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I am doing some complicated series expansions involving integrals and whatnot, but everything is real the whole time (and I specify that using Assumptions at every step). Variants of the expression $$(1-e^2)^{3/2}$$ pop up frequently (0 < e < 1); however, Mathematica keeps representing it as $$I*(-1+e^2)^{3/2}$$ instead, which is messing up some steps further down the line.

Is there a simple reason for this behavior or a way to stop/reverse it?

Edit: Here is the specific example, if it helps:

func1[p_, e_, chi2_] := (p^2 [(-4 e^2 + (-2 + p)^2)/(-6 + p - 
2 e Cos[chi2])]^(1/2))/((-2 + p - 2 e Cos[chi2]) (1 + e Cos[chi2])^2);

func2[p_, e_, chi2_] := Normal[Series[func1[p, e, chi2], {p, Infinity, 3}]];

result= Series[(p)^(3/2)*2*Pi/Integrate[func2[p, e, chi2], {chi2, 0, 2*Pi}, 
 Assumptions -> Element[e | p, Reals] && e > 0 && p > 0 && e < 1], {p, Infinity, 2}]

This gives

I (-1 + e^2)^(3/2) + (3 I (-1 + e^2)^(5/2))/p + (3 I (-1 + e^2)^(3/2) (2 - 8 
e^2 + 6 e^4 - 5 Sqrt[1 - e^2] + 5 e^2 Sqrt[1 - e^2]))/(2 p^2)

$$I (-1 + e^2)^{3/2} + (3 I (-1 + e^2)^{5/2})/p + (3 I (-1 + e^2)^{3/2} (2 - 8 e^2 + 6 e^4 - 5 (1 - e^2)^{1/2} + 5 e^2 (1 - e^2)^{1/2}))/(2 p^2)$$

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2 Answers 2

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You don't say explicitly or give an example of how the form arises. It does not happen automatically for me when I enter (1 - e^2)^(3/2). So it's hard to guess what might prevent it. But here is a way to reverse the factoring out of I:

expr = I*(-1 + e^2)^(3/2);
Simplify[expr,
 Assumptions -> 0 < e < 1,
 ComplexityFunction -> (LeafCount[#] + 100 Count[#, Complex[0, _], Infinity] &)]

(*  (1 - e^2)^(3/2)  *)

On OP's updated example:

Simplify[result, Assumptions -> 0 < e < 1,
 ComplexityFunction -> (LeafCount[#] + 100 Count[#, Complex[0, _], Infinity] &)]

Mathematica graphics

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  • $\begingroup$ It happens if you use E rather than e. @cmm0052, that's one of the pitfalls of rendering things in LaTeX rather than Mma code (as is the preferred convention here). $\endgroup$
    – evanb
    Apr 9, 2017 at 16:14
  • $\begingroup$ @evanb Thanks. I considered whether E was meant, but I assumed the OP meant e since the OP explicitly writes "(0 < e < 1)". $\endgroup$
    – Michael E2
    Apr 9, 2017 at 16:21
  • $\begingroup$ @ebanb -- this doesn't make sense. If the OP meant E, then 0<E<1 is impossible (since E $\approx$ 2.73) $\endgroup$
    – bill s
    Apr 9, 2017 at 17:46
  • $\begingroup$ I have added the specific example. "e" is a variable, but not the one being expanded over. Sorry for the confusion! $\endgroup$
    – cmm0052
    Apr 9, 2017 at 17:53
  • $\begingroup$ @cmm0052 Thanks. I think you changed a Sqrt to a 1/2 power, but forgot to remove the square brackets (typo in your edit). $\endgroup$
    – Michael E2
    Apr 9, 2017 at 17:57
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I often use something like the following to protect subexpressions that I don't want to be changed by simplification:

(1 - e^2)^(3/2) /. e^2 -> 1 - HoldForm[1 - e^2];

This protection can be removed, if necessary using ReleaseHold

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