Re: Why does this lead to an answer with complex numbers?
- To: mathgroup at smc.vnet.net
- Subject: [mg71465] Re: Why does this lead to an answer with complex numbers?
- From: "David W. Cantrell" <DWCantrell at sigmaxi.net>
- Date: Mon, 20 Nov 2006 02:43:46 -0500 (EST)
- Organization: NewsReader.Com Subscriber
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aaronfude at gmail.com wrote: > The expression is > > \!\(FullSimplify[ > Assuming[\[Beta] > 0 && \[Beta] < Pi/2, > Integrate[\(-Log[\@\(1 + x\^2\) - 1/11*x\ ]\), \ x]]]\) Maybe you intended to use x rather than Beta in your assumption. In any event, to answer the question posed in your title: Mathematica will sometimes gives complex answers, even when you wish they were real, and using assumptions may not help. Consider the simple example In:= Assuming[x < 0, Integrate[1/x, x]] Out= Log[x] Since x < 0, Log[x] will be complex. Students in beginning calculus would likely have given the antiderivative as Log[Abs[x]]. Back to your example, the antiderivative given by Mathematica is complex, as you noted, but the imaginary part is merely _constant_, namely - 11 Log/(2 Sqrt) I. Thus, if you just add 11 Log/(2 Sqrt) I to Mathematica's result, you have an antiderivative which is real for all real x. David