Re: a challenging definite integral!
- To: mathgroup at smc.vnet.net
- Subject: [mg79355] Re: a challenging definite integral!
- From: chuck009 <dmilioto at comcast.com>
- Date: Wed, 25 Jul 2007 02:12:46 -0400 (EDT)
Hi Dimitris. Your approach is to find an antiderivative but mine is to evaluate it via a suitable complex contour. Via parts, the problem reduces to evaluating:
Integrate[(1+2x)/(1+Sqrt[x] Exp[x]), {x,0,Infinity}]
One approach might then be to study the complex function:
f[z]=(1+2z)/(1+Exp[z]Sqrt[z]); Log[z]=ln(r)+it; 0<t<2pi
This in itself is a very interesting function to study! It has an infinite number of poles which are located on the sheets of the Lambert W function which can be found by solving:
Exp[2z]z=1
Note that it has a pole on the real axis at W[2]/2 and the others at W[i,2]/2 (i odd) and Conjugate[W[i,2]]/2. Note these poles are the branch points for the Log version. One possible approach is to consider the plot:
p1 = ParametricPlot3D[{r*Cos[t], r*Sin[t],
Re[(1 + 2*z)/(1 + Exp[z]*Exp[(1/2)*(Log[r] + I*t)]) /.
z -> r*Exp[I*t]], Red}, {r, 0.0001, 5}, {t, 0, 2*Pi},
Lighting -> False, PlotPoints -> {50, 50}]
trace1 = ParametricPlot3D[Evaluate[
Append[cfunction /. {z -> r*Exp[I*t]} /. t -> 0,
{Yellow, Thickness[0.005]}]], {r, 0, 5}]
net = Show[{p1, trace1}]*real2 = Fold[ClipGraphics3D, net,
{z <= 8, z >= -8, Sqrt[x^2 + y^2] <= 5}]
The contour would then be along the branch-cut at t=0, around the pole at z=W[2]/2 Exp[2 pi i] and the origin, and then around a larger circle enclosing the remaining poles. I don't think that would work though, however . . . I think I may attempt a numerical calculation of such a contour for say the integral from 0 to 50 just to see if everything "adds" up. :)