The Limit of a Complex Integral

  • Thread starter Bachelier
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  • #1
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Though it is not homework I posted this here, hopefully it'll get more action. Thanks

given [tex]\int_{\Lambda} \frac{e^{i w}}{w^2 + 1} \mathrm{d}w [/tex] where ##w \in \mathbb{C}## with ##Im(w) \geqslant 0## and ##|w| = \xi##

want to evaluate the behavior of the Integral as ##\xi \rightarrow \infty##

So I have [tex]\left|\int_{\Lambda} \frac{e^{i w}}{w^2 + 1} \mathrm{d}w \right| \leqslant \int_{\Lambda} \left|\frac{e^{iw}}{w^2 + 1} \right| \mathrm{d}w [/tex]

we have ##|exp(iw)| \leqslant 1##

so we're left with the denominator. We can get:

## \left| \frac{1}{w^2 + 1} \right| \leqslant \frac{1}{\xi^2 - i^2}##

Hence the integral is:

[tex]\left|\int_{\Lambda} \frac{e^{i w}}{w^2 + 1} \mathrm{d}w \right| \leqslant \frac{ \pi \xi}{\xi^2 + 1} [/tex]

Thus it goes to ##0## if ##\xi## goes to ##\infty##
 
Last edited:

Answers and Replies

  • #2
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What are you integrating over? By saying |w| > 0 IMw > 0 are you indicating we are integrating over a semicircle?
 
  • #3
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What are you integrating over? By saying |w| > 0 IMw > 0 are you indicating we are integrating over a semicircle?

Exactly, my region ##\Lambda## is where the Imaginary part of ##w## is positive. So we are in the upper half-circle.
 
  • #4
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And I am going to let ##\xi## explode, to use a Physicist term. :smile:
 
  • #5
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Nevermind. You're not enclosing the real axis.
 
  • #6
jbunniii
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Your estimate and conclusion look fine to me.
 

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