Chapter 8: Techniques of Integration - Practice Exercises - Page 519: 64

$${\text{The improper integral converges}}$$

$$\eqalign{ & \int_0^\infty {{e^{ - u}}\cos udu} \cr & \cr & {\text{By the definition of the improper integrals, we have }} \cr & \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx \cr & {\text{then}}{\text{,}} \cr & \int_0^\infty {{e^{ - u}}\cos udu} = \mathop {\lim }\limits_{b \to \infty } \int_0^b {{e^{ - u}}\cos udu} \cr & \cr & {\text{Integrate by parts method}} \cr & \,\,\,\,\,\,\,t = {e^{ - u}},\,\,\,\,dt = - {e^{ - u}}du \cr & \,\,\,\,\,\,\,dv = \cos udu,\,\,\,\,v = \sin u \cr & \,\,\,\,\,\int {{e^{ - u}}\cos udu} = {e^{ - u}}\sin u - \int {\sin u\left( { - {e^{ - u}}du} \right)} \cr & \,\,\,\,\,\int {{e^{ - u}}\cos udu} = {e^{ - u}}\sin u + \int {{e^{ - u}}\sin udu} \cr & \,\,\,\,\,{\text{Integrate by parts method again}} \cr & \,\,\,\,\,\,\,\,t = {e^{ - u}},\,\,\,\,dt = - {e^{ - u}}du \cr & \,\,\,\,\,\,\,dv = \sin udu,\,\,\,\,v = - \cos u \cr & \,\,\,\,\,\int {{e^{ - u}}\cos udu} = {e^{ - u}}\sin u + \left( { - {e^{ - u}}\cos u - \int {\left( { - \cos u} \right)\left( { - {e^{ - u}}} \right)} du} \right) \cr & \,\,\,\,\,\int {{e^{ - u}}\cos udu} = {e^{ - u}}\sin u - {e^{ - u}}\cos u - \int {{e^{ - u}}\cos u} du \cr & \,\,\,\,\,2\int {{e^{ - u}}\cos udu} = {e^{ - u}}\sin u - {e^{ - u}}\cos u \cr & \,\,\,\,\,\int {{e^{ - u}}\cos udu} = \frac{1}{2}{e^{ - u}}\sin u - \frac{1}{2}{e^{ - u}}\cos u \cr & \cr & {\text{Then}}{\text{,}} \cr & \mathop {\lim }\limits_{b \to \infty } \int_0^b {{e^{ - u}}\cos udu} = \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{2}{e^{ - u}}\sin u - \frac{1}{2}{e^{ - u}}\cos u} \right)_0^b \cr & = \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{2}{e^{ - b}}\sin b - \frac{1}{2}{e^{ - b}}\cos b} \right) - \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{2}{e^{ - 0}}\sin \left( 0 \right) - \frac{1}{2}{e^{ - \left( 0 \right)}}\cos \left( 0 \right)} \right) \cr & = \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{2}{e^{ - b}}\sin b - \frac{1}{2}{e^{ - b}}\cos b} \right) + \frac{1}{2} \cr & {\text{Use product property of limits}} \cr & = \mathop {\lim }\limits_{b \to \infty } \frac{1}{2}{e^{ - b}}\mathop {\lim }\limits_{b \to \infty } \left( {\sin b - \cos b} \right) + \frac{1}{2} \cr & \cr & {\text{Evaluate the limit when }}b \to \infty \cr & = \left( {\frac{1}{2}{e^{ - \infty }}} \right)\mathop {\lim }\limits_{b \to \infty } \left( {\sin b - \cos b} \right) + \frac{1}{2} \cr & = \frac{1}{2} \cr & \cr & {\text{Then}}{\text{, the improper integral converges.}} \cr} $$