.How would you evaluate this integral?

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In summary, the integral can be evaluated by using various methods such as using complex analysis, showing the function is odd and using symmetry, converting it to a power series, or adding two equivalent integrals to simplify it to a basic integral. The value of the integral is approximately pi.
  • #1
Doom of Doom
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[tex]\int_{0}^{2\pi} \frac{dx}{1+e^{sin(x)}}[/tex]

How would you evaluate this integral? Where do you even start?
 
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  • #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?
 
  • #3
hotcommodity said:
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?

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.
 
  • #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.
 
  • #5
Doom of Doom said:
[tex]\int_{0}^{2\pi} \frac{dx}{1+e^{sin(x)}}[/tex]

How would you evaluate this integral? Where do you even start?

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.
 
  • #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.
 
  • #7
I made it into a power series and got the value to be accurate to .0000001 and it does look like pi. :)
 
  • #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.
 
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  • #9
holy crap so many different ways to do one integral. i like doom of dooms the best
 

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