tag:blogger.com,1999:blog-7379110960796014170.post5235693161569961385..comments2023-11-03T05:13:28.920-04:00Comments on Clueless Fundatma: Mathematica Integral FunSachin Shanbhaghttp://www.blogger.com/profile/08932887228149182854noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-7379110960796014170.post-23584222528721711802012-01-07T08:52:49.975-05:002012-01-07T08:52:49.975-05:00Yeah, the last statement would give you an answer ...Yeah, the last statement would give you an answer with epsilon as a parameter. As you say, one would have to take the limit to evaluate the integral. That is essentially what I meant by the last line "We could isolate the discontinuity by integrating close to zero (using "epsilon"), and <b> perhaps take the limit later </b>."Sachin Shanbhaghttps://www.blogger.com/profile/08932887228149182854noreply@blogger.comtag:blogger.com,1999:blog-7379110960796014170.post-6792573220683933952012-01-07T02:33:59.656-05:002012-01-07T02:33:59.656-05:00Consider wrapping that last step in a Limit as eps...Consider wrapping that last step in a Limit as epsilon->0. Otherwise there does not seem to be any hint given to Mathematica that you are assuming epsilon is small.Anonymousnoreply@blogger.com