Question

I am having trouble trying to understand this problem: http://www.wolframalpha.com/input/?i=int+from+0+to+infinity%3A+e^%28-1%2Fx%29%2Fx^2

Answer 1

1/0 not allowed.

Answer 2

Heres the issue:

Answer 3

http://www.wolframalpha.com/input/?i=int+from+0.1+to+infinity%3A+e^%28-1%2Fx%29%2Fx^2

Answer 4

If I integrate I get: http://www.wolframalpha.com/input/?i=int+e^%28-1%2Fx%29%2Fx^2

Answer 5

And then If I plug in the upper bound I get 1: http://www.wolframalpha.com/input/?i=e^%28-1%2Fx%29+for+x+%3D+infinity

Answer 6

put 1/x =t then change the limits and dx=-dt/x^2
http://www.wolframalpha.com/input/?i=int+from+infinty+to+0%3A+e^%28-x%29

Answer 7

So how do I deal with 1 - indeterminant I guess is the question

Answer 8

Then you plug the lower bound and get 0, since\[\lim_{x \to 0^+} e^{-1/x} = \lim_{u \to -\infty} e^u = 0.\]

Answer 9

Indeterminate forms need to be dealt with analytically as Krebante showed u (but you can figure out via Wolfram as well)

Answer 10

The lim as x->0 is only 0 from the high side

Answer 11

The high side is the only one that matters, because 0 is the lower bound of the integral. The function is only taking values inside the interval \[[0, \infty),\]so you can forget about negative numbers.

Answer 12

That's the definition of an improper integral: http://en.wikipedia.org/wiki/Improper_integral
(notice the limits are always one-sided limits).

Answer 13

Ok

Answer 14

So then with that knowledge I dont see the need for any substitutions beyond solving the integral...

Answer 15

I solve it to get e^(-1/x)

Answer 16

then subtract lim->0from high (0) from lim -> infinity from low ( = 1)

Answer 17

1 - 0 = 1

Answer 18

Yes, that's correct. You don't need to use any substitution in this case.

Answer 19

So is that like a theorem? improper integrals only use 1 sided limits? and the upper bound is the lim from the low, the bottom bound is the lim from the high?

Answer 20

It's not a theorem, it's the definition.

Answer 21

Ok thanks so much!!

Answer 22

You are very welcome.