Quiz: calculating a definite integral
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01-03-2014, 02:50 PM
Post: #37
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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?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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