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Quiz: calculating a definite integral
01-03-2014, 08:39 PM
Post: #40
RE: Quiz: calculating a definite integral
(01-03-2014 05:38 PM)Bunuel66 Wrote:  This seems to show that having the equality is not enough for keeping it directly after integrating.
The problem I see is that \(u=-\frac{1}{x}\) is not defined for \(x=0\). The Taylor-series of \(\exp(u)\) is not defined for \(u=-\infty\).

Cheers
Thomas
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RE: Quiz: calculating a definite integral - Thomas Klemm - 01-03-2014 08:39 PM



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