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Using complex exponentials

  1. May 6, 2009 #1
    ∫_0^∞ e^-x sin (ax) dx

    integrating from 0 to infinity for e^-x sin (ax)

    i was wondering that if i use complex exponentials for this will it be the same as solving for ∫_0^∞ e^-x cos (ax) dx. Will it make any difference if there is sin instead of cos because i know how to solve for cos (ax) but i dont know if it makes any difference if the cos is replaced by a sin.. does it make any difference?

    cheers
     
  2. jcsd
  3. May 6, 2009 #2

    Dick

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    It makes a difference in the sense that the answer is different, yes. The technique is pretty much the same, if that's what you mean.
     
  4. May 6, 2009 #3
    yup, because if i use cos (ax) i can replace it with e ^ i(ax) because i can use the complex exponential there. But what do i use for sin (ax).. i will just do what i did for cos (ax)

    ∫_0^∞ e^-x. e^i(ax) ..... (((since ∫_0^∞ e-x (cos ax + isin ax) dx)))

    Therefore we get:

    ∫_0^∞ e^-(1-ia)x dx

    then i can go ahead from there to get 1/ 1 + a^2.... now the only thing i am confused with is that this is for the complex exp. for cos (ax) what will it be for sin (ax)
     
  5. May 7, 2009 #4

    Cyosis

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    Are you claiming that [itex]\cos ax+i \sin ax=\cos a x[/itex]? This is of course not true. Try [itex]\sin x=Im(e^{i x})[/itex].
     
  6. May 7, 2009 #5

    HallsofIvy

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    You can't just "replace it with e^i(ax)". e^(iax)= cos(x)+ i sin(x). You can replace cos(x) with e^i(ax) and then just take the real part of the answer. And, of course, with sin(ax) you can replace it with e^i(ax) and take only the imaginary part of the answer.

     
  7. May 7, 2009 #6
    ahhh i get it.. sweet.. thank u
     
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