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Integrating e^(e^(x))

  • Thread starter Doonami
  • Start date
  • #1
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Good old complex analysis. I'm trying to evaluate a line integral which looks like this

[tex]\oint[/tex]e (z + [1[tex]/[/tex]z]) for |z| = 1

So I guess I'm dealing with a circle with a radius 1, so I've parameterised:

z = eit

I need to sub this in to my formula of:

[tex]\int[/tex]c f(z)dz = [tex]\int[/tex]f(z(t)) z'(t)dt

(this is from [0,2pi]

However, when I go to sub that in I get an integral of an exponential to the power of an exponential. Can anyone suggest how to do that?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
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Welcome to PF!

Good old complex analysis. I'm trying to evaluate a line integral which looks like this

[tex]\oint[/tex]e (z + [1[tex]/[/tex]z]) for |z| = 1

So I guess I'm dealing with a circle with a radius 1, so I've parameterised:

z = eit

I need to sub this in to my formula of:

[tex]\int[/tex]c f(z)dz = [tex]\int[/tex]f(z(t)) z'(t)dt

(this is from [0,2pi]

However, when I go to sub that in I get an integral of an exponential to the power of an exponential. Can anyone suggest how to do that?
Hi Doonami ! Welcome to PF! :smile:

Hint: go for the obvious … substitute u = 1/z (and be very careful about the limits of integration). :wink:

And cryptic hint: Then compare it with the derivative of the integral. :smile:
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
955
Yes. z= eit so ez+ 1/z becomes
[tex]e^{z+ 1/z}= e^{e^{it}+ e^{-it}}= e^{\frac{e^{2it}+ 1}{e^{it}}[/tex]
If you let u= eit then du= ieitdt so -idu/u= dt.
 

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