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Trigonometric nastiness

  1. Apr 13, 2007 #1

    dextercioby

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    1. The problem statement, all variables and given/known data

    Compute [tex] \int_{0}^{\infty} \frac{\tan x}{x} {} dx [/tex]

    2. Relevant equations



    3. The attempt at a solution

    Got no idea, obviously the derivation differs from the [itex] \int_{-\infty}^{\infty} \frac{\sin x}{x} {} dx [/itex] one. The result i'm supposed to get is [itex] \frac{\pi}{2} [/itex].
     
  2. jcsd
  3. Apr 13, 2007 #2
    i would do integration by parts on the 1/x and the tan(x) then work from there.

    probably let tan(x) equal sin(x)/cos(x) and use substitution integration to solve that part

    then when you've solved the integration, sub in infinity and just logic it out from there i guess.

    sorry if its not much help, i cant think straight at the moment. (halfway thought working on my own integration problem)
     
  4. Apr 13, 2007 #3

    Gib Z

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    No point, that integral is not expressible in terms of elementary functions. Must be some property with the bounds thatll help then, rewrite as [tex]\lim_{a\to\infty} \int^a_0 \frac{\tan x}{x} dx[/tex] and try u= a - x
     
  5. Apr 13, 2007 #4

    Tom Mattson

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    I'm surprised that the interval even converges, since the integrand is so badly behaved. But if it is supposed to come out to [itex]\pi/2[/itex], then I suppose you can do a numerical approximation using the Maclaurin series for sin(x) and cos(x).
     
  6. Apr 13, 2007 #5

    VietDao29

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    Errr... Why do I have a feeling that this integral does not converge. >__<
     
  7. Apr 13, 2007 #6

    Dick

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    Probably because in the strictest sense, it doesn't. It's an improper integral. And so treating it numerically would be a disaster. It may have a shadowy existence via a contour integral argument.
     
  8. Apr 13, 2007 #7

    Gib Z

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    Maclaurin/Taylor series approach does not work, tried very hard.
     
  9. Apr 14, 2007 #8
    As Dick alluded to, if the integral has a value, the argument's going to have to use complex analysis, as for [itex]\int_{0}^\infty \frac{\sin x}{x} dx[/itex]. Of course, this at least appears to be a lot more complicated thanks to the integrand's infinitely many poles on the real line. For example, the sum of the residues on the positive real line diverges! Maple seems to agree with the [itex]\frac{\pi}{2}[/itex] value, though.
     
    Last edited: Apr 14, 2007
  10. Apr 14, 2007 #9

    Gib Z

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    You know its not a good sign when dexter has a problem with an integral ...God Help Us!
     
  11. Apr 17, 2007 #10

    dextercioby

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    So nobody could help me, right ?
     
  12. Apr 17, 2007 #11

    Gib Z

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    I did some numerical analysis of the graph of the function in question, and i found that at all the points the function is equal to zero, k pi, the area from kpi - 1 and kpi +1, is always negative and decreasing. I chose a difference of 1, but any value less than pi/2 will work.

    So All I can say is that the integral [tex]\int_{\frac{\pi}{2}}^{\infty} \frac{\tan x}{x} dx[/tex] is less than zero, and very badly approximated to be -3.775239041.

    I also approximated the area from 0 to 1.57 ( I couldn't do exactly pi/2 for obvious reasons) is 5.5861702.

    So your integral should be the addition of the two integrals, and I got 1.810931159. I looks like its slowly getting to pi/2, but what I just did was completely useless. O well, I tried.
     
  13. Apr 17, 2007 #12

    Dick

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    Did you try defining it as a contour integral? Data said that the sum of the poles on the real line diverged. Just looking at the integral I would guess they would actually form an alternating series.
     
  14. Apr 17, 2007 #13

    Gib Z

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    Yup It does :) I haven't done Contour Integration yet though, So I can't tell you the specifics of the series, just that its alternating and converges to [itex]\frac{\pi}{2}[/itex] :) Sorry dex.
     
  15. Apr 17, 2007 #14

    Dick

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    Well, I'm doubting that now. But if it does, then fill us in. Because I'm currently stumped. It's funny if integral of sin(x)/x and tan(x)/x are equal 0 to infinity.
     
  16. Apr 17, 2007 #15

    Gib Z

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    Well this is kind of hard for me to explain but heres what I saw.

    At every point [itex]k\cdot \pi, k\in \mathbb{Z}[/itex], f(x) is zero as we know. The integral from [itex]k\pi - z, k\pi + z, z < \pi/2[/itex] is always negative, and decreases as k increases. So What I meant was that the areas from k*pi -x, to kpi, summed with k*pi + x, produce an alternating series.
     
  17. Apr 18, 2007 #16

    dextercioby

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    I can't see any possible contour that would allow me to compute the integral.
     
  18. Apr 18, 2007 #17

    Dick

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    I don't either - nor can I see a way to manipulate Si(x). How do you know the integral is pi/2? Where did the problem come from?
     
  19. Apr 19, 2007 #18

    dextercioby

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    It came from my mind, i just fooled around with Maple and it gave pi/2. I looked up the integral in the bible G & R and the value of pi/2 was confirmed.
     
  20. Apr 19, 2007 #19

    Dick

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    I would post a query on the sci.math newsgroup. There's a lot giant brains over there.
     
  21. Apr 20, 2007 #20
    Sorry to burst your bubble but

    This integral ,the way you wrote it,is said to be DIVERGENT.
    That's why nobody could help you.
    Another ,more trivial, example is:
    [tex]\int_{\pi}^{2\pi}\tan x dx[/tex]
    This is divergent too.:smile:
     
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