1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration with limit of zero giving infinity - help please

  1. Mar 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Integral of ∫1/x^2 (or ∫x^-2) between 1 and 0.

    3. The attempt at a solution
    I can integrate it no problem to give me -1/x or x^-1, but when I put it between the limits of 1 and 0 I get ∞-1 which is just ∞.

    Is this right or do I need to use L'Hopital's rule. If so, how? I'm unsure of how it really works.


    Steven Thomas.
  2. jcsd
  3. Mar 9, 2015 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    When you try to put in ##x=0## in ##-\frac 1 x## you don't get ##\infty^{-1}##. What you get is attempted division by ##0## which just means the integral is divergent.
  4. Mar 9, 2015 #3
    Sorry that is infinity minus 1, not infinity to the power of minus 1, which in turn equals infinity.

    Is this an acceptable complete integration, or should I apply L'Hopital's rule?
  5. Mar 9, 2015 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You don't have an indeterminate form like ##\frac 0 0## so there is nothing to apply L'Hospital's rule to. Just say the integral is divergent.
  6. Mar 9, 2015 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    This is an improper integral, so is customarily defined through a limiting process. In particular, since the integrand blows up at ##x = 0## we have
    [tex] \int_0^1 \frac{1}{x^2} \, dx = \lim_{a \to 0+} \int_a^1 \frac{1}{x^2} \, dx \; \Leftarrow \:\text{definition!}[/tex]
    What do you get when you compute the limit, if anything?
  7. Mar 10, 2015 #6
    I got 1/a - 1

    As a-> infinity I get negative infinity. Is that the answer? Or is the answer that it doesn't diverge?

    Also, how do you insert the equation so nicely on this forum? I have the feeling I'm going to be posting a few like this. I do have another one actually if you'd be so kind as to give me a hand please ray.


    Steven Thomas.
  8. Mar 10, 2015 #7
    My other one was integrating the product of x and e^-x between 0 and infinity. I can do the indefinite integral which gives me:

    -x.e^-x - e^-x
    between 0 and ∞, I just can't evaluate the first term when I use the infinity limit because I get negative infinity multiplied by zero.... :oldconfused:

    Steven Thomas
  9. Mar 10, 2015 #8


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    This is one of the indeterminate forms you should recognize from studying L'Hopital's Rule:


    If you're going to be doing a lot of similar integrals, you should review this material.
  10. Mar 10, 2015 #9
    So then I'm pretty sure that the -1/a term goes to minus infinity and hence my whole integral does.

    As for my -x.e^-x, do I take it as f(x) is -x and let g(x) = e^x as then I'd have f(x)/g(x)? Differentiating those would give me -1/e^x so it would tend to 0?
  11. Mar 10, 2015 #10

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Why do you care what happens as ##a \to \infty##? This issue involves ##a \to 0##.

    BTW: I used LaTeX; Somewhere in this Forum there is a tutorial on how to use LaTeX. Sorry: I don't recall what it is called or where it is, exactly, but you can set up a search to find it. (I did not need the tutorial, since I had already been using LaTeX for more than 20 years before joining the Forum.)
  12. Mar 10, 2015 #11
    That's what I got, but don't forget the other part. Also, the right way to write it is $$\int_{0}^{\infty}xe^{-x}dx = \lim_{M \rightarrow\infty}\int_{0}^{M}xe^{-x}dx$$ for which I got
    $$\lim_{M \rightarrow\infty}[\frac{-(x+1)}{e^{x}}]_{0}^{M} = \lim_{M \rightarrow\infty}[\frac{-(M+1)}{e^{M}} - \frac{-(0+1)}{e^{0}}]$$

    & then as you pointed out, you use L'Hopital's rule on the first summand to get the answer.

    edit: below the reply window at the bottom of the screen there's a button labelled "Latex preview". If you click it the Latex editor should expand, and then on the right there's a link to the how-to guide. It loads really slowly (on my computer anyway) so I googled the link below instead. Maybe you'll have better luck with that link. Just remember to include $$ on each end of th Latex code.
    Last edited: Mar 10, 2015
  13. Mar 10, 2015 #12
    Thanks for everyones help, I've managed to sort these and the other ones that I had a problem with because of the help y'all supplied.

    I'll also try that latex thing (I did wonder why it was asking if I'd like to preview some rubber :p) next time I post. Thanks all.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads - Integration limit zero Date
Change integration limits for cylindrical to cartesian coord Feb 7, 2018
Both limits of integration change to zero. Jul 8, 2013