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How can I integrate this by parts? It seems to become recursive. I'm familiar with the classical solution, and cannot use that here due to the constraints of this class.

Here's the integral (to infinity). https://i.sstatic.net/bKnDo.png

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

I'm having trouble finding the indefinite integral first.

Thanks!

Vincenzo Tibullo
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2567655222
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  • There is no closed form for the indefinite integral. Are you allowed to use gamma function? – desos Dec 14 '14 at 19:22
  • Unfortunately I'm not :/

    Here's what a friend sent me. It looks like he may be wrong, and I'm not certain of what he's doing either.

    http://i.imgur.com/mgvR0Pv.jpg

     – 2567655222 Dec 14 '14 at 19:24
  • Looks like he has a sign error somewhere. – desos Dec 14 '14 at 19:32
  • Your friend apparently integrated by parts once and then back. The terms didn't cancel so I suppose it's a sign error. – andrepd Dec 14 '14 at 19:34

2 Answers2

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The antiderivative of $e^{-x^2}$ function is the error function. It is not an elementary function, and because of Liouville's theorem we are not able to express the error function in term of elementary functions. Liouville's theorem charaterise the class of functions for which we could find an elementary primitive function. Risch algorithm is a tool inspired be Liouville's theorem.

What you are looking for is the Gaussian integral. You could find more then one proof at wikipedia, or at ProofWiki.

user153012
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There is is no elementary function for the indefinite gaussian integral. The best you can do is represent it through a Taylor series, and then compute it numerically with the desired precision.

Yet calculating the definite integral is straightforward (wikipedia has several proofs).

$$\int_{-\infty}^{+\infty}e^{-x^2}dx=\sqrt\pi$$

As the integrand is an even function:

$$\int_{0}^{+\infty}e^{-x^2}dx=\frac{\sqrt\pi}{2}$$

CherryDT
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andrepd
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