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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?

Sukhyun Park
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    The answer isn't $\sqrt{\pi/e}$. Unless you share your method with us, we cannot tell where your error lies. – Angina Seng Jan 01 '20 at 04:07
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    Until you [edit] the question to show us just how you got your answer we can't possibly explain why it might be wrong. – Ethan Bolker Jan 01 '20 at 04:07
  • I edited context. and I solved $\frac{1}{2} \sqrt{\frac{\pi}{e}}$ – Sukhyun Park Jan 01 '20 at 05:01
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    Are you bound to not use the values of the error function? – an4s Jan 01 '20 at 05:04
  • I want an exact solution, not an approximate one. – Sukhyun Park Jan 01 '20 at 05:35
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    Your mistake lies in where you change the Cartesian coordinate to the polar coordinate. Try to plot the two areas and find the differences. Admittedly, your method is right if you integrate from $0$ to $\infty$. – Drinzjeng Triang Jan 01 '20 at 05:39
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    The conditions $x \geq 1$ and $y \geq 1$ are not equivalent to $x^2 + y^2 \geq 1$ (they only imply that, but not the other way around, e.g., $x = \sqrt{2/3}$ and $y = \sqrt{2/3}$). There is no elementary exact solution. The number just is what it is. You're hoping for too much if you expect the integral should have an elementary formula. Since $\int_0^\infty e^{-x^2},dx = \sqrt{\pi}/2$ you could rewrite your integral as $\sqrt{\pi}/2 - \int_0^1 e^{-x^2},dx$, but this is not really "progress" since there is no elementary formula for $\int_0^1 e^{-x^2},dx$. – KCd Jan 01 '20 at 05:50
  • Oh, I see. The bound is wrong in two dimensions ... And it's a pity that there's no exact solution. – Sukhyun Park Jan 01 '20 at 06:05
  • @SukhyunPark Jacks' expression will be exact, because complex analytic functions will have Taylor series expansions which converge (and can be integrated) everywhere in the complex plane. – mathreadler Jan 01 '20 at 20:10

2 Answers2

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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 $$ \frac{\sqrt{\pi}}{2}-\int_{0}^{1}e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}-\sum_{n\geq 0}\frac{(-1)^n}{n!(2n+1)}.$$

Jack D'Aurizio
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  • Hmm, how to get the sum? Is it some well known thing? Gamma function? – mathreadler Jan 01 '20 at 19:15
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    @mathreadler: just consider the Maclaurin series of $e^{-x^2}$ and perform a termwise integration over $(0,1)$. $$\int_{0}^{+\infty}e^{-x^2},dx = \frac{\sqrt{\pi}}{2}$$ is fairly well-known and related to $\Gamma(1/2)=\sqrt{\pi}$, if you like. – Jack D'Aurizio Jan 01 '20 at 19:17
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 .

Felix Marin
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