Quiz: calculating a definite integral
01-03-2014, 05:38 PM (This post was last modified: 01-03-2014 06:21 PM by Bunuel66.)
Post: #38
 Bunuel66 On Vacation Posts: 29 Joined: Jan 2014
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.

Regards
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