Re: Keeping it real

*To*: mathgroup at smc.vnet.net*Subject*: [mg120287] Re: Keeping it real*From*: Heike Gramberg <heike.gramberg at gmail.com>*Date*: Sat, 16 Jul 2011 05:43:48 -0400 (EDT)*References*: <201107150120.VAA23674@smc.vnet.net>

If you use Refine with appropriate Assumptions, you get a slightly better result: Block[{a, c0, d0}, Block[{$Assumptions = 0 < a < d0 && 0 < a < d0 + c0 z, pos}, int = Expand@ Refine[2*a* Integrate[((-d0/z^2)/(c0 + d0/z))/ Sqrt[-a^2 + z^2 (c0 + d0/z)^2], z]] ]] output: -2 I Log[-((2 d0 (-I a + Sqrt[-a^2 + (d0 + c0 z)^2]))/( d0 + c0 z))] + ( 2 a Log[(2 d0 ((-a^2 + d0 (d0 + c0 z))/Sqrt[-a^2 + d0^2] + Sqrt[-a^2 + (d0 + c0 z)^2]))/z])/Sqrt[-a^2 + d0^2] Note that the second Log-term is now real (provided the parameters satisfy the assumptions made). The first Log-term can be simplified a bit further by using ComplexExpand. Since I want to leave the second Log as it is, I just extract the first Log and apply ComplexExpand: result = Block[{a, c0, d0, int}, Block[{$Assumptions = 0 < a < d0 && 0 < a < d0 + c0 z, pos}, int = Expand@ Refine[2*a* Integrate[((-d0/z^2)/(c0 + d0/z))/ Sqrt[-a^2 + z^2 (c0 + d0/z)^2], z]]; pos = Position[int, -2 I Log[___]][[1]]; ReplacePart[int, pos -> Expand@Simplify@ComplexExpand[Extract[int, pos]]]] ] Output: 2 Arg[I a - Sqrt[-a^2 + (d0 + c0 z)^2]] - 2 I Log[2 d0] + ( 2 a Log[(2 d0 ((-a^2 + d0 (d0 + c0 z))/Sqrt[-a^2 + d0^2] + Sqrt[-a^2 + (d0 + c0 z)^2]))/z])/Sqrt[-a^2 + d0^2] Which is real except for the constant term -2 I Log[2 d0]. Since an indefinite integral is unique up to an arbitrary constant, you can add 2 I Log[2 d0] to get a real solution. Heike. On 15 Jul 2011, at 02:20, 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 >

**References**:**Keeping it real***From:*amannuc <amannuc@yahoo.com>