Category: Science > Math
Asked by: real_madras-ga
List Price: $5.00
30 May 2006 10:28 PDT
Expires: 29 Jun 2006 10:28 PDT
Question ID: 733691
How do you integrate log(z - e^(ix))?
|There is no answer at this time.|
From: dgfinck-ga on 02 Jun 2006 20:59 PDT
I x Integrate[Log[z - E ], x] = -I 2 I x -- x + x Log[-E + z] - 2 I x E 2 I x Log[1 - ----] -x z I z (--- - ----------------- - 2 z z I x E PolyLog[2, ----] z ----------------) z Answer provided by: http://integrals.wolfram.com/ Go there, enter "Log[z-E^(I*x)]" and click "Compute" to see the result in a better (non-ASCII) format. Regards, df
From: postlp60-ga on 04 Jun 2006 17:46 PDT
using the taylor series you get e^ix = cosx +isinx, this may make is easier to integrate. Also trying using intergrator.com. I swear by it. Hope this helps. Preston
From: berkeleychocolate-ga on 06 Jun 2006 11:18 PDT
Would you please clarify this question? You could mean log(z - exp( i Re(z))), but then you would have to describe the contour. Or you could simply have made a mistake using z and x and want the indefinite integral of log( x - exp (ix)). Or ... ?
If you feel that you have found inappropriate content, please let us know by emailing us at firstname.lastname@example.org with the question ID listed above. Thank you.
|Search Google Answers for|