# Indefinate integral

1. Aug 17, 2011

### Shannabel

1. The problem statement, all variables and given/known data
find the integral from (2 to infinity) of 1/(1+e^x)

2. Relevant equations

3. The attempt at a solution

lim(t-->infinity) integral from (2 to t) of 1/(1+e^x)
1/1+e^x = (1+e^x-e^x)/(1+e^x)
= 1-e^x(1+e6x)

when you solve the integral:
x-ln(1+e^x) between 2 and t
t-ln(1+e^t)-(2-ln(1+e^2))

here we're confused, because when you plug in t=infinity, you end up with a limit that doesn't exist... help?

2. Aug 17, 2011

### HallsofIvy

Well, first, of course, you don't find a limit by "pluggin in t= infinity". This of the form "infinity- infinity" so you should be able to use L'Hopital's rule.

3. Aug 17, 2011

### Shannabel

then i should hve 1-(e^x)/(1+e^x)?
and use l'hospital's rule again and have
-(e^x)/(e^x)
=-1?

4. Aug 17, 2011

### dynamicsolo

You are confronted at this point with the limit

$$\lim_{t \rightarrow \infty} t - \ln | 1 + e^{t} | - 2 + \ln | 1 + e^{2} |$$

Plainly, that t - 2 is going to be a problem (and pulling the logarithms into a single term isn't much better). Try writing the other two terms as logarithms, e.g. , $$t \rightarrow \ln(e^{t}) ,$$ and write the entire limit expression as a single logarithm. With a little manipulation of the ratio in the argument of the logarithm, you'll find you won't even need l'Hopital...

Last edited: Aug 17, 2011
5. Aug 17, 2011

### Ray Vickson

In order to get a finite result for int_{x=2..infinity} f(x) dx (f(x) > 0 for all x >= 2) we need f(x) to go to zero quickly enough as x --> infinity. Does this happen for your f(x)?

RGV

6. Aug 17, 2011

### vela

Staff Emeritus
No. You need a quotient to apply the Hospital rule, so you'd need to rearrange your expression first.

I think the most straightforward way to evaluate the limit is to use
$$t-\log(1+e^t) = t-\log[e^t(1+e^{-t})]$$
Use the properties of log to simplify your expression and then let t go to infinity.