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I am getting an imaginary unit in the output after Abs and saying Complex Expand, why? Then, when I integrate in x and z, I miss beta. I don't get why beta vanishes in the definite integral.

IntegrandSerie[beta_, x_, z_] = 
ComplexExpand@
Abs[Series[(1 - (1 - I)/2*Sqrt[beta/2]*
      Cosh[(1 - I)*z*
         Sqrt[beta/2]]/(Sinh[(1 - I)/2*Sqrt[ beta/2]]))*(dWNorm[
     x]) + (Z0 I beta dWNorm1/
      2)*(Sinh[(1 - I) Sqrt[
        beta/2] z]/((1 - I) Sqrt[
         beta/2] Cosh[(Sqrt[beta/2] (1 - I)/2)] - 
       2 Sinh[Sqrt[beta/2] (1 - I)/2]))*(S[x]), {beta, 0, 1}]]^2
IntegralSerie = 
Integrate[IntegrandSerie[beta, x, z], {x, -1, 1}, {z, -1/2, 1/2}]
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  • $\begingroup$ Look up the TargetFunctions option of ComplexExpand[]. $\endgroup$ – J. M. will be back soon Apr 11 '17 at 15:26
  • $\begingroup$ @J.M. I have checked and inserted TargetFunctions->{Abs, Arg} but still I get that imaginary unit. $\endgroup$ – Andrea G Apr 11 '17 at 15:34
  • $\begingroup$ I suppose it may be because it's a series. Try ComplexExpand[Abs[s], {_SeriesData}] for your series s. I think without the extra argument, it converts Abs[s] to Sqrt[s^2]. $\endgroup$ – Michael E2 Apr 11 '17 at 16:06
  • $\begingroup$ @MichaelE2 what is the part {_SeriesData}? And what do you mean by extra argument? Shall I substitute something in SeriesData or just copy? $\endgroup$ – Andrea G Apr 11 '17 at 16:28
  • 1
    $\begingroup$ @MichaelE2 Mmmm, I am a bit new to Mathematica.Regarding my case, without entering to deep in the documentation, shall I substitute for s my expression and for SeriesData the {beta,0,1}? $\endgroup$ – Andrea G Apr 11 '17 at 16:39

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