Fine analysis!

I just want to add that the asymptotic behavior of the involute (t-->oo) can be obtained directly using
MultiSeries:-asympt
(for t-->-oo  one has to change t to -t). Of course, MultiSeries:-series(...,t=infinity) is also possible.

@Markiyan Hirnyk 

What should be in your opinion the Maple's answer for your test problem

pdsolve([diff(u(t, x), t, t) = diff(u(t, x), x, x), u(t, 0) = 0, u(t, Pi) = 0]);

Same question in an ideal CAS.

@David1 

I am afraid that Groebner wil not help you very much for this problem. The degree of the polynomials would be huge and also the coefficients.
But, do you really need all the solutions?  You may add constraints to select some of them and then use the Direct Search package https://www.maplesoft.com/applications/view.aspx?SID=87637

It's a formal result.
It is easy to give sufficient conditions for the coefficients to have standard solutions. No CAS will do this in the near future.
Anyway, it is much better than nothing.

P.S. There are plenty of real bugs in Maple. Why not focus on these?

@Markiyan Hirnyk 

This is not really a trick. And it is hard to speak about bugs in this context, see http://mathworld.wolfram.com/ConstantProblem.html

Testzero:=testeq:
M^(-1);
Error, (in rtable/Power) singular matrix

It is an interesting and important problem.
Unfortunately the post is not easy to read because you decided to keep the notations (and the example) from the cited question.
This reminds me of a joke about Cauchy: when dealing with three similar quantities such as a,b,c they were denoted: NO2, ∑^t_1,2   and ∏⁷_p,q .

@digerdiga 

'''sol(x)'''  cannot work because the first eval evaluates sol(X)  giving infinity  ( = min([]) ).
In '''sol'''(x), the outer '...' is for the evaluation of the argument, the second is for eval itself and the third is for the final result which must be unevaluated, to be used by eval(%, X = 1).

@acer 

hmm is simply wrong (bug).
Actually, appending assuming alpha>0, beta>0, hmm remains unevaluated.

@Markiyan Hirnyk 
But you have considered alpha=beta (=1) and the integral is divergent in this case.

@_Maxim_ 

But isn't this normal?
The result of 
Digits:=30: evalf[10](expr);
is and should be the same as:
evalf[10](expr): evalf[30](%);

@Markiyan Hirnyk 

Why? It can be computed by hand using the convolution formula.

@_Maxim_ 

It works, but by chance, for Digits=30 the coefficient of the 1/s term is 0.0  instead of  x.y*10^(-...).

evalf[10]( evalf[40](series(f(1.*s), s = 0, 1)) );
        (1.*10^(-40)*I)/s+O(s)

@Ian Jones 

You must have typos; it is not validated numerically.

@Carl Love 

I mean this:

Download digits.mw