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.