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Quiz: calculating a definite integral
01-03-2014, 05:38 PM (This post was last modified: 01-03-2014 06:21 PM by Bunuel66.)
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RE: Quiz: calculating a definite integral
Quote: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?

Well, let's consider the very same approach with the function exp(-1/x). Using a similar schema we end up with a serie 1-1/x+1/2!x²-1/3!x³...... who gives the same as exp(-1/x) for x<0 and x>0. Then, if we integrate left and right, 1/x will become log(x) which is not defined for x<0 while the integral of exp(-1/x) does exist for x<0.
This seems to show that having the equality is not enough for keeping it directly after integrating.

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RE: Quiz: calculating a definite integral - Bunuel66 - 01-03-2014 05:38 PM

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