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Show that $\int_{-\infty}^{\infty} \exp(-x^2)\, dx = \sqrt{\pi}$.

I believe it is necessary to use the Fubini's Theorem and the Change of Variables Theorem. I guess the variable shift function can be $g: (s,v)\times (s,2\pi) \rightarrow \mathbb{R^2}$, such that $g(r,\theta)=(r\cos\theta,r\sin\theta)$.

I've seen this question here, but none explains why g fits the assumptions of the theorem. As, for example, why $g$ is a diffeomorphism of class $C^1$, or why $g((s,v)\times (s,2\pi))$ is compact and $J$-measurable.

PS.: I know that Change of Variables Theorem It has other versions with different hypotheses, but these are the book I follow.

Ivo Terek
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Let DC
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    Several proofs. – Lucian Jun 21 '15 at 21:44
  • @Lucian I dont know that problem It was one of the "classics". Thank you. – Let DC Jun 21 '15 at 21:48
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    Let the domain of your change of variable function $g$ be a compact interval of the form $[r,R]\times [\epsilon,2\pi]$, where $\epsilon,r$ are small and positive, and $R\to\infty$. The restriction of $g$ to that compact set is injective, and no problems arise. Actually, if you use the domain $[0,R]\times[0,2\pi]$ the non-injectivity only happens in a Jordan null set, and can thus be ignored. – Jyrki Lahtonen Jun 21 '15 at 21:51
  • @LetDC, I'm a bit busy now, sorry. I'm glad that good people already showed up. If I understood the question right, the problem is not with the computation, but why the theorem applies. Jyrki's comment already explains it.. (it could be converted into an answer, IMO) – Ivo Terek Jun 21 '15 at 21:56
  • @IvoTerek Yes, the other comments helped me. But thank you, and good luck in their occupations. – Let DC Jun 21 '15 at 22:40

1 Answers1

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Identify it as the square root of the product of 2 integrals which can be combined into a double integral of $exp[-(x^2+y^2)]$. Then switch to polar coordinates to evaluate. That's how Gauss did it (I think).

BruceZ
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