       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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