Answer
\[\frac{1}{4}\]
Work Step by Step
\[\begin{align}
& \int_{0}^{\infty }{\int_{0}^{\infty }{xy{{e}^{-\left( {{x}^{2}}+{{y}^{2}} \right)}}dx}dy} \\
& \int_{0}^{\infty }{y\left[ \int_{0}^{\infty }{x{{e}^{-\left( {{x}^{2}}+{{y}^{2}} \right)}}dx} \right]dy} \\
& \text{Solve }\int_{0}^{\infty }{x{{e}^{-\left( {{x}^{2}}+{{y}^{2}} \right)}}dx}\text{ using the definition of improper integrals} \\
& \int_{0}^{\infty }{x{{e}^{-\left( {{x}^{2}}+{{y}^{2}} \right)}}dx}=\underset{b\to \infty }{\mathop{\lim }}\,\int_{0}^{b}{x{{e}^{-\left( {{x}^{2}}+{{y}^{2}} \right)}}dx} \\
& \underset{b\to \infty }{\mathop{\lim }}\,\int_{0}^{b}{x{{e}^{-\left( {{x}^{2}}+{{y}^{2}} \right)}}dx}=-\frac{1}{2}\underset{b\to \infty }{\mathop{\lim }}\,\int_{0}^{b}{\left( -2x \right){{e}^{-\left( {{x}^{2}}+{{y}^{2}} \right)}}dx} \\
& =-\frac{1}{2}\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-\left( {{x}^{2}}+{{y}^{2}} \right)}} \right]_{0}^{b} \\
& =-\frac{1}{2}\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-\left( {{b}^{2}}+{{y}^{2}} \right)}}-{{e}^{-\left( {{0}^{2}}+{{y}^{2}} \right)}} \right] \\
& =-\frac{1}{2}\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-\left( {{b}^{2}}+{{y}^{2}} \right)}}-{{e}^{-{{y}^{2}}}} \right] \\
& \text{Evaluate the limit when }b\to \infty \\
& =-\frac{1}{2}\left[ {{e}^{-\infty }}-{{e}^{-{{y}^{2}}}} \right] \\
& =\frac{1}{2}{{e}^{-{{y}^{2}}}} \\
& \text{Then,} \\
& \int_{0}^{\infty }{y\left[ \int_{0}^{\infty }{x{{e}^{-\left( {{x}^{2}}+{{y}^{2}} \right)}}dx} \right]dy}=\int_{0}^{\infty }{y\left[ \frac{1}{2}{{e}^{-{{y}^{2}}}} \right]dy} \\
& =\frac{1}{2}\int_{0}^{\infty }{y{{e}^{-{{y}^{2}}}}dy} \\
& =-\frac{1}{4}\int_{0}^{\infty }{\left( -2y \right){{e}^{-{{y}^{2}}}}dy} \\
& =-\frac{1}{4}\left[ {{e}^{-{{y}^{2}}}} \right]_{0}^{\infty } \\
& =-\frac{1}{4}\left[ {{e}^{-{{\left( \infty \right)}^{2}}}}-{{e}^{0}} \right] \\
& =-\frac{1}{4}\left[ {{e}^{-\infty }}-{{e}^{0}} \right] \\
& =-\frac{1}{4}\left[ 0-1 \right] \\
& =\frac{1}{4} \\
\end{align}\]
