Answer
$$\frac{1}{2}$$
Work Step by Step
$$\eqalign{
& \int_0^\infty {{e^{ - x}}\cos xdx} \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ then}} \cr
& \int_0^\infty {{e^{ - x}}\cos xdx} = \mathop {\lim }\limits_{b \to \infty } \int_0^b {{e^{ - x}}\cos xdx} \cr
& {\text{Integrate by tables}} \cr
& \int {{e^{au}}\cos budu} = \frac{{{e^{au}}}}{{{a^2} + {b^2}}}\left( {a\cos bu + b\sin bu} \right) + C \cr
& \int {{e^{ - x}}\cos x} dx = \frac{{{e^{ - x}}}}{{{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}}}\left( { - \cos x + \sin x} \right) + C \cr
& \int {{e^{ - x}}\cos x} dx = \frac{{{e^{ - x}}}}{2}\left( {\sin x - \cos x} \right) + C \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{b \to \infty } \int_0^b {{e^{ - x}}\cos xdx} = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{e^{ - x}}}}{2}\left( {\sin x - \cos x} \right)} \right]_0^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{e^{ - b}}}}{2}\left( {\sin b - \cos b} \right) - \frac{{{e^{ - 0}}}}{2}\left( {\sin 0 - \cos 0} \right)} \right] \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{e^{ - b}}}}{2}\left( {\sin b - \cos b} \right) + \frac{1}{2}} \right] \cr
& {\text{Evaluate }}b \to \infty \cr
& = \frac{{{e^{ - \infty }}}}{2}\left( {\sin \infty - \cos \infty } \right) + \frac{1}{2} \cr
& = 0 + \frac{1}{2} \cr
& = \frac{1}{2} \cr} $$
