Complicated double integral

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  • #26
hunt_mat
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##\frac 1 k (\cosh(ka)-1)##
Go figure.
This solution doesn't look right to me.
 
  • #27
PeroK
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This solution doesn't look right to me.
I figured as much. DIY
 
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  • #28
hunt_mat
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I figured as much. DIY
I did, or what I thought looked sensible. My solution doesn't look right to you, and yours doesn't look right to me.
 
  • #29
hunt_mat
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Thank you for your help thought PeroK.
 
  • #30
Vanadium 50
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This has been painful to watch. PeroK has been giving excellent advice.

  • First, split the integral into two pieces, one where the delta function is zero everywhere and one where it is not.
  • Do the inner integral. The first part (above) is zero and the second part (above) sets s = -a. The only q left should be inside the sinh.
  • Set a new variable r = q + a. Set you limits in terms of r.
  • Do the outer (and only remaining) integral. I believe you will have only one a left.
 
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  • #31
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Thank you.
 
  • #32
hunt_mat
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This has been painful to watch. PeroK has been giving excellent advice.

  • First, split the integral into two pieces, one where the delta function is zero everywhere and one where it is not.
  • Do the inner integral. The first part (above) is zero and the second part (above) sets s = -a. The only q left should be inside the sinh.
  • Set a new variable r = q + a. Set you limits in terms of r.
  • Do the outer (and only remaining) integral. I believe you will have only one a left.
Actually I did that, and I posted the function. Why was that incorrect?
 
  • #33
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You can do it pretty easily just be inspecting the integral limits and looking at the interval on which the delta function is nonzero. More specifically, changing the lower bound on the outer integral to ##-a## projects out the integration interval on which the delta function is "satisfied".
 

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