@mimi990 

fun11 is actually a vector (of dimension 1) containing the integral. So, fun11[1]  is the integral itself.
The subsop converts the definite integral into an indefinite one.

@9009134 

This function is also non-periodic. 

In general, if two continuous nonconstant periodic functions f, g : R --> R have incommensurable periods (i.e. their ratio is irrational) then f + g is non-periodic.

I don't know what you mean by minimal period in this case; d is not periodic. The periods of the two terms are incommensurable, there is no common period.

@Kitonum 

It's impossible to get uniform approximations this way. E.g. for t=40800*Pi..40800*Pi+3  the difference is very large.

@mmcdara 

This is another function. OP says f(t)=t^2 for t in [0,2*Pi]. It has discontinuous periodic extension.
Yours is continuous.

@mmcdara 

Sorry, but you are confusing "Fourier series" with "Fourier transform".

@mmcdara 

But sum(1/u, u in L) is nonsense in Maple (just try it for L:=[1,2]). Any such nonsense will produce infinity due to the empty catch clause. I think that you meant add instead.

@mmcdara 

The standard periodic extension of a function f defined in [0,T) is simply x |--> f(x - floor(x/T)*T).
It's not about distributions here. In the theory of Fourier series, the integrability is in [0,T). I think that you have in mind the Fourier transform.

@Zeineb 

Have you checked for all b?

@tomleslie 

Nice, vote-up!
I'd suggest to insert in your code

nE,nV := NumberOfEdges(T), NumberOfVertices(T);
# Produce all subgraphs of G which have the same number of edges and vertices as the "test" sub-graph T
U:=choose(Edges(G),nE): nops(%);
U1:=select(t -> (nops(`union`(t[]))=nV), U): nops(%);
gL:= Graph~(U1): nops(%);

in order to reduce the number of candidates.
 

@Carl Love 

The Squeezing Theorem cannot be used here.
If   limit(liminf(f(x,y), y=0), x=0)  and  limit(limsup(f(x,y), y=0), x=0)  exist and are equal,
it does not imply the existence of limit( limit(f(x,y), y=0), x=0); and actually it does not exist.

BTW, it would be nice to have liminf and limsup in Maple.

@acer 

A very nice and professional patch (a single line of code added). I hope to be included in the next Maple update.

@9009134 

Maple cannot help much here because the only transforms are integrations by parts for generic functions, and  some other functions (e.g. I_0, I_2)  just appear from nowhere. These are easier to do by hand, but as I said above, some integrals are missing from some terms. [The final result could be correct though, if the δu etc are chosen to be 0 on the boundary (so, these terms become 0), but these aspects must be examined carefully].

In conclusion, what is needed here is (proper) maths, not Maple. 

@9009134 

The problem is that you are using nonstandard and confusing notations.
I understand that δu  is simply a function of (r,theta). But why do you use N_rr if it's not a derivative? It should be simply N.

What about the formula?

It seems that Omega0 is the rectangle [r_1, r_2] x [theta_1, theta_2]. But then in the first term after "=",  an integral wrt theta is missing.

Any help will be difficult if you don't try to use rigour, proper math notations and all needed details.

Mathematica and Maple syntaxes are quite different.
In order to use confortably Maple a good idea is to read first an introductory guide or at least a tutorial.

Your problem contains a system of recurrent equations. In Maple you must use P(t) instead of P[t] etc and call rsolve  [ or you may define 4 recursive procedures].
But I'd suggest to start with a tutorial.

@Carl Love 

For my solution I wanted a minimal modification of OP's code.

Of course using mod is faster but msolve is more general.
For example,
Sol:= (b,r)-> `if`(igcd(b,10000)=1, r/b mod 10000, FAIL):
is not correct, while the corresponding msolve works.

BTW, it's strange the answer given by msolve for

msolve(a*b = 2391, 10000);

    {a = 2391/b, b = b}, {a = a, b = 2391/a}

It should solve the problem, but it does not; here 2391/b  must be interpreted mod 10000 and only when it makes sense.