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Integrating along the imaginary axis...

 
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Dec16-10, 12:40 PM   #1
 

Integrating along the imaginary axis...


I'm really confused with how to prove this result...could anybody help please?

Let I_{1} be the line segment that runs from iR (R>0) towards a small semi-circular indentation (to the right) at zero of radius epsilon (where epsilon >0) and I_{2} a line segment that runs from the indentation to -iR.

Define

f(z)=\frac{e^{2\pi iz^{2}/m}}{1-e^{2\pi iz}}

Prove that

I_{1}+I_{2}=-i\intop_{\varepsilon}^{R}e^{-2\pi iy^{2}/m}dy

How can I do this?
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Dec16-10, 01:53 PM   #2
 
Quote by Cairo View Post
Define

[tex]f(z)=\frac{e^{2\pi iz^{2}/m}}{1-e^{2\pi iz}}[/tex]

Prove that

[tex]I_{1}+I_{2}=-i\intop_{\varepsilon}^{R}e^{-2\pi iy^{2}/m}dy[/tex]

How can I do this?
First, draw out the contour. It's clear you're going to have to work from the definition of the complex line integral. So, how would you parametrize I_1 and I_2?
Dec16-10, 02:07 PM   #3
 
Would it be kiR for -oo<k<00 ?
Dec16-10, 02:11 PM   #4
 

Integrating along the imaginary axis...


I_1 is a segment running from iR to [itex]i\varepsilon[/itex] right? So it wouldn't make sense for your parameter to range over the reals.
Dec16-10, 02:34 PM   #5
 
Hmmmm...

I'm not sure how to proceed here then. Would it not be valid to have kiR where
oo<k<epsilon?

This would give the line segment, right?
Dec16-10, 02:40 PM   #6
 
I'm talking nonsense! Forgive me!

Would it be ik where R<k<epsilon?
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