1. Not finding help here? Sign up for a free 30min 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!

Where am i screwing up? Improper Integral

  1. Oct 19, 2007 #1
    the SM shows a different method but i figure i should arrive at the same answer either way. what step am i screwing up on? thanks!

    [tex]\int_{ - \infty}^{\infty}\frac {x^{2}dx}{9 + x^6}[/tex]

    [tex]\int_{ - \infty}^{0}\frac {x^{2}dx}{9 + x^6} + \int_{0}^{\infty}\frac {x^{2}dx}{9 + x^6}[/tex]

    [tex]u = x^{3}[/tex]
    [tex]du = 3x^{2}dx[/tex]

    [tex]\frac {1}{3}[\lim_{t\rightarrow - \infty}\int_{t}^{0}\frac {du}{9 + u^{2}} + \lim_{t\rightarrow \infty}\int_{0}^{t}\frac {du}{9 + u^{2}}][/tex]

    [tex]3[\lim_{t\rightarrow - \infty}\int_{t}^{0}\frac {du}{1 + (\frac {u}{3})^{2}} + \lim_{t\rightarrow \infty}\int_{0}^{t}\frac {du}{1 + (\frac {u}{3})^{2}}][/tex]

    [tex]3[\lim_{t\rightarrow - \infty}\int_{t}^{0}\frac {du}{1 + (\frac {u}{3})^{2}} = \arctan{0} - \arctan{\frac {t^{3}}{3}] +
    3[\lim_{t\rightarrow \infty}\int_{0}^{t}\frac {du}{1 + (\frac {u}{3})^{2}} = \arctan{\frac {t^{3}}{3} - \arctan{0}][/tex]
     
    Last edited: Oct 19, 2007
  2. jcsd
  3. Oct 19, 2007 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The integral isn't singular at x=0. You don't have to split it up. Just do it.
     
  4. Oct 20, 2007 #3
    you don't need to go from u back to x and then plug-in t. since the limits of u are the same limits for x.

    I think your integral should be divided by 1/9...since it's:
    [tex]\int \frac{dx}{a^2+x^2}=\frac{1}{a}arctan\frac{x}{a}[/tex]

    other than that I don't see a problem. what is the solution from SM?
     
  5. Oct 20, 2007 #4
    my first u-sub is where i got the 1\3 ... then i pulled out a 1\9, but if i pull out that 1\9 it becomes 3.

    since i pulled it out from the denominator, simplifying the complex fraction gives me 3 as my constant infront, or no?

    [​IMG]
     
    Last edited: Oct 20, 2007
  6. Oct 20, 2007 #5
    If you pull out a 9 from the denominator, wouldn't there be a 1/27 at the front?
    In fact, you need not get the 9 out at all. As bob pointed out, there is a standard procedure for dealing with such integrals.
     
  7. Oct 20, 2007 #6
    ah damnit. i been pulling out a 1\9 for some odd reason. i pull out a 9 which gives me the (x\a)^2, lol. whew finally.
     
  8. Oct 20, 2007 #7
    no I think it goes like this:
    [tex]\frac{1}{9+x^2}=\frac{1}{9}\frac{9}{9+x^2}=\frac{1}{9}\frac{1}{\frac{9+x^2}{9}}=\frac{1}{9}\frac{1}{1+(\frac{x}{3})^2}[/tex]
    and doing a substitution when you integrate that you have to multiply by 3 so you'd get 1/3 * arctan x/3

    what do you get when you plug in for t? I get the same answer as the book.
     
    Last edited: Oct 20, 2007
  9. Oct 20, 2007 #8
    i get that answer now too! w00t :-] thanks again bob and neutrino!
     
  10. Oct 20, 2007 #9
    is it ok if i quit using T and just have my integral going from x to 0 etc? it won't be "improper" in terms of notation to do so?
     
  11. Oct 20, 2007 #10
    do you mean just replacing t by x and have no limit? or have x -> infinity?

    [tex]3[\lim_{x\rightarrow - \infty}\int_{x}^{0}\frac {du}{1 + (\frac {u}{3})^{2}} = \arctan{0} - \arctan{\frac {x^{3}}{3}] +
    3[\lim_{x\rightarrow \infty}\int_{0}^{x}\frac {du}{1 + (\frac {u}{3})^{2}} = \arctan{\frac {x^{3}}{3} - \arctan{0}][/tex]

    that would just lead to confusion, is the x in the limit the same as the original x in the problem?
     
  12. Oct 20, 2007 #11
  13. Oct 20, 2007 #12
    o whops ><

    you still need the limit sign with t:

    [tex]\frac{1}{9}[\lim_{t\rightarrow - \infty}\int_{t}^{0}\frac {du}{1 + (\frac {u}{3})^{2}} = \arctan{0} - \lim_{t\rightarrow - \infty}\arctan{\frac {t^{3}}{3}] +
    \frac{1}{9}[\lim_{t\rightarrow \infty}\int_{0}^{t}\frac {du}{1 + (\frac {u}{3})^{2}} = \lim_{t\rightarrow - \infty}\arctan{\frac {t^{3}}{3} - \arctan{0}][/tex]

    I was taught that you could do it but just to remember that by pluggin in you don't mean infinity is a number.

    [tex]\frac{1}{9}(-\arctan\frac{-\infty^3}{3})+\frac{1}{9}\arctan\frac{\infty^3}{3}=\frac{1}{9}(-\arctan (-\infty)+\arctan\infty)=\frac{1}{9}(\frac{\pi}{2}+\frac{\pi}{2})[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Where am i screwing up? Improper Integral
Loading...