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May 14th, 2012, 10:50 AM   #1
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another somewhat challenging contour integral?.

Here's another integral I ran across.



I thought perhaps a rectangular contour may be in order, but turns out maybe a semi-circular one after all?.

I tried rewriting it as

cosh has an infinite number of poles .

is a pole of as well as for cosh(x).

Anyone have a good method for this one?. I am not certain I am doing it correctly due to cosh and the rational part having poles at the same place. Namely, .

Here is a link to a similar problem:

http://129.81.170.14/~vhm/papers_html/final21.pdf

The residue at they get .

I get close, but I do not know from where they get that residue. They do not explain...just show it.
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May 15th, 2012, 10:37 AM   #2
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Re: another somewhat challenging contour integral?.



For k=2 I get f.i.




http://www.wolframalpha.com/input/?i=Re ... a%29%7D%5D
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May 15th, 2012, 11:44 AM   #3
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Re: another somewhat challenging contour integral?.

...
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May 15th, 2012, 01:39 PM   #4
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Re: another somewhat challenging contour integral?.

If we make the sub in the original integral, we get:



cosh and both have a pole at x=i and cosh at

For i the residue is



Multiply by the and we get

The other 0's of cosh are at . Where k=0 would correspond to i, the one we just found.

So, for the other poles for k=1,2,3,....

the residues are at



Taking the sum, beginning at 1, and summing them up we get:



Now, do not forget to add the k=0 for i from before and we get:



Since this is an even function, divide by 2 and get
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