Determine whether $\int_{1}^{\infty }e^{-x^{2}}dx$ converges or diverges.

Solution By definition, $\int_{1}^{\infty }e^{-x^{2}}dx=\lim\limits_{b\rightarrow \infty }\int_{1}^{b}e^{-x^{2}}dx$ converges if the limit exists and equals a real number. Since $e^{-x^{2}}$ has no antiderivative, we use the Comparison Test for Improper Integrals. We proceed as follows: For $x \ge 1,$ $$\begin{eqnarray*} \begin{array}{rcl@{\qquad}l} x^{2} &\geq & x \\[3pt] -x^{2} &\leq & -x \\[3pt] 0 &\lt&e^{-x^{2}}\leq e^{-x} & {\color{#0066A7}{\hbox{ Since }{e>1}\hbox{, if }{a\leq b}\hbox{, then }{e}^{a}{\leq e}^{b}.}} \end{array} \end{eqnarray*}$$

Figure 28 illustrates this.

Based on the Comparison Test, if $\int_{1}^{\infty }e^{-x}dx$ converges, so does $\int_{1}^{\infty }e^{-x^{2}}dx.$ We investigate $\int_{1}^{\infty }e^{-x}dx.$ $$\begin{eqnarray*} \int_{1}^{\infty }e^{-x}dx :\quad\!\!\!\! \lim\limits_{b\rightarrow \infty }\int_{1}^{b}e^{-x}dx &=& \lim\limits_{b\rightarrow \infty }\!\big[-e^{-x}\big] _{1}^{b}=\lim\limits_{b\rightarrow \infty }[-e^{-b}+e^{-1}] =\lim\limits_{b\rightarrow \infty }\left[ \dfrac{1}{e}-\dfrac{1}{e^{b}} \!\right] \\[5pt] &=& \lim\limits_{b\rightarrow \infty }\dfrac{1}{e}-\lim\limits_{b \rightarrow \infty }\dfrac{1}{e^{b}}=\dfrac{1}{e} \end{eqnarray*}$$

Since $\int_{1}^{\infty }e^{-x}dx$ converges, we conclude that $\int_{1}^{\infty}e^{-x^{2}}dx$ converges.