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MathGroup Archive 2006

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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
  • References: <ejosmm$n3k$1@smc.vnet.net>

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[90]:= Assuming[x < 0, Integrate[1/x, x]]

Out[90]= 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[30]/(2 Sqrt[30]) I.

Thus, if you just add 11 Log[30]/(2 Sqrt[30]) I to Mathematica's result,
you have an antiderivative which is real for all real x.

David


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