Re: Integrals of Fourier Series

• Subject: [mg3207] Re: [mg3179] Integrals of Fourier Series
• From: jpk at apex.mpe.FTA-Berlin.de (Jens-Peer Kuska)
• Date: 18 Feb 1996 01:11:02 -0600
• Approved: usenet@wri.com
• Distribution: local
• Newsgroups: wri.mathgroup
• Organization: Wolfram Research, Inc.
• Sender: daemon at wri.com

```
Dear George Oster,

the situation is not so simple as it seems.
a) there is no finite algorithm to deal with
products of infinite sums
b) to leave (Sum[_,_]*Sum[_,_]) unevaluatet you must
supply different a summation index.

The rest is easy

(* my version of Your series with a named index *)
In[1]:=
u[x_,n_]:=
Sum[A[n]*Sin[n*Pi*x/L],{n,1,Infinity}]

(* Do the derivative *)
In[2]:=
u2=D[u[x,n],{x,2}]

Out[2]=
2   2          n Pi x
n  Pi  A[n] Sin[------]
L
Sum[-(-----------------------),
2
L

{n, 1, Infinity}]

(* calculate the square and replace one summation index *)
In[3]:=
sqru=u2*(u2 /. n->k)

Out[3]=
2   2          k Pi x
k  Pi  A[k] Sin[------]
L
Sum[-(-----------------------),
2
L

{k, 1, Infinity}]

2   2          n Pi x
n  Pi  A[n] Sin[------]
L
Sum[-(-----------------------),
2
L

{n, 1, Infinity}]

(* Collect the product of two sum's in a double sum *)
In[10]:=
sqru2=sqru /. Literal[Sum[an_,iter1_]*Sum[ak_,iter2_]] :> Sum[an*ak,iter1,iter2]

Out[10]=
2  2   4               k Pi x      n Pi x
k  n  Pi  A[k] A[n] Sin[------] Sin[------]
L           L
Sum[-------------------------------------------, {n, 1, Infinity},
4
L

{k, 1, Infinity}]

(* Change the integration rule to map into a sum *)
In[20]:=
Unprotect[Integrate];
Literal[Integrate[Sum[as_,iter__],range_]]:=Sum[Integrate[as,range],iter]
Protect[Integrate];
Integrate[sqru2,{x,0,L}]

Out[23]=
2  2   3
Sum[(k  n  Pi  A[k] A[n] (k Sin[(k - n) Pi] + n Sin[(k - n) Pi] -

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

{n, 1, Infinity}, {k, 1, Infinity}]

There are one notice, for a general rule You have to change

Literal[Integrate[Sum[as_,iter__],x_Symbol]]:=
Sum[Integrate[as,range],iter] /; FreeQ[{iter},x]

Literal[Integrate[Sum[as_,iter__],interval_List]]:=
Sum[Integrate[as,range],iter] /; FreeQ[{iter},First[interval]]

For multidimensional integrals the pattern's become more complicated.

Hope that helps

Jens

```

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