Re: Simple integral over special functions---HOW?
- To: mathgroup@smc.vnet.net
- Subject: [mg12648] Re: [mg12565] Simple integral over special functions---HOW?
- From: Bob Hanlon <BobHanlon@aol.com>
- Date: Sat, 30 May 1998 17:36:39 -0400
Since the general results are known, the easiest method is to use Gradshteyn & Ryzhik 7.132 ( note condition that 2 Re lambda > Abs[Re[mu]] )and modify Integrate (see notebook below). Bob Hanlon ____________________________ Notebook[{ Cell[BoxData[{ FormBox[\(Unprotect[Integrate]; \), TraditionalForm], FormBox[ RowBox[{ RowBox[{ RowBox[{\(\[Integral]\_\(-1\)\%1\), RowBox[{ FractionBox[ RowBox[{ SubscriptBox[ TagBox["P", LegendreP], \(n_?OddQ\)], "(", "x_", ")"}], \(\@\(1 - x_\^2\)\)], \(\[DifferentialD]x_\)}]}], ":=", "0"}], ";"}], TraditionalForm], FormBox[ RowBox[{ RowBox[{ RowBox[{\(\[Integral]\_\(-1\)\%1\), RowBox[{ FractionBox[ RowBox[{ SubscriptBox[ TagBox["P", LegendreP], \(2\ n_?IntegerQ + 1\)], "(", "x_", ")"}], \(\@\(1 - x_\^2\)\)], \(\[DifferentialD]x_\)}]}], ":=", "0"}], ";"}], TraditionalForm], FormBox[ RowBox[{ RowBox[{ RowBox[{\(\[Integral]\_\(-1\)\%1\), RowBox[{ RowBox[{\(\((1 - x_\^2)\)\^\[Lambda]_\), " ", RowBox[{ SubsuperscriptBox[ TagBox["P", LegendreP], "\[Nu]_", \(\[Mu]_: 0\)], "(", "x_", ")"}]}], \(\[DifferentialD]x_\)}]}], ":=", \(\(\[Pi]\ 2\^\[Mu]\ \(\[CapitalGamma](\[Lambda] + \[Mu]\/2 + 1)\)\ \(\[CapitalGamma](\[Lambda] - \[Mu]\/2 + 1) \)\)\/\(\(\[CapitalGamma](\[Lambda] + \[Nu]\/2 + 3\/2)\)\ \(\[CapitalGamma](\[Lambda] - \[Nu]\/2 + 1)\)\ \(\[CapitalGamma](\(-\(\[Mu]\/2\)\) + \[Nu]\/2 + 1)\)\ \(\[CapitalGamma](\(-\(\[Mu]\/2\)\) - \[Nu]\/2 + 1\/2) \)\)\)}], ";"}], TraditionalForm], FormBox[\(Protect[Integrate]; \), TraditionalForm]}], "Input"], Cell[BoxData[ FormBox[ RowBox[{\(f(\[Nu]_, \[Mu]_: 0, \[Lambda]_)\), ":=", RowBox[{"Evaluate", "[", RowBox[{"Module", "[", RowBox[{\({x}\), ",", RowBox[{\(\[Integral]\_\(-1\)\%1\), RowBox[{ RowBox[{\(\((1 - x\^2)\)\^\[Lambda]\), " ", RowBox[{ SubsuperscriptBox[ TagBox["P", LegendreP], "\[Nu]", "\[Mu]"], "(", "x", ")"}]}], \(\[DifferentialD]x\)}]}]}], "]"}], "]"}]}], TraditionalForm]], "Input"], Cell[BoxData[ \(TraditionalForm\`\(?? f\)\)], "Input"], Cell[BoxData[ \(TraditionalForm\`f(\[Nu], \[Mu], \[Lambda])\)], "Input"], Cell[BoxData[ \(TraditionalForm\`f(\[Nu], \[Lambda])\)], "Input"], Cell[BoxData[ \(TraditionalForm\`f(\[Nu], \(-\(1\/2\)\))\)], "Input"], Cell[BoxData[ \(TraditionalForm\`Simplify[f(2\ n + 1, \(-\(1\/2\)\))]\)], "Input"], Cell["which is zero.", "Text"], Cell[BoxData[{ FormBox[ RowBox[{\(n /: IntegerQ[n] = True\), ";", RowBox[{\(\[Integral]\_\(-1\)\%1\), RowBox[{ RowBox[{\(1\/\@\(1 - x\^2\)\), " ", RowBox[{ SubscriptBox[ TagBox["P", LegendreP], \(2\ n + 1\)], "(", "x", ")"}]}], \(\[DifferentialD]x\)}]}]}], TraditionalForm], FormBox[\(Clear[n]; \), TraditionalForm]}], "Input"], Cell[BoxData[{ FormBox[ RowBox[{\(n /: OddQ[n] = True\), ";", RowBox[{\(\[Integral]\_\(-1\)\%1\), RowBox[{ RowBox[{\(1\/\@\(1 - x\^2\)\), " ", RowBox[{ SubscriptBox[ TagBox["P", LegendreP], "n"], "(", "x", ")"}]}], \(\[DifferentialD]x\)}]}]}], TraditionalForm], FormBox[\(Clear[n]; \), TraditionalForm]}], "Input"], Cell[BoxData[ FormBox[ RowBox[{"Table", "[", RowBox[{ RowBox[{\(\[Integral]\_\(-1\)\%1\), RowBox[{ RowBox[{\(1\/\@\(1 - x\^2\)\), " ", RowBox[{ SubscriptBox[ TagBox["P", LegendreP], "n"], "(", "x", ")"}]}], \(\[DifferentialD]x\)}]}], ",", \({n, 0, 12}\)}], "]"}], TraditionalForm]], "Input"] }, FrontEndVersion->"Macintosh 3.0", ScreenRectangle->{{0, 832}, {0, 604}}, WindowSize->{534, 577}, WindowMargins->{{Automatic, 86}, {Automatic, 2}}, ShowCellLabel->False, MacintoshSystemPageSetup->"\<\ 00<0001804P000000]P2:?oQon82n@960dL5:0?l0080001804P000000]P2:001 0000I00000400`<300000BL?00400@0000000000000006P801T1T00000000000 00000000000000000000000000000000\>" ] In a message dated 5/24/98 2:09:53 AM, morrison@phyast.nhn.ou.edu wrote: >Hi. I have repeatedly run into trouble trying to get Mathematica to >evaluate analytically simple integrals involving special functions. For >example, the following integral has a simple analytic form: > Integrate[LegendreP[n,x]/Sqrt[1-x^2],{x,-1,+1}] When I enter the >above into Mathematica, it returns the integral unevaluated unless I >specify a value for n. Figuring that the problem was that Mathematica >didn't realize that n is a non-negative integer, I did the following: > n/: IntegerQ[n] = True; > Integrate[LegendreP[n,x]/Sqrt[1-x^2], {x,-1,+1}, Assumptions -> >n >= 0] >Again, Mathematica returned the integral unevaluated, along with the >assumption. >It's as though the information that n is an integer in the Global` >context doesn't get communicated to the Integrate command. Can anyone >tell me (a) whether the above method fully specifies n as an integer in >all such situations (I have other failures where function definitions >seem unaware of such a specification) and if not, how to properly >specify a symbol as an integer and (b) how to make Integrate evaluate >integrals such as the example above?