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?
\[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]
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