## Thomas' Calculus 13th Edition

$${\text{The improper integral converges }}$$
\eqalign{ & \int_{ - \infty }^\infty {\frac{{2dx}}{{{e^x} + {e^{ - x}}}}} \cr & {\text{By the definition of the improper integrals, we have }} \cr & \int_{ - \infty }^\infty {\frac{{2dx}}{{{e^x} + {e^{ - x}}}}} = \mathop {\lim }\limits_{a \to - \infty } \int_a^c {\frac{{2dx}}{{{e^x} + {e^{ - x}}}}} + \mathop {\lim }\limits_{b \to \infty } \int_c^b {\frac{{2dx}}{{{e^x} + {e^{ - x}}}}} \cr & {\text{taking }}c = 0 \cr & \int_{ - \infty }^\infty {\frac{{2dx}}{{{e^x} + {e^{ - x}}}}} = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{{2dx}}{{{e^x} + {e^{ - x}}}}} + \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{{2dx}}{{{e^x} + {e^{ - x}}}}} \cr & \cr & {\text{Integrating }}\int {\frac{{2dx}}{{{e^x} + {e^{ - x}}}}} \cr & \,\,\,{\text{multiply the numerator and denominator by }}{e^x} \cr & = \int {\frac{{2{e^x}dx}}{{{e^{2x}} + 1}}} \cr & \,\,\,\,{\text{Let }}u = {e^x},\,\,\,\,du = {e^x}{\text{ }} \cr & \,\,\,\int {\frac{{2{e^x}dx}}{{{e^{2x}} + 1}}} = \int {\frac{{2du}}{{{u^2} + 1}}} \cr & = 2{\tan ^{ - 1}}u + C \cr & = 2{\tan ^{ - 1}}\left( {{e^x}} \right) + C \cr & \cr & {\text{Then}}{\text{,}} \cr & \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{{2dx}}{{{e^x} + {e^{ - x}}}}} = \mathop {\lim }\limits_{a \to - \infty } \left( {2{{\tan }^{ - 1}}\left( {{e^x}} \right)} \right)_a^0 \cr & = 2\mathop {\lim }\limits_{a \to - \infty } \left( {{{\tan }^{ - 1}}\left( {{e^0}} \right) - {{\tan }^{ - 1}}\left( {{e^a}} \right)} \right) \cr & = 2\mathop {\lim }\limits_{a \to - \infty } \left( {{{\tan }^{ - 1}}\left( 1 \right) - {{\tan }^{ - 1}}\left( {{e^a}} \right)} \right) \cr & {\text{Evaluate the limit when }}a \to - \infty \cr & = 2\left( {{{\tan }^{ - 1}}\left( 1 \right) - {{\tan }^{ - 1}}\left( {{e^{ - \infty }}} \right)} \right) \cr & = 2{\tan ^{ - 1}}\left( 1 \right) \cr & = \frac{\pi }{2} \cr & \cr & \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{{2dx}}{{{e^x} + {e^{ - x}}}}} = \mathop {\lim }\limits_{b \to \infty } \left( {2{{\tan }^{ - 1}}\left( {{e^x}} \right)} \right)_0^b \cr & = 2\mathop {\lim }\limits_{b \to \infty } \left( {{{\tan }^{ - 1}}\left( {{e^b}} \right) - {{\tan }^{ - 1}}\left( {{e^0}} \right)} \right) \cr & = 2\mathop {\lim }\limits_{b \to \infty } \left( {{{\tan }^{ - 1}}\left( {{e^b}} \right) - {{\tan }^{ - 1}}\left( 1 \right)} \right) \cr & {\text{Evaluate the limit when }}b \to \infty \cr & = 2\left( {{{\tan }^{ - 1}}\left( \infty \right) - {{\tan }^{ - 1}}\left( 1 \right)} \right) \cr & = 2\left( {\frac{\pi }{2} - \frac{\pi }{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{{2dx}}{{{e^x} + {e^{ - x}}}}} = \frac{\pi }{2} + \frac{\pi }{2} \cr & \int_{ - \infty }^\infty {\frac{{2dx}}{{{e^x} + {e^{ - x}}}}} = \pi \cr}