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Re: Keeping it real

  • To: mathgroup at smc.vnet.net
  • Subject: [mg120286] Re: Keeping it real
  • From: Sebastian Hofer <sebhofer at gmail.com>
  • Date: Sat, 16 Jul 2011 05:43:37 -0400 (EDT)
  • Reply-to: comp.soft-sys.math.mathematica at googlegroups.com

On Friday, July 15, 2011 3:23:35 AM UTC+2, amannuc wrote:
> I am facing perhaps the "age-old" question of limiting the indefinite
> integrals that Mathematica returns. In particular, I am trying to
> avoid complex numbers. I have tried using "Assumptions" in the Integrate
> command to no avail.
>
> The specific integral is:
>
> SetAttributes[{c0, d0}, {Constant}] (* Probably redundant. See Block
> function *)
> Simplify[Block[{a, c0, d0},
>   2.0 * a *
>    Integrate[
> ((-d0/z^2)/(c0 + d0/z))/Sqrt[-a^2 + z^2 (c0 + d0/z)^2],
>     z,
> Assumptions -> {z \[Element] Reals, a \[Element] Reals,
>       c0 \[Element] Reals, d0 \[Element] Reals}]]]
>
>
>
> Admirably, Mathematica returns an answer, although with complex
> numbers. I verified the answer is correct by taking its derivative.
> The returned answer is:
> (0. - 2. I) Log[-((2 d0 (-I a + Sqrt[-a^2 + (d0 + c0 z)^2]))/(
>     d0 + c0 z))] + ((0. + 2. I) a Log[(
>    2 d0 ((I (-a^2 + d0 (d0 + c0 z)))/Sqrt[a^2 - d0^2] +
>       Sqrt[-a^2 + (d0 + c0 z)^2]))/z])/Sqrt[a^2 - d0^2]
>
> Note all the "I"s. This is a fairly simple answer returned. I doubt
> that is the only solution. When I evaluate it for cases of interest, I get
> complex numbers, which can't be right in my particular case which is
> based on a physical problem.
>
>
> Thanks for any insights.
>
> --
> Tony Mannucci

The integrand will be complex whenever -a^2 + z^2 (c0 + d0/z)^2<0, as the square root in the denominator will then yield complex values. It is therefore not surprising that the general solution of the integral has the given form. I haven't checked explicitly, but I'm pretty sure that the result given by Mathematica will reduce to real values whenever above condition is violated.

Best regards,
Sebastian


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