Quiz: calculating a definite integral

[Thomas Klemm]

(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