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Re: Why does this lead to an answer with complex numbers?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg71466] Re: Why does this lead to an answer with complex numbers?
*From*: "dimitris" <dimmechan at yahoo.com>
*Date*: Mon, 20 Nov 2006 02:43:47 -0500 (EST)
*References*: <ejosmm$n3k$1@smc.vnet.net>
I don't understand what exactly you want; the variable of integration
is x and you have assumptions for \[Beta].
Anyway let's check your integral.
Here is the integrand.
f[x_] := -Log[Sqrt[1 + x^2] - (1/11)*x]
Here is its plot.
Plot[f[x], {x, -10, 10}]
Here is contour plots of the real and the imaginary part.
Show[GraphicsArray[Block[{$DisplayFunction = Identity},
(ContourPlot[#1[f[x + I*y]], {x, -3, 3}, {y, -3, 3}, ContourShading
-> False, Contours -> 50, PlotPoints -> 100,
FrameLabel -> TraditionalForm /@ {Re[z], Im[z]}, PlotLabel ->
HoldForm[#1[f[z]]]] & ) /@ {Re, Im}]], ImageSize -> 600]
Here is 3D plots of the real and imaginaty part.
Show[GraphicsArray[Block[{$DisplayFunction = Identity},
(Plot3D[#1[f[x + I*y]], {x, -3, 3}, {y, -3, 3}, PlotPoints -> 40,
AxesLabel -> TraditionalForm /@ {Re[z], Im[z],
HoldForm[#1[f[z]]]}] & ) /@ {Re, Im}]], ImageSize -> 800]
Clearly, there is branch cuts connecting the branch points
{-Infinity,-1} and {1,Infinity}.
Here is the indefinite integral of f[x].
F[x_] = FullSimplify[Integrate[f[x], x]]
x - x*Log[-(x/11) + Sqrt[1 + x^2]] -
(1/(8*Sqrt[30]))*(11*(4*ArcTan[(2*Sqrt[30]*x)/11] +
4*ArcTan[2*Sqrt[30]*Sqrt[1 + x^2]] +
I*(2*Log[900*(121 + 120*x^2)^2] - Log[(121 + 120*x^2)*(-121 - 122*x^2 +
22*x*Sqrt[1 + x^2])] - Log[(121 + 120*x^2)*(121 + 122*x^2 + 22*x*Sqrt[1
+ x^2])])))
Here is contour plots of the real and the imaginary part of F[x]
Show[GraphicsArray[Block[{$DisplayFunction = Identity},
(ContourPlot[#1[F[x + I*y]], {x, -3, 3}, {y, -3, 3}, ContourShading ->
False, Contours -> 50, PlotPoints -> 400,
FrameLabel -> TraditionalForm /@ {Re[z], Im[z]}, PlotLabel ->
HoldForm[#1[F[z]]]] & ) /@ {Re, Im}]], ImageSize -> 600]
Here is 3D plots of the real and imaginary part of F[x]
Show[GraphicsArray[Block[{$DisplayFunction = Identity},
(Plot3D[#1[F[x + I*y]], {x, -3, 3}, {y, -3, 3}, PlotPoints ->
{100,40},Mesh->False,
AxesLabel -> TraditionalForm /@ {Re[z], Im[z], HoldForm[#1[F[z]]]}] & )
/@ {Re, Im}]], ImageSize -> 800]
Applying the fundamental theorem of calculus (we can here!) we have
e.g.
FullSimplify[Limit[F[x], x -> 6, Direction -> 1] - Limit[F[x], x -> -6,
Direction -> -1]]
N[%]
12 - (11*ArcTan[(12*Sqrt[30])/11])/Sqrt[30] - 6*Log[4441/121]
-12.438696582264026
The respective result from Mathematica is also
Integrate[f[x], {x, -6, 6}]
N[%]
12 - (11*ArcTan[(12*Sqrt[30])/11])/Sqrt[30] - 6*Log[4441/121]
-12.438696582264026
As a final check
NIntegrate[f[x], {x, -6, 6}]
-12.438696582260306
Best Regards
Dimitris
aaronfude at gmail.com wrote:
> The expression is
>
> \!\(FullSimplify[
> Assuming[\[Beta] > 0 && \[Beta] < Pi/2,
> Integrate[\(-Log[\@\(1 + x\^2\) - 1/11*x\ ]\), \ x]]]\)
>
> Thanks!
>
> Aaron Fude.
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