Contour integration

  • Thread starter Wiseman101
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  • #1

Homework Statement



By an appropriate choice of contour, validate the following relationship by evaluation,
attachment.php?attachmentid=25455&stc=1&d=1272478598.jpg



Homework Equations



attachment.php?attachmentid=25456&stc=1&d=1272479130.png


The Attempt at a Solution


Any help on this one would be much appreciated.
 

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Answers and Replies

  • #2
Cyosis
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I'm afraid copy pasting the relevant equations don't constitute as a attempt at a solution. What type of contour are you thinking of and do you know how to find the poles?

Edit: I believe the answer should be [itex]\pi/2[/itex]
 
Last edited:
  • #3
gabbagabbahey
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I'd start by finding all the singularities of the integrand...where are they? What type of singularities are they?
 
  • #4
Put in z in place of the x's. The singularities are when the bottom line is zero, so theres a singularity at z = i and z = -i. The contour im thinking would be a semi circle on the positive side of the axis so -i isnt needed so z = i is one simple pole. I dont know how to handle the z^4 though.
 
  • #5
Cyosis
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That is correct so far. You handle the z^4 term in the same manner, [itex](z^2+1)(z^4+1)=0[/itex] gives you [itex](z^2+1)=0[/itex] and [itex](z^4+1)=0[/itex]. Solve for z.
 
  • #6
Ok, so i split ([tex]z^{2}[/tex] + 1) into (z+i)(z-i) to get the first pole. So am i right in spliting ([tex]z^{4}[/tex] + 1) into ([tex]z^{2}[/tex] + i)([tex]z^{2}[/tex] - i)? Do i then split each of those again?
 
  • #7
Cyosis
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Don't need to split them further, just solve for z now.
 
  • #8
Is z = [tex]\sqrt{i}[/tex] the other simple pole? I tried to do it out using that but i didnt get out [tex]\frac{\pi}{6}[/tex] as the final answer
 
  • #9
Cyosis
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You should have a total of three poles within the contour, [itex]\sqrt{i}[/itex] is one of them.

As for not getting pi/6, read post #2.
 
  • #10
Got it! Thanks a lot for all your help.
 

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