How can an improper integral have a finite value How is it possible for an improper integral to have a finite value? For example, an integral that has infinity as its upper bound and has the horizontal asymptote y=0. I understand that one can arrive at a finite value by taking the limit. However, when thinking about this visually, if a function goes on until infinity, then how can the area under it be finite? What is the physical difference between taking the limit and summing the area under the graph until infinity?
 A: As @eyeballfrog notes in the comments, you may find it helpful to think of the intuition for $$\sum_{n=1}^\infty 2^{-n}=1.$$ Visually, this says that if you fill a glass half full, then fill half of the empty space remaining, then fill half of the empty space remaining, etc. you will approach filling one glass. Despite infinitely many fills, we have a finite volume in the limit because intuitively each fill introduces sufficiently less volume than the preceding fill (with vanishing volume in the limit).
A continuous analogue of the above series is the improper integral
$$\int_0^\infty e^{-x}dx=1.$$
Again, as with cup filling, though integrating over an infinite interval, $e^{-x}$ vanishes sufficiently quickly in the limit for the integral to be convergent.
The above improper integral has an integrand with a horizontal asymptote. An improper integral whose integrand has a vertical asymptote may be thought of much the same way; just think of the integrand as a function with a horizontal asymptote that has been rotated 90 degrees.
