Quiz: calculating a definite integral

[Bunuel66]

(01-03-2014 08:39 PM)Thomas Klemm Wrote:  

(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

Don't get the point, \(\exp(-\infty)\)=0. The serie is converging whatever the sign of x (more and more slowly as you're closing to 0-....). Then we have two expressions who provides similar values whatever the sign of x, and after integration we have a new set of expressions with one which is no more defined on one side (x<0). And as you mention, this is not exactly a Taylor serie in the sense that the sum is not using the derivatives of u(x). The problem is maybe a little bit more subtle (at least for me) than it seems ;-(...

Regards