Re: Integrating by parts....II

• To: mathgroup at smc.vnet.net
• Subject: [mg8786] Re: Integrating by parts....II
• From: Paul Abbott <paul at physics.uwa.edu.au>
• Date: Thu, 25 Sep 1997 12:26:20 -0400
• Organization: University of Western Australia
• Sender: owner-wri-mathgroup at wolfram.com

```Adrean Webb wrote:

> Can Mathematica still solve the function below?
>
>         f[n] = Integrate[Cos[x]*E[-x^2]*HermiteH[n,x],{x,-Inf,Inf}]

Yes.  You can compute the set of integrals,

Integrate[Cos[x]*E[-x^2]*HermiteH[n,x],{x,-Inf,Inf}]

for arbitrary n using the Hermite generating function.  See the Notebook
below for details. (Select from Notebook[  down to ] inclusive and paste
into a new Mathematica Notebook).

Cheers,
Paul

____________________________________________________________________
Paul Abbott                                   Phone: +61-8-9380-2734
Department of Physics                           Fax: +61-8-9380-1014
The University of Western Australia
Nedlands WA  6907                     mailto:paul at physics.uwa.edu.au
AUSTRALIA                             http://www.pd.uwa.edu.au/~paul

God IS a weakly left-handed dice player
____________________________________________________________________

Notebook[{
Cell[TextData[{
"Using the Hermite generating function, ",
Cell[BoxData[
\`\(\[ScriptCapitalH](x, z) ==
\[Sum]\+\(n = 0\)\%\[Infinity]\(\(\( H\_n\)(x)\)\
z\^n\)\/\(n!\),
\)\)]]
}], "Text",
Evaluatable->False,
AspectRatioFixed->True],

Cell[BoxData[
\`\(\[ScriptCapitalH](x_, z_) = \[ExponentialE]\^\(2\ x\ z - z\^2\);
\)\)], "Input",
AspectRatioFixed->True],

Cell[CellGroupData[{

Cell[BoxData[
\`\(\(\[Integral]\_\(-\[Infinity]\)\%\[Infinity]
\[ExponentialE]\^\(-x\^2\)\ \(cos(x)\)\
\(\[ScriptCapitalH](x, z)\) \[DifferentialD]x // Factor\)
//
ComplexExpand\) // TrigReduce\)], "Input",
AspectRatioFixed->True],

Cell[BoxData[
\(TraditionalForm\`\(\ at \[Pi]\ \(cos(z)\)\)\/\ at \[ExponentialE]\%4\)],
"Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
FormBox[
RowBox[{"%",
FormBox[\(+\(O(z)\)\^5\),

Cell[BoxData[
FormBox[
InterpretationBox[
RowBox[{
\(\ at \[Pi]\/\ at \[ExponentialE]\%4\), "-",
\(\(\ at \[Pi]\ z\^2\)\/\(2\ \ at \[ExponentialE]\%4\)\), "+",
\(\(\ at \[Pi]\ z\^4\)\/\(24\ \ at \[ExponentialE]\%4\)\), "+",
InterpretationBox[\(O(z\^5)\),
SeriesData[ z, 0, {}, 0, 5, 1]]}],
SeriesData[ z, 0, {
Times[
Power[ E,
Rational[ -1, 4]],
Power[ Pi,
Rational[ 1, 2]]], 0,
Times[
Rational[ -1, 2],
Power[ E,
Rational[ -1, 4]],
Power[ Pi,
Rational[ 1, 2]]], 0,
Times[
Rational[ 1, 24],
Power[ E,
Rational[ -1, 4]],
Power[ Pi,
Rational[ 1, 2]]]}, 0, 5, 1]], TraditionalForm]],
"Output"]
}, Open  ]]
}
]

```

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