Answer
$$a = \frac{\pi }{2}$$
Work Step by Step
$$\eqalign{
& \int_0^{ + \infty } {\frac{1}{{{x^2} + {a^2}}}} = 1 \cr
& {\text{Using the definition of improper integrals}} \cr
& \mathop {\lim }\limits_{b \to + \infty } \int_0^b {\frac{1}{{{x^2} + {a^2}}}} = 1 \cr
& {\text{Integrate}} \cr
& \mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right)} \right]_0^b = 1 \cr
& \frac{1}{a}\mathop {\lim }\limits_{b \to + \infty } \left[ {{{\tan }^{ - 1}}\left( {\frac{b}{a}} \right) - {{\tan }^{ - 1}}\left( {\frac{0}{a}} \right)} \right] = 1 \cr
& \frac{1}{a}\mathop {\lim }\limits_{b \to + \infty } \left[ {{{\tan }^{ - 1}}\left( {\frac{b}{a}} \right)} \right] = 1 \cr
& {\text{Evaluate the limit}} \cr
& \frac{1}{a}\left[ {{{\tan }^{ - 1}}\left( {\frac{{ + \infty }}{a}} \right)} \right] = 1 \cr
& \frac{1}{a}\left( {\frac{\pi }{2}} \right) = 1 \cr
& {\text{Solve for }}a \cr
& a = \frac{\pi }{2} \cr} $$
