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MathGroup Archive 1996

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Re: Integrals of Fourier Series

  • To: mathgroup at smc.vnet.net
  • Subject: [mg3255] Re: Integrals of Fourier Series
  • From: Paul Abbott <paul at earwax.pd.uwa.edu.au>
  • Date: Wed, 21 Feb 1996 02:15:09 -0500
  • Organization: University of Western Australia
  • Sender: owner-wri-mathgroup at wolfram.com

George Oster wrote:

> Suppose I want to substitue a Fourier Series into an integral:
> 
> u[x_] := Sum[A[n] Sin[n Pi x/L], {n, 1, Infinity}]
> 
> Integrate[(u''[x])^2, {x, 0, L}]
> 
> This has an easy analytical solution that I can't get Mma to find, > because Mma doesn't know that Sum and Integrate commute, and that 
> Sin[n Pi] = 0 for all integer n.
> 
> How to do this?

Instead of working with the Sum, just focus on the summand:

	Remove[u]

	u[n_][x_] = Sin[n Pi x/L];

The generic (cross-product) term of the integrand is

	u[n]''[x] u[m]''[x]

	 2  2   4     m Pi x      n Pi x
	m  n  Pi  Sin[------] Sin[------]
	                L           L
	---------------------------------
	                4
	               L

and the integral can be evaluated to yield

	Integrate[%, {x, 0, L}]

	  2  2   3
	(m  n  Pi  (m Sin[(m - n) Pi] + n Sin[(m - n) Pi] - 
 
	      m Sin[(m + n) Pi] + n Sin[(m + n) Pi])) / 
 
	      3   2    2
	  (2 L  (m  - n ))

When m!=n,

	% /. Sin[n_ Pi] -> 0

	0

Taking the limit as m->n:

	Limit[%%, m->n] /. Sin[n_ Pi] -> 0

	 4   4
	n  Pi
	------
	    3
	 2 L

one finds that the doubly-infinite summation reduces to

	Sum[a[n]^2 %, {n, 1, Infinity}]

	     4   4     2
	    n  Pi  a[n]
	Sum[------------, {n, 1, Infinity}]
	           3
	        2 L

_________________________________________________________________ 
Paul Abbott
Department of Physics                       Phone: +61-9-380-2734 
The University of Western Australia           Fax: +61-9-380-1014
Nedlands WA  6907                         paul at physics.uwa.edu.au 
AUSTRALIA                           http://www.pd.uwa.edu.au/Paul
_________________________________________________________________

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