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Integration by parts

  1. Dec 25, 2009 #1
    What's up with this

    [tex]
    \int_{-\infty}^\infty \sin{x}\frac{1}{x}dx=\pi
    [/tex]

    Now I try integration by parts
    [tex]
    \int_{-\infty}^\infty \sin{x}\frac{1}{x}dx=[-\cos{x}\frac{1}{x}]_{-\infty}^\infty-\int_{-\infty}^\infty \cos{x}\frac{1}{x^2}dx = -\int_{-\infty}^\infty \cos{x}\frac{1}{x^2}dx = \infty
    [/tex]

    Why don't I get the same result?
     
  2. jcsd
  3. Dec 25, 2009 #2

    mathman

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    There is a singularity at x=0 which cannot be ignored. To get a clearer picture, carry out the original integral (0,∞) and double it (the integrand is an even function). The integration by parts will now have two ∞'s.
     
  4. Dec 25, 2009 #3
    Yes that works, but what did I do wrong in my initial calculation? I just used integration by parts.
     
  5. Dec 25, 2009 #4
    so how did you do the first and last integrals? how do you know the first integral equals pi and how did you integrate cos/x^2
     
  6. Dec 25, 2009 #5
    Here's a bunch of ways for evaluating int sin(x)/x: http://www.mathlinks.ro/viewtopic.php?t=197640

    The last integral is obviously divergent because the integrand diverges as 1/x^2 as x goes to zero. If you're trying to give me a hint, then I think I missed it.
     
  7. Dec 25, 2009 #6
    um the integrand in

    [tex]\int_{-\epsilon}^{\epsilon}\delta(x)dx[/tex]

    diverges as x goes to zero as well the integral is 1.

    now someone might call b.s. on me because really that's a definition but still my point is you can't reason just cause there's a divergence at some point that the integral around that point diverges (or can you?)
     
  8. Dec 26, 2009 #7

    HallsofIvy

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    The "delta function" is not an integrable function.
     
  9. Dec 26, 2009 #8
    So what's wrong with the way I use integration by parts? :-)
     
  10. Dec 26, 2009 #9

    tiny-tim

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    Because cosx/x is discontinuous at x = 0 (it's ∞ at 0+, and -∞ at 0-) :wink:
     
  11. Dec 26, 2009 #10
    So is there some general rule for when integration by parts works and when it fails?
     
  12. Dec 26, 2009 #11

    tiny-tim

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    integration by parts always works.

    what went wrong in this case was you got ∞ - ∞, which is an indeterminate form :wink:
     
  13. Dec 26, 2009 #12
    What do you mean? I got
    [tex]
    \int_{-\infty}^\infty \sin{x}\frac{1}{x}dx=\pi = -\int_{-\infty}^\infty \cos{x}\frac{1}{x^2}dx = \infty

    [/tex]

    Somehow it must be wrong to write
    [tex]
    \int_{-\infty}^\infty \sin{x}\frac{1}{x}dx= -\int_{-\infty}^\infty \cos{x}\frac{1}{x^2}dx

    [/tex]

    but I don't see why, because I just used the formula for integration by parts.
     
  14. Dec 26, 2009 #13

    tiny-tim

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    No, you also got [-cosx/x]-∞, = [-cosx/x]-∞0 + [-cosx/x]0, = ∞ + ∞.
     
  15. Dec 27, 2009 #14
    I guess you are right. I never thought about that. What I have is

    [tex]
    \left[\frac{\cos{x}}{x}\right]_{-\infty}^\infty=\int_{-\infty}^\infty\frac{d}{dx}\frac{\cos{x}}{x}dx = \lim_{a\rightarrow\infty}\int_{c}^a\frac{d}{dx}\frac{\cos{x}}{x}dx+\lim_{b\rightarrow-\infty}\int_{b}^c\frac{d}{dx}\frac{\cos{x}}{x}dx
    [/tex]

    where [itex] c[/itex] is any real constant. This clearly diverges for any [itex]c[/itex] ...
     
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