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Contour integral with exponential in the denominator

  • Thread starter jncarter
  • Start date
  • #1
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Homework Statement


Use the contour integral
[itex]\int_{C}\frac{e^{pz}}{1+e^z}dz[/itex]​
to evaluate the real integral
[itex]\int^{\infty}_{- \infty}\frac{e^{px}}{1+e^x}dx[/itex]​
0<p<1
The contour is attached.
It is a closed rectangle in the positive half of the complex plane. It height is 2i∏.


Homework Equations


[itex]\oint f(z)dz = 2 \pi i \sum Res[f(z)] [/itex]
[itex] Res[f(z=z_{0})] = (m-1)! \frac{d^{m-1}}{dz^{m-1}}(z-z_{0})^{m}f(z)|_{z=z_{0}}[/itex]
where m is the degree of the pole.



The Attempt at a Solution


I found that there was a simple pole at z = i∏, so I must use the residue theorem to find the value of the complex integral.
[itex] Res[f(z=i \pi)] = (z-i \pi) \frac{e^{pz}}{1+e^z}|_{z=i \pi}[/itex]
In the past I've had to fiddle with the denominator to get the z-z0 terms to cancel out, but in those cases it involved something nice and simple. I have no idea what to do with this equation.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
The residue is defined as the LIMIT of your expression as z->i*pi, it's not the value of the expression at z=i*pi (unless you do your magic cancellation). Try using l'Hopital to find the limit.
 

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