Delayed evaluation of integral limits
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Ian Bell
Feb 13, 2015, 5:00:05 PM2/13/15
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I am trying to do this integration
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which should turn into -log(1-z), but I can't figure out how to make it do that. Manually we can do indefinite integral to get -log(z-1), substitute limits to get -log(z-1)-(-log(-1)), combine to get -log(1-z). But sympy insists on the result being imaginary. I see why, but the simplification should remove the log(-1), right?
Ian
which should turn into -log(1-z), but I can't figure out how to make it do that. Manually we can do indefinite integral to get -log(z-1), substitute limits to get -log(z-1)-(-log(-1)), combine to get -log(1-z). But sympy insists on the result being imaginary. I see why, but the simplification should remove the log(-1), right?
Ian
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