The solution in the question provided by Integrate
can be simplified by eliminating ConditionalExpression
but otherwise cannot be simplified much. It is, however, possible to broaden the applicability of the solution substantially. To begin, note that c1
is a quadratic polynomial in ω^2
divided by cubic polynomial in ω^2
. Consequently, the integral can be evaluated by closing the contour of integration at large Abs[ω]
in the upper half of the complex ω
plane and then summing the residues of the enclosed poles of c1
. Unfortunately, Mathematica 11.3 seems to have difficulty recognizing which poles actually are enclosed. As noted in the question,
s = Integrate[c1, {ω, -∞, ∞}];
is too long to be reproduced here. However, the structure of s
can be presented succinctly by
rts = Cases[s, _Root, Infinity] // Union
which returns as Root
functions the three solutions for ω^2
from Denominator[c1] == 0
. Replacing them in s
by
ss = s /. Root[_, n_] -> rt[n]
(* ConditionalExpression[
-((I Pi Γ ((4 g2^2 + κ1 κ2)^2 (Sqrt[rt[2]] + Sqrt[rt[3]]) + 16 rt[1] (rt[2]
Sqrt[rt[3]] + Sqrt[rt[2]] rt[3]) + Sqrt[rt[1]] ((4 g2^2 + κ1 κ2)^2 +
4 (8 g2^2 - κ1^2 - κ2^2) Sqrt[rt[2]] Sqrt[rt[3]] + 16 rt[2] rt[3])))/
(16 Sqrt[rt[1]] (Sqrt[rt[1]] + Sqrt[rt[2]]) Sqrt[rt[2]] (Sqrt[rt[1]] +
Sqrt[rt[3]]) (Sqrt[rt[2]] + Sqrt[rt[3]]) Sqrt[rt[3]])),
Im[Sqrt[rt[1]]] < 0 && Im[Sqrt[rt[2]]] < 0 && Im[Sqrt[rt[3]]] < 0] *)
The condition given is that the three poles Sqrt[rt[_]]
lie in the lower half of the complex plane. But, what about the other three poles, given by - Sqrt[rt[_]]
? Now consider a specific solution, obtained by direct integration.
Integrate[c1 /. {g1 -> 1, g2 -> 7, κ1 -> 3, κ2 -> 4, Γ -> 5}, {ω, -∞, ∞}]
N[%]
(* (10220 Pi)/5247 *)
(* 6.11913 *)
But, applying this to the general solution yields
s /. {g1 -> 1, g2 -> 7, κ1 -> 3, κ2 -> 4, Γ -> 5}
(* Undefined *)
Why Undefined
is returned can be seen from
Sqrt@N[rts /. {g1 -> 1, g2 -> 7, κ1 -> 3, κ2 -> 4, Γ -> 5}]
(* {0. + 2.4901 I, 7.06613 - 1.75495 I, 7.06613 + 1.75495 I} *)
The first and third roots are in the upper half plane, and the second in the lower half. To obtain the correct solution, discard the ConditionalExpression
and choose the poles in the upper half plane.
N[s[[1]] /. Sqrt[Root[z_, n_]] :> Sign[Im[Sqrt[Root[z, n]]]] Sqrt[Root[z, n]]
/. {g1 -> 1, g2 -> 7, κ1 -> 3, κ2 -> 4, Γ -> 5}] // Chop
(* 6.11913 *)
(FullSimplify
cannot obtain the compact exact value, (10220 Pi)/5247
in a reasonable amount of time, but the corresponding numerical values agree to any number of significant figures.) Note that the substitution above does not work for Im[Sqrt[Root[z, n]]] == 0
. Although modest changes would accommodate purely real poles, they appear to be rare for c1
. To be precise, I have found none despite diligently searching.
Root
objects are better behaved. Bronze age math versus space age math... $\endgroup$c1
, particularly those on the real axis? $\endgroup$Denominator
ofc1
. The square roots of the threeRoot
functions appearing in the result ofIntegrate
are the values of the zeroes. Probably, they cannot be simplified except in special cases. If two zeroes lie on the real axis, probably the Principal Value of the integral should be taken. I should add that the integration path assumed byIntegrate
in obtaining the answer that it returns is not necessarily correct. $\endgroup$Residue
function. What do you mean by the integration path assumed byIntegrate
is not correct? What should be my approach then? $\endgroup$