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You are not logged in. #1 20110720 17:02:43
A tough integral!The problem begins here: To evalute this we need the poles which are the zeroes of the denominator. This is the first obstacle. None of the packages can factor the 12th degree denominator. They can do it numerically but this is a case where the exact answer is needed. Without getting too heavy into it the reason they can not is because they all are set by default to the field of integers. So when we try to factor we get, That is not good enough. When we use Factor[x^1210 x^10+37 x^842 x^6+26 x^48 ,Extension>{I}] We get: This is much better, we can now get the poles exactly, they may be long and ugly but they will be exact. The roots ( poles ) are: There are 4 groups of these, consisiting of the conjugates and the negations. We take only the poles that are in the upper half of the contour. Finding the residues of the six poles that count, for that we use the residue command. Summing the residues: This method evaluates integrals from ∞ to ∞, ours is from 0 to ∞, since the integral is symmetric the answer is: This is how a human does it: http://www.mat.uniroma2.it/~tauraso/AMM/AMM11148.pdf In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. 