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Crazy integral

  1. Nov 1, 2007 #1
    [tex]\int_{0}^{2\pi} \frac{dx}{1+e^{sin(x)}}[/tex]

    How would you evaluate this integral? Where do you even start?
     
  2. jcsd
  3. Nov 1, 2007 #2
    Well, we know that [tex]\int \frac{dx}{1 + x^{2}} = arctan(x) + C [/tex].

    How can we think of [tex] e^{sin(x)} [/tex] as a term that's been squared?
     
  4. Nov 2, 2007 #3

    Gib Z

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    No idea :(

    I would have let u=sin x, so the integral becomes [tex]\int \frac{1}{1+e^u} \frac{du}{\sqrt{1-u^2}}[/tex] then did integration by parts.
     
  5. Nov 2, 2007 #4
    I was completely wrong about using [tex]\int \frac{dx}{1 + x^{2}} = arctan(x) + C [/tex].

    I tried evaluating the integral using my TI-89, but it wouldn't evaluate it unless I put in endpoints.
     
  6. Nov 2, 2007 #5
    I would try the following. Put [tex]z=\exp\left(i x\right)[/tex]. The integral then becomes:

    [tex]\oint\frac{1}{1+\exp\left(\frac{z-z^{-1}}{2i}\right)}\frac{dz}{iz}[/tex]

    where the contour integral is over the unit circle in the complex plane. Next, use the residue theorem.
     
  7. Nov 2, 2007 #6
    Actually, I think I found a way to do this without using complex analysis.
    The answer is pi.

    Let [tex]f(x) = \frac{1}{1+e^{sin(x)}}[/tex] and [tex]g(x)=f(x-\pi)-\frac{1}{2} = \frac{1}{1+e^{sin(x-\pi)}}-\frac{1}{2}[/tex].

    I will now show that g(x) is an odd function on the interval [tex]\left[-\pi,\pi\right][/tex]. For this to be true, I need g(x)=-g(-x), or g(x)+g(-x)=0.

    Using the fact that sin(x-pi)=-sin(x) and sin(-x-pi)=sin(x),

    [tex]g(x)+g(-x)= [/tex]

    [tex]=\frac{1}{1+e^{sin(x-\pi)}}-\frac{1}{2}+\frac{1}{1+e^{sin(-x-\pi)}}-\frac{1}{2}[/tex]

    [tex]=\frac{1}{1+e^{-sin(x)}}+\frac{1}{1+e^{sin(x)}}-1[/tex]

    [tex]=\frac{1+e^{sin(x)}+1+e^{-sin(x)}}{(1+e^{-sin(x)})(1+e^{sin(x)})}-1[/tex]

    [tex]=\frac{2+e^{sin(x)}+e^{-sin(x)}}{1+e^{sin(x)}+e^{-sin(x)}+1}-1[/tex]

    [tex]=1-1=0[/tex]

    Therefore, g(x) is odd, which implies that [tex]\int_{0}^{2\pi}g(x+\pi) dx=0[/tex].


    Because [tex]f(x)=g(x+\pi)+\frac{1}{2}[/tex],

    [tex]\int_{0}^{2\pi}f(x) dx[/tex] simply becomes [tex]\int_{0}^{2\pi}\frac{1}{2} dx[/tex], which is obviously pi.
     
  8. Nov 2, 2007 #7
    I made it into a power series and got the value to be accurate to .0000001 and it does look like pi. :)
     
  9. Nov 2, 2007 #8
    What happens here is that the integral is equal to: [tex]\int_{0}^{\pi} \frac{dx}{1+e^{sin(x)}}+[/tex] [tex]\int_{0}^{\pi} \frac{dx}{1+e^{-sin(x)}}[/tex]

    The second integral then becomes: [tex]\int_{0}^{\pi}\frac{e^{sinx}}{1+e^{sinx}}[/tex]

    Thus adding the integrals reduces to [tex]\int_{0}^{\pi}dx[/tex] as Doom of Doom has already figured out.
     
    Last edited: Nov 2, 2007
  10. Nov 3, 2007 #9
    holy crap so many different ways to do one integral. i like doom of dooms the best
     
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