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April 17th, 2010, 01:02 PM   #1
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Double integrad into geometric series problem

Hey all,

I have a double integrand 1/(1-xy) [0,1], [0,1] . My task is to show it is equivalent to series 1/n^2 where n=1 , goes to infinity. I know you have to turn it into a geometric series, which involves 1/1-u and u=xy but I don't know much else. Thanks in advance.
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April 18th, 2010, 06:22 AM   #2
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For -1 ? u < 1, 1/(1 - u) = 1 + u + u + u + . . ., a geometric series.
Hence for -1 ? xy < 1, 1/(1 - xy) = 1+ xy + xy + xy + . . ., and you should be able to evaluate the double integral of each term in that series quite easily.
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