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Help with definite integral solved analytically

POSTED BY: Anuj Malik
5 Replies

You also need a space between the I and the \[Theta]. [It's running a long time, and I have to go to a meeting, so I can't say if you need more fixes.]

POSTED BY: Michael Rogers

With change variable we have:

$\int_{-\phi }^{\phi } \frac{\sqrt{1+e^{2 i \theta }-2 e^{i \theta } \cos (\phi )}}{1+e^{i \theta } \alpha } \, d\theta =\int_{\cos (\phi )-i \sin (\phi )}^{\cos (\phi )+i \sin (\phi )} -\frac{i \sqrt{1+x^2-2 x \cos (\phi )}}{x+x^2 \alpha } \, dx$

 HoldForm[
   Integrate[Sqrt[
     1 + E^(2 I \[Theta]) - 2 E^(I \[Theta]) Cos[\[Phi]]]/(
     1 + E^(I \[Theta]) \[Alpha]), {\[Theta], -\[Phi], \[Phi]}] == 
    Integrate[-((I Sqrt[1 + x^2 - 2 x Cos[\[Phi]]])/(
      x + x^2 \[Alpha])), {x, Cos[\[Phi]] - I Sin[\[Phi]], 
      Cos[\[Phi]] + I Sin[\[Phi]]}]] // TeXForm

then:

 IntegrateChangeVariables[
  Inactive[Integrate][-((I Sqrt[1 + x^2 - 2 x Cos[\[Phi]]])/(
    x + x^2 \[Alpha])), {x, Cos[\[Phi]] - I Sin[\[Phi]], 
    Cos[\[Phi]] + I Sin[\[Phi]]}], t, t == x - Exp[I*\[Phi]], 
  Assumptions -> 0 < \[Alpha] < 1 && 0 < \[Phi] < Pi]
  (*    Can't compute. Weakness!!!  *)

  IntegrateChangeVariables[
    Inactive[Integrate][-((I Sqrt[1 + x^2 - 2 x Cos[\[Phi]]])/(
      x + x^2 \[Alpha])), x], t, t == x - Exp[I*\[Phi]], 
    Assumptions -> 
     0 < \[Alpha] < 1 && 0 < \[Phi] < Pi] // FullSimplify(*OK*)

  (*Then:*)

  W = Integrate[-((
     I Sqrt[t (t + 2 I Sin[\[Phi]])])/((E^(I \[Phi]) + 
        t) (1 + (E^(I \[Phi]) + t) \[Alpha]))), t]

  (* -((2 I E^(-I \[Phi]) Sqrt[
    t] (-((E^(I \[Phi]) + \[Alpha]) Sqrt[1 + E^(I \[Phi]) \[Alpha]]
          ArcTan[(Sqrt[t] Sqrt[-1 - E^(-I \[Phi]) \[Alpha]])/(
          Sqrt[1 + E^(I \[Phi]) \[Alpha]] Sqrt[t + 2 I Sin[\[Phi]]])]) +
       E^(I \[Phi]) \[Alpha] Sqrt[-1 - E^(-I \[Phi]) \[Alpha]]
        ArcTanh[(E^(-I \[Phi]) Sqrt[t])/Sqrt[t + 2 I Sin[\[Phi]]]] + 
      E^(I \[Phi]) Sqrt[-1 - E^(-I \[Phi]) \[Alpha]]
        Log[Sqrt[t] + Sqrt[t + 2 I Sin[\[Phi]]]]) Sqrt[
    t + 2 I Sin[\[Phi]]])/(\[Alpha] Sqrt[-1 - E^(-I \[Phi]) \[Alpha]]
     Sqrt[t (t + 2 I Sin[\[Phi]])]))*)

     FullSimplify[(Limit[W, 
           t -> (x - Exp[I*\[Phi]] /. x -> Cos[\[Phi]] + I Sin[\[Phi]] // 
              ComplexExpand), 
           Assumptions -> 0 < \[Alpha] < 1 && 0 < \[Phi] < Pi, 
           Direction -> -1] - 
          Limit[W, 
           t -> (x - Exp[I*\[Phi]] /. x -> Cos[\[Phi]] - I Sin[\[Phi]] // 
              ComplexExpand), 
           Assumptions -> 0 < \[Alpha] < 1 && 0 < \[Phi] < Pi, 
           Direction -> -1]) // PowerExpand, 
       Assumptions -> 0 < \[Alpha] < 1 && 0 < \[Phi] < Pi] // Expand


       (* \[Pi] + \[Pi]/\[Alpha] - (\[Pi] Sqrt[
         1 + \[Alpha]^2 + 2 \[Alpha] Cos[\[Phi]]])/\[Alpha]*)

$\int_{-\phi }^{\phi } \frac{\sqrt{1+e^{2 i \theta }-2 e^{i \theta } \cos (\phi )}}{1+e^{i \theta } \alpha } \, d\theta =\pi +\frac{\pi }{\alpha }-\frac{\pi \sqrt{1+\alpha ^2+2 \alpha \cos (\phi )}}{\alpha }$

Regards M.I.

POSTED BY: Mariusz Iwaniuk
Posted 2 months ago

You have a missing ]

enter image description here

POSTED BY: Ehud Behar

yes by mistake, I not give space

POSTED BY: Anuj Malik

Yes I need to solve this integration

POSTED BY: Anuj Malik
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