How do I solve $\int _1^{\infty} e^{-x^2} dx$

Question

How do I solve

$$\int _1^{\infty} e^{-x^2} dx?$$

I solved like this.

let $I = \int _1^{\infty} e^{-x^2} dx$

then $I^2 = \int _1^{\infty} \int _1^{\infty} e^{-x^2 -y^2} dxdy$

If I change the cartetian coordinate to the polar coordinate.

$I^2 = \int _0^{\frac{\pi}{2}} \int _1^{\infty} e^{-r^2} rdrd\theta$

$= \int _0^{\frac{\pi}{2}} [\frac{e^{-r^2}}{-2}] _1^{\infty} d\theta$

$= \int _0^{\frac{\pi}{2}} \frac{1}{2e} d\theta$

$= \frac{1}{2e} \frac{\pi}{2}$

and then,

$ I = \frac{1}{2} \sqrt\frac{\pi}{e}$

I got The solution is $\frac{1}{2}\sqrt{ \frac{\pi }{ e} }$

but in the integral calculator the answer is $0.13940279...$.(https://www.integral-calculator.com/)

What did I do wrong?

Answer 1

Nope. If you consider $I^2$ you get the integral of $e^{-(x^2+y^2)}$ over the region $x,y\geq 1$, not the region $x^2+y^2\geq 1$.
Your integral simply equals $$ \frac{\sqrt{\pi}}{2}-\int_{0}^{1}e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}-\sum_{n\geq 0}\frac{(-1)^n}{n!(2n+1)}.$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{1}^{\infty}\expo{-x^{2}}\dd x & = {\root{\pi} \over 2} \pars{{2 \over \root{\pi}}\int_{1}^{\infty}\expo{-x^{2}}\dd x} \\[3mm] & = \bbox[#ffc,15px,border:1px groove navy]{{\root{\pi} \over 2}\,\mrm{erfc}\pars{1}}\ \approx 0.1394 \end{align}

See erfc Function in DLMF web page .