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Weird integral

  1. Aug 21, 2007 #1
    Consider integrating sin(t)/t from 0 to infinity via the Taylor expansion method. If you do that then you will get infinite terms in the series which clearly is wrong as that integral equals pi/2. Can anyone tell me why using the Taylor expansion method dosen't work? It always dosen't work when integrating to infinity. But other methods like complex integral does work. Why?

    Taylor expansion use to be a sure thing for me every time but now my innocence has been broken.
     
  2. jcsd
  3. Aug 21, 2007 #2

    matt grime

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    It's probably a uniform convergence thing - or you're using an invalid Taylor expansion - or possibly both. In general you can't interchange limits and integrals.

    What does
    "It always dosen't work when integrating to infinity"
    mean?
     
  4. Aug 21, 2007 #3
    Just renormalize...

    If you're brave, you'll just work out how to sum divergent series. I believe some little known physicist once did that for a special and got some award out of it or something. :tongue:
     
  5. Aug 21, 2007 #4
    Taylor expansion expands a function in terms of polynomials. Integrate any polynomial from 0 to infinity and you get infinity hence "It always dosen't work when integrating to infinity". So a series with infinity in every term is not pleasant to say the least. In the case of sint/t, the terms are oscillating between negative and positive so the integration of the taylor polynomial on face value is 0 as you might like to subtract infinities after grouping the terms together, two by two. However infinities shouldn't be subtracted to the integral is not 0 but pi/2 when evaluated using another method.
     
  6. Aug 22, 2007 #5

    matt grime

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    You don't evaluate integrals over the range 0 to infinity just by putting in infinity and 0.
     
  7. Aug 23, 2007 #6

    dextercioby

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    To the OP: What's the radius of convergence for the MacLaurin series expansion of [itex]\frac{\sin x}{x} [/itex] ?

    Daniel.
     
  8. Aug 23, 2007 #7
    what are you talking about? dimensional regularization?
     
  9. Aug 23, 2007 #8
    infinity. Or for all values of x. That is after we remove the singularity at x=0.
     
  10. Aug 23, 2007 #9
    Its got to do with putting the limit as x->infinity. You are suggesting the limit doesn't exist?
     
  11. Aug 23, 2007 #10
    I was thinking of Feynman and QED renormalisation...
     
  12. Aug 23, 2007 #11
    I think

    [tex]
    \lim_{R\to\infty} \lim_{N\to\infty}\int\limits_{-R}^{R} \Big(\sum_{k=0}^{N} a_k x^k\Big) dx
    [/tex]

    should be giving correct answer. But it is probably not possible to calculate it reasonably.
     
  13. Aug 23, 2007 #12
    Indeed -- after taking the N->inf limit the function is what Mathematica called SinIntegral, i.e. the integral of sinc. It's a special function, so no expression in terms of "normal" functions. It's pretty much for this purpose of calculating these random integrals that complex contour integration was pursued.
     
  14. Aug 23, 2007 #13
    there is another way to do this integral (that does not involve contour integration). define

    [tex]
    f(x, \alpha) = \frac{\sin (x)}{x} e^{-\alpha x}
    [/tex]

    where you recover your function when [tex]\alpha = 0[/tex]. then differentiate past the integral sign with respect to [tex]\alpha[/tex]. i'll leave the rest to you.
     
  15. Aug 24, 2007 #14

    dextercioby

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    Good idea, but it still doesn't give the right value for the integral.



    [tex] \frac{d}{d\alpha }\int_{0}^{\infty }\frac{\sin x}{x}e^{-\alpha x}dx=\int_{0}^{\infty }\left( -\left( \sin x\right) e^{-\alpha x}\right) \,dx= \frac{1}{\alpha ^{2}+1}\lim_{x\rightarrow \infty }\left( e^{-\alpha x}\cos x+\alpha e^{-\alpha x}\sin x-1\right) =-\frac{1}{\alpha ^{2}+1} [/tex]

    [tex]\int_{0}^{\infty }\frac{\sin x}{x}e^{-\alpha x}dx=-\int \frac{d\alpha }{\alpha ^{2}+1}=\allowbreak -\arctan \alpha +C [/tex]
     
  16. Aug 24, 2007 #15

    Gib Z

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    It does. The constant turns out to be pi/2, but the way I do that is not very rigorous...
     
  17. Aug 24, 2007 #16

    Gib Z

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    Ok Here it goes: Ignoring x between -1 and 1 :( maybe someone can fill that in later for me, but anyway: The original integral is less than the negative of its derivative, since its all values in the integrand are being divided by x.

    So basically it becomes [tex] \frac{1}{\alpha^2 +1} > -\arctan \alpha + C[/tex]

    Take the limit as alpha approaches infinity, the LHS approaches infinity, hence
    [tex] \lim_{\alpha \to \infty} -\arctan \alpha + C = 0[/tex].

    Knowing that as x approaches pi/2 on the left, it goes to infinity, the limit of x as x goes to infinity for arctan x is pi/2.

    Ie C = pi/2.

    I know there are many holes in the argument, but it works.
     
  18. Aug 24, 2007 #17
    that is correct. good job.
     
  19. Aug 24, 2007 #18
    to be clear:

    [tex]f(x, \alpha) = -\tan^{-1} (\alpha) + C[/tex]
    [tex]\lim_{\alpha \rightarrow \infty} f(x, \alpha) = \lim_{\alpha \rightarrow \infty}[/tex] [tex]\int_0^{\infty} \frac{\sin(x)}{x} e^{-\alpha x} dx = 0[/tex]

    but also

    [tex]\lim_{\alpha \rightarrow \infty} f(x, \alpha) = -\tan^{-1} (\alpha) + C = -\frac{\pi}{2} + C[/tex]
    [tex]\Rightarrow C = \frac{\pi}{2}[/tex]
    [tex]\Rightarrow f(x, \alpha) = \frac{\pi}{2} - \tan^{-1} (\alpha)[/tex]

    and

    [tex]f(x, 0) = \frac{\pi}{2}[/tex]
     
  20. Aug 24, 2007 #19

    dextercioby

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    I see it now, nice job, quetzalcoatl9 !
     
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