Contour integral with exponential in the denominator

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SUMMARY

The discussion focuses on evaluating the contour integral \(\int_{C}\frac{e^{pz}}{1+e^z}dz\) to compute the real integral \(\int^{\infty}_{- \infty}\frac{e^{px}}{1+e^x}dx\) for \(0 PREREQUISITES

  • Complex analysis, specifically contour integration
  • Residue theorem and its applications
  • Understanding of poles and residues in complex functions
  • Familiarity with l'Hôpital's rule for limit evaluation
NEXT STEPS
  • Study the residue theorem in detail, focusing on its application in complex integrals
  • Learn about evaluating limits using l'Hôpital's rule in complex analysis
  • Explore examples of contour integrals involving exponential functions
  • Review the properties of poles and their significance in complex function theory
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Students and professionals in mathematics, particularly those studying complex analysis, as well as anyone interested in advanced integration techniques and their applications in evaluating real integrals.

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Homework Statement


Use the contour integral
\int_{C}\frac{e^{pz}}{1+e^z}dz​
to evaluate the real integral
\int^{\infty}_{- \infty}\frac{e^{px}}{1+e^x}dx​
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


\oint f(z)dz = 2 \pi i \sum Res[f(z)]
Res[f(z=z_{0})] = (m-1)! \frac{d^{m-1}}{dz^{m-1}}(z-z_{0})^{m}f(z)|_{z=z_{0}}
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.
Res[f(z=i \pi)] = (z-i \pi) \frac{e^{pz}}{1+e^z}|_{z=i \pi}
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.
 
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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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