Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^\infty {x{e^{ - {x^2}}}dx} \cr
& {\text{definition of improper integral}} \cr
& \int_{ - \infty }^\infty {x{e^{ - {x^2}}}dx} = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {x{e^{ - {x^2}}}dx} + \mathop {\lim }\limits_{b \to \infty } \int_0^b {x{e^{ - {x^2}}}dx} \cr
& {\text{evaluate the integrals}} \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left. {\left( {{e^{ - {x^2}}}} \right)} \right|_a^0 - \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left. {\left( {{e^{ - {x^2}}}} \right)} \right|_0^b \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left( {{e^{ - {{\left( 0 \right)}^2}}} - {e^{ - {a^2}}}} \right) - \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left( {{e^{ - {b^2}}} - {e^{ - {{\left( 0 \right)}^2}}}} \right) \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left( {1 - {e^{ - {a^2}}}} \right) - \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left( {{e^{ - {b^2}}} - 1} \right) \cr
& {\text{evaluate the limits}} \cr
& = - \frac{1}{2}\left( {1 - {e^{ - {{\left( { - \infty } \right)}^2}}}} \right) - \frac{1}{2}\left( {{e^{ - {{\left( { - \infty } \right)}^2}}} - 1} \right) \cr
& = - \frac{1}{2}\left( {1 - 0} \right) - \frac{1}{2}\left( {0 - 1} \right) \cr
& = - \frac{1}{2} + \frac{1}{2} \cr
& = 0 \cr} $$
