Answer
$${\text{The integral converges for }}a < 1{\text{ to }}\frac{a}{{{a^2} + 1}}$$
Work Step by Step
$$\eqalign{
& \int_0^\infty {{e^{ax}}\cos x} dx \cr
& {\text{Use the definition of improper integrals}} \cr
& \int_0^\infty {{e^{ax}}\cos x} dx = \mathop {\lim }\limits_{b \to \infty } \int_0^b {{e^{ax}}\cos x} dx \cr
& {\text{Integrating by parts, we obtain}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{e^{ax}}}}{{{a^2} + 1}}\left( {a\cos x + \sin x} \right)} \right]_0^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{e^{ab}}}}{{{a^2} + 1}}\left( {a\cos b + \sin b} \right) - \frac{{{e^0}}}{{{a^2} + 1}}\left( {a\cos 0 + \sin 0} \right)} \right] \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{e^{ab}}}}{{{a^2} + 1}}\left( {a\cos b + \sin b} \right) - \frac{a}{{{a^2} + 1}}} \right] \cr
& {\text{*For }}a = 1 \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{e^b}}}{2}\left( {a\cos b + \sin b} \right) - \frac{1}{2}} \right] \cr
& = \infty {\text{ }}\left( {{\text{Diverges}}} \right) \cr
& {\text{*For }}a < 1 \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{e^{ - b}}}}{{{a^2} + 1}}\left( {a\cos b + \sin b} \right) - \frac{a}{{{a^2} + 1}}} \right] \cr
& = 0\left( {a\cos b + \sin b} \right) - \frac{a}{{{a^2} + 1}} \cr
& = - \frac{a}{{{a^2} + 1}} \cr
& {\text{*For }}a > 1 \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{e^b}}}{2}\left( {a\cos b + \sin b} \right) - \frac{1}{2}} \right] \cr
& = \infty {\text{ }}\left( {{\text{Diverges}}} \right) \cr
& {\text{The integral converges for }}a < 1,{\text{ then}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{e^{ - ab}}}}{{{a^2} + 1}}\left( {a\cos b + \sin b} \right) - \frac{{\left( { - a} \right)}}{{{a^2} + 1}}} \right] \cr
& = \frac{0}{{{a^2} + 1}}\left( {a\cos b + \sin b} \right) - \frac{{\left( { - a} \right)}}{{{a^2} + 1}} \cr
& = \frac{a}{{{a^2} + 1}} \cr
& {\text{The integral converges for }}a < 1{\text{ to }}\frac{a}{{{a^2} + 1}} \cr} $$
