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Quiz: calculating a definite integral
01-03-2014, 02:50 PM
Post: #37
RE: Quiz: calculating a definite integral
(01-03-2014 01:29 PM)Bunuel66 Wrote:  Great, but then what is the justification for this expansion if it is not based on a Taylor one?
It looks a lot like the expansion of exp(u) with u replaced by xlog(x). My view is that it is the Taylor expansion of exp(u) around 0 with u taking the value given by xlog(x), which is numerically correct. But I still miss the justification for getting an equality when integrating on both sides.
We start with \(x^{-x}=\exp(-x\log(x))\), substitute \(u=-x\log(x)\), use the Taylor-series of \(\exp(u)\) and plug \(u\) back into that expression. We just have to make sure that \(u\) is defined for all \(x \in [0, 1]\).
Both sides are equal. So integrating both sides yields the same result.

Where exactly is the problem?
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RE: Quiz: calculating a definite integral - Thomas Klemm - 01-03-2014 02:50 PM

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