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A serious Fourier integral

  1. Apr 30, 2010 #1
    1. The problem statement, all variables and given/known data

    The integral is

    [tex]\int _{-\pi }^{\pi }\!{\frac {{{\rm e}^{x\cos \left( \theta \right) }} \cos \left( n\theta \right) }{\pi }}{d\theta}.[/tex]

    3. The attempt at a solution

    I've tried a couple of alternative methods huddling in my mind to solve this integral, but none of them worked. Actually using De Moivre's formula and integrationa by parts this can be written as

    [tex]{\frac {{{\rm e}^{x+in\pi }}-{{\rm e}^{-x-in\pi }}}{\pi }}+\int _{-\pi }^{\pi }\!{\frac {x\sin \left( \theta \right) {{\rm e}^{x\cos\left( \theta \right) +in\theta}}}{in\pi }}{d\theta}.[/tex]

    where we hit the second integral which, in its indefinite form, cannot be described by the elementary functions.

    Any help will be appreciated!

    AB
     
  2. jcsd
  3. Apr 30, 2010 #2

    gabbagabbahey

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    Is it safe to assume that [itex]n[/itex] is an integer and [itex]x[/itex] has no [itex]\theta[/itex] dependence?

    If so, just begin by defining [itex]I_n(x)\equiv\frac{1}{\pi}\int_{-\pi}^{\pi}{\rm e}^{x\cos\theta}\cos\left(n\theta\right)}d\theta[/itex], then use integration by parts once, along with the trig identity [itex]\sin(a)\sin(b)=\frac{1}{2}\left[\cos(a-b)-\cos(a+b)\right][/itex] to derive a recurrence relation. The resulting infinite series is very well known:wink:
     
    Last edited: Apr 30, 2010
  4. Apr 30, 2010 #3
    Yes and do you have any idea then?

    AB
     
  5. Apr 30, 2010 #4

    gabbagabbahey

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    See my edited post above :smile:
     
  6. Apr 30, 2010 #5

    Would you mind elaborating the integration by parts part a little more?

    Thanks for your time.

    AB
     
  7. Apr 30, 2010 #6

    gabbagabbahey

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    Well, I assume you know how to differentiate [itex]\rm{e}^{x\cos\theta}[/itex] and integrate [itex]\cos(n\theta)d\theta[/itex], so....
     
  8. Apr 30, 2010 #7
    Okay, I get this recurrence equation in the end:

    [tex]\frac{x}{2n}(I_{n-1}-I_{n+1})=I_n[/tex]

    and this is..?!

    AB
     
  9. Apr 30, 2010 #8

    gabbagabbahey

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    Compare that to the recurrence relation you get for Bessel's functions.
     
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