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I have been working with normal pdfs for awhile. I haven't really thought about it until now, but what purpose does the constant $$\frac{1}{\sqrt{ 2\pi\sigma^2}}$$ serve? I know from my analysis class that integrals like $$\int e^{-x^2}dx $$ don't have elementary functions.

I saw on wikipedia that an arbitrary Gaussian definite integral from negative infinity to infinity evaluates to

$$\sqrt \frac {\pi}{a}$$

My guess is that this crazy constant out front has to do with ensuring that the normal pdf integrates to 1 over its domain. Is that right?

J.G.
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Will E.
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    Yes it is. There are different ways to prove this normalisation constant. But the deep reason why $\pi$ appears remains somewhat elusive. –  Mar 27 '19 at 12:39
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    There are many proofs of the normalisation. Hopefully, at least one will have an "oh, that's why" feel rather than just a "blah blah look what happens" feel. I hope that link works; let me know if it doesn't. – J.G. Mar 27 '19 at 12:40
  • Also see here for proofs: https://math.stackexchange.com/questions/9286/proving-int-0-infty-mathrme-x2-dx-frac-sqrt-pi2 – Hans Lundmark Mar 27 '19 at 15:50
  • Thanks friends! – Will E. Mar 28 '19 at 02:00

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