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Homework Help: 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

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    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?

    http://img201.imageshack.us/img201/821/integrallr0.jpg [Broken]​
     
    Last edited by a moderator: May 3, 2017
  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]
     
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