \left( { - \frac{\pi }{2}} \right) \cr & = \frac{\pi }{2} \cr & \cr & \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{{4dx}}{{{x^2} + 16}}} = \mathop {\lim }\limits_{b \to \infty } \left( {{{\tan }^{ - 1}}\left( {\frac{x}{4}} \right)} \right)_0^b \cr & = \mathop {\lim }\limits_{b \to \infty } \left( {{{\tan }^{ - 1}}\left( {\frac{b}{4}} \right) - {{\tan }^{ - 1}}\left( {\frac{0}{4}} \right)} \right) \cr & = \mathop {\lim }\limits_{b \to \infty } \left( {{{\tan }^{ - 1}}\left( {\frac{b}{4}} \right) - \left( 0 \right)} \right) \cr & = \mathop {\lim }\limits_{b \to \infty } \left( {{{\tan }^{ - 1}}\left( {\frac{b}{4}} \right)} \right) \cr & {\text{Evaluate the limit when }}b \to \infty \cr & = {\tan ^{ - 1}}\left( {\frac{\infty }{4}} \right) \cr & = \frac{\pi }{2} \cr & \cr & {\text{Since both of these integrals converge}},{\text{ the improper integral is}} \cr & \int_{ - \infty }^\infty {\frac{{4dx}}{{{x^2} + 16}}} = \frac{\pi }{2} + \frac{\pi }{2} \cr & \int_{ - \infty }^\infty {\frac{{4dx}}{{{x^2} + 16}}} = \pi \cr} $$