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 _________________________________________________________________ ==== [MESSAGE SEPARATOR] ====