Integrating e^(e^(x))

[Doonami]

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

\ointe (z + [1/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:

\intc f(z)dz = \intf(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?

[tiny-tim]

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

\ointe (z + [1/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:

\intc f(z)dz = \intf(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:

[HallsofIvy]

Yes. z= eit so ez+ 1/z becomes
e^{z+ 1/z}= e^{e^{it}+ e^{-it}}= e^{\frac{e^{2it}+ 1}{e^{it}}
If you let u= eit then du= ieitdt so -idu/u= dt.