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Calculating residue

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


Calculate the residu for the singularity of [tex] \frac{z \sin{z}}{\left( z - \pi \right)^3} [/tex]


Homework Equations


[tex]R \left( a_0 \right) = \frac{1}{2 \pi i} \oint \frac{z \sin{z}}{\left( z - \pi \right)^3} dz[/tex]


The Attempt at a Solution


[tex]\pi [/tex] is an essential singularity so the residue cannot be calculated using the limit.
I have tried calculating the integral around the path from 3-i to 4-i to 4+i to 3+i to 3-i. In principle this should work, but Maple gives me a wrong answer if I try to evaluate.

[tex]\frac{1}{2 \pi i} \left( \int^4_3 \frac{\left( x - i \right) \sin{x - i}}{\left( x - i - \pi \right)^3} dx + \int^1_{-1} \frac{\left( 4 + iy \right) \sin{4 + iy}}{\left( 4 + iy - \pi \right)^3} dy + \int^3_4 \frac{\left( x + i \right) \sin{x + i}}{\left( x + i - \pi \right)^3} + \int^{-1}_1 \frac{\left( 3 + iy \right) \sin{3 + iy}}{\left( 3 + iy - \pi \right)^3} dy \right)[/tex]
 
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Answers and Replies

  • #2
vela
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How did you conclude that [itex]z=\pi[/itex] is an essential singularity?
 
  • #3
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I was wrong on that. Calculating the residue using the limit works fine, but my problem stands. The integral should give the same result.
 
  • #4
vela
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You're missing a factor of i on the y integrals since dz = i dy.
 
  • #5
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It works fine thanks!

Is there a better choice that makes it easier to calculate the integral?
 
  • #6
vela
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Perhaps let x run from 0 to 2 pi to take advantage of the periodicity of sin z, but I'd try to avoid the contour integrals in the first place.
 

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