I perfer 'map': having a sequence for me then it is clear what function
is applied (and it can be changed easily). Once I learned: tuning is the
'final' step - it has to work correctly first.

And thus would write (or similar):

fct:= x -> 1/x * sin(t);
pValues:= [seq(i, i=0.1 .. 1.0, 0.2)];
plot( map(fct, pValues), t=-2*Pi .. 2*Pi);

Then I may have lost some msec. But days (?) later I still see what I did
and may change it.

The original post is far too vague (though no bashing).

Anyway I agree on oscillating integrals. That is a field of its own
(not only for integral transforms). The given link reminded me of
Iserles, Norsett, Olver and the in-linked Evans & Webster studies
some nice method (especially Bakhvalov-Vasileva).

For numerical integration in higher dimensions my feeling is that
'cuba' is not yet stable - but I never looked deeper into that field
(ok, vague as well)

Yes. Are you aware of an easy way to convince Maple NOT to use the same x,y,z after transforming? I hate that.

Int(x^(-5/3) * cos((x-1)*h), x = 1..infinity);
IntegrationTools:-Change(%, x=xi+1, xi);
value(%); simplify(%);

or

Int(x^(-5/3) * cos((x-1)*h), x = 1..infinity);
#IntegrationTools:-Change(%, x=xi+1, xi);
value(%); simplify(%): simplify(%, size);

You will have to study LommelS1 <---> LommelS2
and may consider assumptions on h.

VectorCalculus:-int((z-z0)*z/sqrt((x-x0)^2+(y-y0)^2+(z-z0)^2),
  [x,y,z]=Sphere(<0,0,0>,1), inert);
subs(x0=1/2, y0=1/3, z0=1/4, %);
evalf[10](%);

                             0.8430087463

VectorCalculus:-int((z-1/4)*z/sqrt((x-1/2)^2+(y-1/3)^2+(z-1/4)^2),
  [x,y,z]=Sphere(<0,0,0>,1),inert);
evalf[10](%);
                             0.8430087463

just try again to state your task ... may be someone has an idea ...

Please provide code, not pictures: Do you mean int(x^(-5/3) * cos((x-1)*h), x = 0..infinity)

A variation: use trace or determinant to have the 4th equation

#Determinant(P^%T.M.P) = Determinant(E):
Trace(P^%T.M.P) = Trace(E):
{Eqs, % mod 5};

  {2*a*b+3*c*d, 2*a^2+3*c^2-1, 2*b^2+3*d^2-1, 2*a^2+2*b^2+3*c^2+3*d^2 = 2}

sols:=[msolve(%, 5)]; #nops(%);

{seq( eval(P, sols[j]), j=1..nops(sols))}; # map('q -> q mod 5', %);

   [2    2]  [2    2]  [2    3]  [2    3]  [3    2]  [3    2]  [3    3]  [3    3]
  {[      ], [      ], [      ], [      ], [      ], [      ], [      ], [      ]}
   [1    4]  [4    1]  [1    1]  [4    4]  [1    1]  [4    4]  [1    4]  [4    1]

I think that is at least difficult: Maple would be friendly for 'irreduc'
and 'factor' for x^2 - Pi^2, but not for Pi^1 instead (the help says it
is "over the field implied by the coefficients present").

If using indexed RootOf: is there a way to find out which are conjugated
(over the Reals)?

This is why I have doubts that using a CAS is the way to learn Math ...

This is why I have doubts that using a CAS is the way to learn Math ...

What - for example - should be the result if having all points with integer coordinates in a square like [0 ... 4]^2 (i.e. for an equal spaced grid)?

remark on the initial question:

Though multiplication and division are a bit different here is a simple
example why it makes sense to leave needed precision to the user:

  restart; Digits:=10; # the default

  n:= 1234567890;

  a:=0.5678;
  b:=0.1234;
 
  x:= n+a;
  y:= n+b;

Now x-y should be the same as a-b and if expecting 5 decimals then the
chosen Digits = 10 should enough:

  a-b;
                                0.4444

However the 'n' does not cancel out in numerics: 

  x-y;
                                  1.

It can not be 'reconstructed' from x and y. 

It even becomes worse if using '5 Digits':

  evalf[5](x-y);
                                  0.

As user you may know/estimate what is needed, depending on magnitude
and operation, thus 10 + 5 Digits. Already for *defining* x,y. Here.

... Z/(2) x Z/(2) ?

Edited: https://en.wikipedia.org/wiki/Nilpotent_group#Examples gives references that any Abelian group is nilpotent. And I am sure that the authors are aware of the fundamental theorem of finite abelian groups. So it may be a trivial error in formulation of the task (group/commutator = 0 ?). PS: I never liked group theory that much

@velimir74 : are you sure that in your eq1 there is a 'plus' in the numerator? The other two have multiplication.