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As described in title, there is an expression like

X=(0.03 ((0.00565508 - 
      4.25452 I) + (0.149772 - 55.6062 I) \[Beta] + \[Beta]^2 - (0. + 
       0.0765116 I) \[Omega] - I \[Beta] \[Omega]))/((236.578 + 
    3092.07 \[Beta] + 0.223032 \[Beta]^2 + \[Beta]^3 + 
    8.50904 \[Omega] + 111.212 \[Beta] \[Omega] + 
    0.0765116 \[Omega]^2 + \[Beta] \[Omega]^2) (0.0765116 + 
    0.0381921/(1 + 186.323 (55.6062 + \[Omega])^2))) - (
 0.000793435 ((0.00073138 - 0.555136 I) + 
    0.00998334 \[Beta] - (0. + 0.00998334 I) \[Omega]))/((236.578 + 
    3092.07 \[Beta] + 0.223032 \[Beta]^2 + \[Beta]^3 + 
    8.50904 \[Omega] + 111.212 \[Beta] \[Omega] + 
    0.0765116 \[Omega]^2 + \[Beta] \[Omega]^2) (0.0732601 - 
    I (55.6062 + \[Omega])) (0.0765116 + 0.0381921/(
    1 + 186.323 (55.6062 + \[Omega])^2)))

And then Integrate[X \[Rho][\[Omega]], \[Omega]] cannot output. But if assigning one arbitrary value to \beta like Integrate[X \[Rho][\[Omega]] /. {\[Beta] -> .1 I + .2}, \[Omega]], the result can be outputted immediately

(0. + 0.784195 I) (1. \[Omega] - (0.0000262829 + 
      2.20158*10^-6 I) ArcTan[0.273419/(
     55.5062 + 1. \[Omega])] + (0.00158507 + 0.0017454 I) ArcTan[
     0.0732601/(
     55.6062 + 1. \[Omega])] - (3.11538 + 6.99225 I) ArcTan[0.0897/(
     55.6062 + 1. \[Omega])] + (1.78695 + 69.8793 I) ArcTan[0.273419/(
     55.7061 + 1. \[Omega])] - (1.10079*10^-6 - 0.0000131414 I) Log[
     3081.02 + 111.012 \[Omega] + 1. \[Omega]^2] + (0.000872699 - 
      0.000792533 I) Log[
     3092.05 + 111.212 \[Omega] + 1. \[Omega]^2] + (0.76366 - 
      0.756408 I) Log[
     3092.06 + 111.212 \[Omega] + 1. \[Omega]^2] - (34.9397 - 
      0.893475 I) Log[3103.25 + 111.412 \[Omega] + 1. \[Omega]^2]

Where \[Rho][\[Omega]]=136.5 - 2 (55.6062 + \[Omega]).

So how to integrate with \beta unassigned?

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    $\begingroup$ Change the definition to \[Rho][\[Omega]_] := 136.5 - 2 (55.6062 + \[Omega]) . But it remains unlikely that the integral is evaluable, because integrand includes possible singularities depending on \[Beta] $\endgroup$ Commented Jul 11, 2022 at 7:00

1 Answer 1

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And then Integrate[X [Rho][[Omega]], [Omega]] cannot output.

Use exact input. It is always better to use exact input with exact functions.

X=SetAccuracy[X,Infinity];
ρ[ω]=136.5-2 (55.6062+ω)//Rationalize
Integrate[X ρ[ω],ω]

Gives

1/2500I ((
  675539944105574375 ((1001712246397957120 + 1319736636244501 I) + 
     18014398509481984 ω))/6207358442523020510609484546048
.....

Output too large....

V 13.1

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