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How is integral calculated?

$$\int_ 0^\infty e^{-x^2}~dx $$

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    $\begingroup$ You can't do it in terms of a finite number of elementary functions, no matter what the method. $\endgroup$ – DonAntonio Apr 9 '13 at 18:04
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    $\begingroup$ If the integral is from $0$ to $\infty$, then the solution is easy though. $\endgroup$ – Lord Soth Apr 9 '13 at 18:05
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    $\begingroup$ This has been asked many times before, e.g. this. $\endgroup$ – J. M. is a poor mathematician Apr 9 '13 at 18:09
  • $\begingroup$ Question edited (added 0 to $\infty$) $\endgroup$ – TheNotMe Apr 9 '13 at 18:13
  • $\begingroup$ Make the change of variables $t^2=u$ and then use gamma function. $\endgroup$ – Mhenni Benghorbal Apr 9 '13 at 18:34
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You can, however, write it as an infinite series and integrate term by term.

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The standard way to do this integral is to evaluate its square. That is, let

$$I = \int_0^{\infty} dx \: e^{-x^2}$$

Then

$$I^2 = \int_0^{\infty} dx \: e^{-x^2} \int_0^{\infty} dy \: e^{-y^2} = \int_0^{\infty} dx \:\int_0^{\infty} dy \: e^{-(x^2+y^2)}$$

Now the trick is to convert to polar coordinates: $dx dy = r dr d\theta$, $r \in [0,\infty)$, $\theta \in [0, 2 \pi)$:

$$I^2 = \int_0^{\infty} dr \,r\, e^{-r^2} \int_0^{2 \pi} d\theta = 2 \pi \frac{1}{2} = \pi$$

Therefore

$$I = \sqrt{\pi}$$

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Neither. The function has no antiderivative that can be expressed in terms of elementary functions, so both those methods fall down.

If you wish to evaluate the definite integral over a particular interval, though, you've got a shot.

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See Gaussian integral and Error function. It is $\sqrt \pi $ on $(- \infty, +\infty) $ . Indefinite could not be expressed in elementary functions.

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