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Quiz: calculating a definite integral
01-03-2014, 12:47 PM
Post: #35
RE: Quiz: calculating a definite integral
(01-03-2014 11:42 AM)Bunuel66 Wrote:  From a strict mathematical standpoint, I'm a bit puzzled as the Taylor expansion used seems a bit unjustified
This is not the Taylor-series expansion:
1-x log(x)+\frac{1}{2} x^2 log^2(x)-\frac{1}{6} x^3 log^3(x)+\frac{1}{24} x^4 log^4(x)-\frac{1}{120} x^5 log^5(x)+\frac{1}{720} x^6 log^6(x)+O(x^7)
The coefficients are not constant but dependent on \(x\).
Quote:If somebody has some light....
The question is: can we switch integration and summation?
As far as I can see the function is well behaved even at \(x=0\) so what can possibly go wrong?
It follows from \(\lim_{x\to 0}x\log(x)=0\).

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RE: Quiz: calculating a definite integral - Thomas Klemm - 01-03-2014 12:47 PM

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