Re: Integrating special functions
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- Subject: [mg131594] Re: Integrating special functions
- From: Roland Franzius <roland.franzius at uos.de>
- Date: Tue, 10 Sep 2013 03:33:06 -0400 (EDT)
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Am 07.09.2013 11:59, schrieb Herman:
> Dear All,
>
> I would like to integrate the following function with Legendre polynomial and Gamma function. I am open to any suggestions.
>
> ass = {s > 0, \[Alpha] > 0};
>
> \[Phi][s_, x_, \[Alpha]_] := (-1)^s Sqrt[(s \[Alpha])/
> Gamma[1 + 2*s]] LegendreP[s, s, Tanh[\[Alpha] x]]
>
> \[Phi]1[s_, x_, \[Alpha]_] := D[[\[Phi][s, x, \[Alpha]], x]]
>
> a3 = Table[
> Integrate[-\[ImaginaryI] x \[Phi][s, x, a] \[Phi]1[s, x,
> a], {x, -\[Infinity], \[Infinity]}, Assumptions -> \[Alpha] > 0,
> s > 0]
> would it be possible to get a closed form of the integration a3?
>
I don't want to invest an hour to solve a somehow trivial but a bit
tedious problem.
So I only will give you some hints.
The associated Legrendre function P[n,n,x] is something like (1-x^2)^n
for integer n. It represents the extreme form of the angular momentum
wave density concentrated at the equator of sphere, rotating wave part
e^(i n phi) suppressed, if the argument, the Tanh-function varies in
(-1,1).
LengendreP[n,n,Tanh[x]] -> constant * Sech[x]^n ;/ n>0 && n\in Integers
The product of a function and its derivative is the derivative of its square
f'[x] f[x] <-> D[ 1/2 f[x]^2, x]
The integral can be integrated by parts because
Limit[ x Sech[x]^n], x->+-oo ] ==0
So you problem is only to integrate
Integrate[ x D[1/2 Sech[a x]^(2n),x ],{x,-oo,oo}]
or
- 1/2 Integrate[Sech[a x]^(2n) ,{x,-oo,oo}]
--
Roland Franzius