sigmaxi, the former integral may be wrong,
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作者: paciii ® 02/18/2009, 23:29:45 作者简介 电邮作者 修改

the correct way may be: Integrate the two terms together: expand it and get: intg [(-16n^4-16n^3-4n^2+4n+1)/(64n^6+96n^5+48n^4+24n^3+12n^2+6n+2)]dn, n=0,inf It's integrable, but I can't remeber how to do it. And there may exist better way of integrating it. The expression is at 1/n^2, so definitely convergence.

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