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Entropy, denoted as H, is:

$$ H = -\int_a^b p\ln(p) dx $$

where the range a to b is some arbitrary boundary and where p is given by the classic:

$$ p(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left( \frac{x-\mu}{\sigma}\right)^2} $$

Here is what I have tried so far to analytically solve this integral:

1) Integration by parts (too messy to write all out here), but quickly realized I needed a fancy substitution, thus leading me to:

2) Convert from x-space (where the range is a to b) to t-space (where the range is $\alpha$=$\frac{a-\mu}{\theta}$ to $\beta$=$\frac{b-\mu}{\theta}$) and, in general, $t$=$\frac{x-\mu}{\theta}$.

I think I'm on the right track (the $\theta$ in the denominator nicely cancels using the substitution approach) but I got stuck going from x-space to t-space and back again (i.e., I'm in xt-space purgatory and I want out!). Can anyone help me make more headway into #2 or perhaps suggest an alternative approach?

warship
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    Are you familiar with the error function? I expect your solution will probably be in terms of it. You can't simply integrate $\mathrm{e}^{-x^2}$ over a finite interval and get out an answer in terms of elementary functions. – Mark Mitchison Apr 16 '15 at 23:53
  • @MarkMitchison Thank you for pointing me to this. This has been something that I've been considering, although I'm not so certain how it can be used to solve my particular problem. Perhaps indeed an analytical solution in terms of elementary functions is not possible here, it certainly seems possible when doing the substitutions however (because of the nifty simplifications at the end). – warship Apr 16 '15 at 23:59
  • Substitutions followed by integration by parts should give you a solution in terms of the error function and some other terms. – Mark Mitchison Apr 17 '15 at 00:13
  • While the value you are after is physical quantity, this really is a math question (How do I solve this integral?). – Kyle Kanos Apr 17 '15 at 01:04

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