Does the improper integral $\int_0^\infty e^{-x^2}dx$ converge? I want to show the convergence of the following improper integral $\int_0^\infty e^{-x^2}dx$.
I try to use comparison test for integrals
$x≥0$, $-x ≥0$, $-x^2≥0$ then $e^{-x^2}≤1$. So am ending with the fact that $\int_0^\infty e^{-x^2}dx$ converges if $\int_0^\infty dx$ converges but I don’t  appreciate this. Thanks
Write

$$\int_0^\infty e^{-x^2} \, dx = \int_0^1 e^{-x^2} \, dx + \int_1^\infty e^{-x^2} \, dx$$

The first integral on the right hand side is of a bounded function over a bounded interval, hence finite. For the second notice that we have $x^2 &gt; x$ when $x &gt; 1$. Therefore, $e^{-x^2} &lt; e^{-x}$ for $x &gt; 1$ and it follows that

$$\int_1^\infty e^{-x^2} \, dx &lt; \int_1^\infty e^{-x} \, dx$$
$$= \lim_{x \to \infty} -e^{-x} + e^{-1} = 1/e &lt; \infty$$
A simple answer as @neemy… Remember the function of Normal probability

$$N(\mu,\sigma)=\displaystyle{\int_{-\infty}^{+\infty}{\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\;dx}=\bf{1}}$$

For $\mu=0$ and $\sigma=1$
$$\displaystyle{\int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\;dx}=\bf{1}}$$

Introduce a change of variable with its respective Jacobian
$$x=\sqrt{2}\;u\\dx=\sqrt{2}\;du$$

and result
$$\displaystyle{\int_{-\infty}^{+\infty}{\frac{\sqrt{2}}{\sqrt{2\pi}}e^{-u^2}\;du}=\bf{1}} \Rightarrow \displaystyle{\int_{-\infty}^{+\infty}{e^{-u^2}\;du}=\bf{\sqrt{\pi}}}$$

The Normal is even function, therefore
$$\displaystyle{\int_0^{+\infty}{e^{-u^2}\;du}=\frac{\sqrt{\pi}}{2}}$$

P.D.: Excuse my English, please.