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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

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Please use LaTeX. Is this integral of $\int_0^{\infty}e^{-x^2}\,dx$ (\int_0^{\infty}e^{-x^2}dx), $\int_0^{\infty}(e-x^2)\,dx$ (\int_0^{\infty}(e-x^2)dx$, or what? –  Arturo Magidin Jan 6 '12 at 4:32
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I assume you mean $\int_0^\infty e^{-x^2}\ dx$. Hint: compare to $\int_0^\infty e^{-x}\ dx$. –  Robert Israel Jan 6 '12 at 4:32
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@neemy: Yes the integral $\int_{0}^{\infty} e^{-x^2} \ dx$ converges and it's value is $\sqrt{\pi}$ –  user9413 Jan 6 '12 at 4:34
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There is a wonderful math markup language called LaTeX which this site supports and allows you to communicate math much more clearly and legibly. Learn it; it will help you here and elsewhere. I have edited your question with proper formatting. If the new version is not the question you were trying to ask, just let me know. –  Alex Becker Jan 6 '12 at 4:37
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@Chandrasekhar You probably meant to write the value as $\sqrt{\pi}/2$, not $\sqrt{\pi}$. –  Dilip Sarwate Jan 6 '12 at 13:32

5 Answers 5

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 $x^2 > x$ when $x > 1$. Therefore, $e^{-x^2} < e^{-x}$ for $x > 1$ and it follows that

$$\int_1^\infty e^{-x^2} \, dx < \int_1^\infty e^{-x} \, dx$$ $$= \lim_{x \to \infty} -e^{-x} + e^{-1} = 1/e < \infty$$

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It does, and you can also compute its value:

$\int_{[0, \infty) \times[0, \infty)} e^{-(x^2+y^2)} dx dy = \int_{[0, \infty)}\big(\int_{[0, \infty)} e^{-(x^2+y^2)} dy \big)dx = \int_{[0, \infty)} e ^ {-x^2}\big(\int_{[0, \infty)} e^{-y^2} dy \big)dx =$ $= \int_{[0, \infty)} e ^ {-x^2}dx \ \ \int_{[0, \infty)} e^{-y^2} dy = \big( \int_{[0, \infty)} e ^ {-x^2}dx \big)^2$

Then we can use polar coordinates:

$\int_{[0, \infty) \times[0, \infty)} e^{-(x^2+y^2)} dx dy = \int_0^{\infty} \big(\int_0^{\frac\pi2}re^{-r^2}d\theta)dr = \frac\pi2\int_0^{\infty}re^{-r^2}dr = \frac\pi2 \frac12 = \frac\pi4$

Therefore:

$\int_{[0, \infty)} e ^ {-x^2}dx = \big(\frac\pi4\big)^{\frac12}$

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I'm assuming you're asking about the convergence of $\int_0^\infty e^{-x^2}dx$. The easiest way that I can think of to prove this is to note that $e^{-x^2}$ is continuous and bounded, and hence integrable, on the interval $[0,1]$, and that on the remaining unbounded interval $[1,\infty)$ it is a function everywhere bounded in absolute value by a function $e^{-x}$ that is integrable on that interval (seeing that $\int_1^\infty e^{-x}dx$ converges is a simple calculation, since $e^{-x}$ has an easy antiderivative). Thus, by the comparison test, $\int_0^\infty e^{-x^2}dx$ converges. I think this is the idea you're referring to in your question.

$\int_0^\infty e^{-x^2}dx$ does not exist if $\int_0^\infty dx$ does, since the latter one most certainly does not converge (it equals $\lim_{x \to \infty} x = \infty$). While it is true that $e^{-x^2} \le 1$ on $[0,\infty)$, this fact isn't really helpful, because the comparison test only gives us information when the bounding function is itself integrable (otherwise you get absurdities like claiming that $\int_0^\infty e^{-x}dx$ diverges because $e^{-x} \le x $ and $\int_0^\infty x dx$ diverges.)

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Exercise 3-41 of Spivak's "Calculus on Manifolds" outlines how to prove that $\int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}$ using polar coordinates, a related double integral, and two useful families of regions over which to evaluate the double integral. The description is actually quite long. Ironically, he ends the problem with a quote attributed to Lord Kelvin (William Thomson), who was trying to convey the meaning of "mathematician" to a class: "A mathematician is one to whom "that" (i.e., $\int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}$) is as obvious as that twice two makes four is to you." A bit harsh, I daresay, on Lord Kelvin's part. The reality being conveyed by Spivak is that it's obvious once you've done the necessary work.

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Does Spivak quote the next part? "Liouville was a mathematician." I always thought that was a nice thing to say about someone. –  Dylan Moreland Jan 6 '12 at 6:54
    
@Dylan-Yes. Spivak includes that statement. –  Chris Leary Jan 6 '12 at 18:15

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.

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I didn't downvote, but I comment in order to explain that vote: you assume that a more complicated integral is defined in order to prove that $\int e^{-x^2} \, dx$ is defined. Of course, integrability of the two functions $e^{-x^2}$ and the Gaussian is equivalent by a simple substitution (part of that statement is what you showed). That's begging the question –  t.b. Mar 18 '12 at 23:50

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