Answer
$$\frac{1}{a}$$
Work Step by Step
$$\eqalign{
& \int_0^\infty {{e^{ - ax}}dx} ,{\text{ }}a > 0 \cr
& {\text{definition of improper integral}} \cr
& \int_0^\infty {{e^{ - ax}}dx} = \mathop {\lim }\limits_{b \to \infty } \int_0^b {{e^{ - ax}}dx} \cr
& {\text{evaluate the integral}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left. {\left( { - \frac{1}{a}{e^{ - ax}}} \right)} \right|_0^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( { - \frac{1}{a}{e^{ - 2b}} + \frac{1}{a}{e^0}} \right) \cr
& {\text{simplify}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( { - \frac{1}{a}{e^{ - 2b}} + \frac{1}{a}} \right) \cr
& {\text{evaluate the limit}} \cr
& = - \frac{1}{a}{e^{ - 2\left( \infty \right)}} + \frac{1}{a} \cr
& = 0 + \frac{1}{a} \cr
& = \frac{1}{a} \cr} $$
