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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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