1 - {x^2} < {e^{ - {x^2}}} < \frac{1}{{1 + {x^2}}} \\   {\left( {1 - {x^2}} \right)^n} < {e^{ - n{x^2}}} < \frac{1}{{{{\left( {1 + {x^2}} \right)}^n}}} \\   \int\limits_0^1 {{{\left( {1 - {x^2}} \right)}^n}dx}  < \int\limits_0^\infty  {{e^{ - n{x^2}}}dx}  < \int\limits_0^\infty  {\frac{{dx}}{{{{\left( {1 + {x^2}} \right)}^n}}}}  \\   \frac{{\left( {2n} \right)!!}}{{\left( {2n + 1} \right)!!}} < \frac{1}{{\sqrt n }}\int\limits_0^\infty  {{e^{ - {x^2}}}dx}  < \frac{{\left( {2n - 3} \right)!!}}{{\left( {2n - 2} \right)!!}}\frac{\pi }{2} \\   \frac{n}{{2n + 1}}\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}\frac{1}{{\sqrt n }} < \int\limits_0^\infty  {{e^{ - {x^2}}}dx}  < \sqrt n \frac{{2n - 2}}{{2n - 3}}\frac{{\left( {2n - 1} \right)!!}}{{\left( {2n} \right)!!}}\frac{\pi }{2} \\   \frac{1}{2}\sqrt \pi   < \int\limits_0^\infty  {{e^{ - {x^2}}}dx}  < \frac{1}{{\sqrt \pi  }}\frac{\pi }{2} \\   \frac{\sqrt \pi}{2}   < \int\limits_0^\infty  {{e^{ - {x^2}}}dx}  < \frac{{\sqrt \pi  }}{2} \Rightarrow \int\limits_0^\infty  {{e^{ - {x^2}}}dx}  = \frac{{\sqrt \pi  }}{2} \\
{}