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Improper Integrals

  • Thread starter morris4019
  • Start date
  • #1
Hi, I am currently in Calculus 2 at my local college and I am having trouble wrapping my head around Improper Integrals. The question below I have been working on for awhile and I think i have an answer but was wondering if anyone could confirm if I was thinking about this question the right way:


Homework Statement


Calculate ∫1/(x^2-1)dx from 2 to positive infinity. (Hint: You will need to write the antiderivative as a single logarithm in order to be able to calculate the appropriate limit.)


Homework Equations





The Attempt at a Solution





What I have so far is the following:

First re-writing as a limit->

lim ∫1/(x^2-1)dx from 2 to T
T->infinity

then using partial fractions

lim 1/2∫1/(x-1)dx - 1/2∫1/(x+1)dx from 2 to T
T->infinity

lim 1/2*ln|x-1| - 1/2*ln|x+1| from 2 to T
T->infinity

re-writing as a single natual log

lim 1/2*ln((x-1)/(x+1)) from 2 to T
T->infinity

now subtracting the endpoints

lim 1/2*ln((T-1)/(T+1)) - 1/2*ln(1/3)
T->infinity

now here is where i got a little confused again. T is approaching infinity but because we are taking a limit i can say that the first term is siply 1/2 correct? and the second remains 1/2*ln(1/3)? So i'm getting for my answers (exact and then approx):

1/2 - 1/2*ln(1/3) or approx 1.0493


Now I could be completely wrong and that is why I can't seem to get comfortable with these things. Can anyone shed some light on what I'm doing wrong, or perhaps right?

Thanks
-Mike
 

Answers and Replies

  • #2
25
1
what do you know about taking the limit as x->inf of the ratio of polynomials with the same degree? Hint - has to do with the coefficients

Then what is the Ln of that?
 
  • #3
The limit of a ratio of polylnomials, with the same degree both top and bottom, should just equal the coefficient ratio, correct? so...

lim 2x^2 / 3x^2 = 2/3
x->inf

So my limit near the bottom:

lim 1/2*ln((T-1)/(T+1)) - 1/2*ln(1/3)
T->infinity

should be:

=> 1/2*(ln 1) - 1/2*ln(1/3)
=> 1/2*(0) - 1/2*ln(1/3)
=> -1/2*ln(1/3)

does this look a little better?
 
  • #4
25
1
Yes, That is correct.
 
  • #5
thank you, that makes more sense. I've spoken to a few people about what i did above and no one pointed out that I made that mistake with the natural log.
 

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