Quiz: calculating a definite integral

[Bunuel66]

(01-06-2014 10:28 AM)Thomas Klemm Wrote:  

(01-03-2014 09:04 PM)Bunuel66 Wrote:  Don't get the point, \(\exp(-\infty)\)=0.

The domain of \(\exp(x)\) is \(\mathbb{R}\), but \(-\infty \notin \mathbb{R}\). Thus you can not just plug \(-\infty\) into the Taylor-series of this function and expect everything works. You can calculate \(\lim_{x\to\infty}\exp(x)\) but that's not the same as \(\exp(-\infty)\). This expression is just not defined.

HTH
Thomas

Could have been rewriten as a limit to be more rigorous...;-) That said the serie gives the same value than the function also for x<0. Doesn't seems to be the point. And as you mention this is not a Taylor serie strictly speaking.

Regards.