Answer
$0$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^\infty {x{e^{ - {x^2}}}} dx \cr
& {\text{Using the definition of improper integrals }} \cr
& \int_{ - \infty }^\infty {x{e^{ - {x^2}}}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_b^0 {x{e^{ - {x^2}}}} dx + \mathop {\lim }\limits_{b \to \infty } \int_0^b {x{e^{ - {x^2}}}} dx \cr
& {\text{Integrating}} \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{a \to \infty } \left[ {{e^{ - {x^2}}}} \right]_a^0 - \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - {x^2}}}} \right]_0^b \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{a \to \infty } \left[ {{e^{ - {{\left( 0 \right)}^2}}} - {e^{ - {{\left( a \right)}^2}}}} \right] - \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - {{\left( b \right)}^2}}} - {e^{ - {{\left( 0 \right)}^2}}}} \right] \cr
& {\text{Simplifying}} \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {1 - {e^{ - {{\left( a \right)}^2}}}} \right] - \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - {{\left( b \right)}^2}}} - 1} \right] \cr
& {\text{Evaluate the limits when }}a \to - \infty {\text{ and }}b \to \infty \cr
& = - \frac{1}{2}\left[ {1 - {e^{ - {{\left( { - \infty } \right)}^2}}}} \right] - \frac{1}{2}\left[ {{e^{ - {{\left( \infty \right)}^2}}} - 1} \right] \cr
& = - \frac{1}{2}\left[ {1 - 0} \right] - \frac{1}{2}\left[ {0 - 1} \right] \cr
& = - \frac{1}{2} + \frac{1}{2} \cr
& = 0 \cr
& {\text{Therefore}}{\text{, the integral converges.}} \cr} $$
