principal value
- To: mathgroup at smc.vnet.net
- Subject: [mg67952] principal value
- From: dimmechan at yahoo.com
- Date: Thu, 13 Jul 2006 06:55:29 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Hi specialists.
I have to evaluate the following integral (from an elasticity project)
\!\(Integrate[ArcTan[x]\/\(x - p\), {p, 0, 1}, {x, 0, 1}]\).
Since p take values in the interval {0,1}, the option PrincipalValue
must be used.
\!\(Integrate[ArcTan[x]\/\(x - p\), {p, 0, 1}, {x, 0, 1},
PrincipalValue \
\[Rule] True]\)
\!\(\*FormBox[
RowBox[{\(1\/32\), " ",
RowBox[{"(",
RowBox[{
RowBox[{"32", " ",
TagBox["C",
Function[ {}, Catalan]]}], "+
", \(4\ \(\(log\^2\)(2)\)\), "-", \(Ï?\ \((Ï? + log(16))\)\)}],
")
"}]}], TraditionalForm]\)
symb=N[%]
0.395399
I want also to compute previous numerically. To this end, I work as
follows:
First,
Needs["NumericalMath`"]
$Version
5.2 for Microsoft Windows (June 20, 2005)
\!\(f[x_, p_] := ArcTan[x]\/\(x - p\)\)
cauchy1[g_,x_,p_]:=CauchyPrincipalValue[g,{x,0,{p},1}]
\!\(\(mat1 =
Table[{p,
cauchy1[ArcTan[x]\/\(x - p\), x, p]}, {p, 0.0001, 0.9999,
0.0001}];\)\)
For p=0 the integral of f[x,p] in the interval 0<=x<=1 converges.
Integrate[f[x,0],{x,0,1}]
C
mat2=N[%]
0.915966
For p=1 the integral of f[x,p] in the interval 0<=x<=1 diverges.
Integrate[f[x,1],{x,0,1}]
\!\(\*FormBox[
RowBox[{\(Integrate::"idiv"\), \(\(:\)\(\ \)\), "\<\"Integral of \
\\!\\(TraditionalForm\\`\\(\\(\\(
tan\\^\\(-1\\)\\)\\)(x)\\)\\/\\(x - 1\\)\\) does not converge on
\
\\!\\(TraditionalForm\\`\\({0, 1}\\)\\).
\\!\\(\\*ButtonBox[\\\"Moreâ?¦\\\", \
ButtonStyle->\\\"RefGuideLinkText\\\", ButtonFrame->None, \
ButtonData:>\\\"Integrate::idiv\\\"]\\)\"\>"}], TraditionalForm]\)
\!\(TraditionalForm\`â?«\_0\%1\(\(\( tan\^\(-1\)\)(x)\)\/\(x -
1\)\) \[DifferentialD]x\)
However (in the Hadamard sense)
Integrate[f[x,1],{x,0,1},GenerateConditions\[Rule]False]
\!\(\*FormBox[
RowBox[{
TagBox["C",
Function[ {},
Catalan]], "-", \(1\/8\ Ï?\ \(log(2)\)\)}], TraditionalForm]\)
mat3=N[%]
0.643767
So, I consruct a new list:
mat4=Insert[mat1,{0,mat2},1];
mat5=Insert[mat4,{1,mat3},Length[mat4+1]];
Then, I use the command to evaluate the integral
num=ListIntegrate[mat3, 2]
0.395833
The numerical result is different. What am I loosing something?
P.S. Evaluating an integral in the Hadamard sense, is (in most cases)
of no cost in elasticity theory. Usually, such "losing" of a
singularity, as before in p=1, corresponds to rigid body displacement.