Wikipedia:Reference desk/Archives/Mathematics/2016 July 16

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July 16

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Help with Integral

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Hi all, how to solve the integral  ? 31.154.81.56 (talk) 07:51, 16 July 2016 (UTC)[reply]

The problem is the term   so try using the substitution   to simplify. Dmcq (talk) 08:01, 16 July 2016 (UTC)[reply]
How should I use it, if there's no multiplication by the derivative  ? 31.154.81.56 (talk) 13:24, 16 July 2016 (UTC)[reply]
I don't understand your question, there is a multiplication by 1/y at one stage, what problem do you have with that? Dmcq (talk) 13:36, 16 July 2016 (UTC)[reply]
As far as I know  , but this is not the case here. Am I wrong? Or maybe you meant something else? 31.154.81.56 (talk) 13:45, 16 July 2016 (UTC)[reply]
You forgot the part   and hence you end up with   instead. Factor the denominator to get   and apply a partial fraction decomposition.--Jasper Deng (talk) 15:25, 16 July 2016 (UTC)[reply]
It took to me some time to understand my mistake. Thank you both for your explanations! 31.154.81.0 (talk) 18:57, 17 July 2016 (UTC)[reply]


Lazy people like me will do the partial fraction expansion as follows. The singularity at   gives rise to the term   the coefficient of -1 here is obtained by taking the limit of   times the integrand for y approaching -1, which is trivial to evaluate. The singularity at   will produce the remaining part of the partial fraction expansion, we can obtain this effortlessly by using the fact that the integrand decays like   for large   . This means that the remaining part of the partial fraction expansion must eliminate the asymptotic terms for large   coming from   that decay less fast for large   . For large   , we have:
 

The partial fraction expansion is thus given by:

 

Count Iblis (talk) 17:39, 16 July 2016 (UTC)[reply]

A simpler solution:
 
Ruslik_Zero 13:47, 17 July 2016 (UTC)[reply]

Thank you all for the detailed solutions! It was really helpful for me! 31.154.81.0 (talk) 18:57, 17 July 2016 (UTC)[reply]