@one man 

No need for other examples of animated curves, you already posted dozens.
If you can, show us a real (approximate) parametrization for this surface, this would be useful.

Then it's a bug at least in documentation.

@Axel Vogt 

@one man 

If you really want to help the OP, why don't you state clearly in the worksheet:
the parametization for this surface is given by ...
(in the worksheet there isn't any parametrization, the surface is ploted by implicitplot3d).

In this case, an exact parametrization is:

plot3d([
(2 + cos(t))^(1/4) * signum(cos(u))*sqrt(abs(cos(u))),
(2 + cos(t))^(1/4) * signum(sin(u))*sqrt(abs(sin(u))),
signum(sin(t))*sqrt(abs(sin(t)))],
t=-Pi..Pi, u=-Pi..Pi, style=surface);

One must change the variable.

I have not verified if it agrees with Mathematica in this case.

Ok, I'll reply, even if it's probably a waste of time.
1. About directional limits/series:

2. Maple and Mathematica expansions agree for x>0.

All Maple's answers here are mathematically correct.
(It would be easier to see this for the similar function f(x) = Psi(1/x) - Psi(2/x)).

Facts:

• x = 0 is an essential non-isolated singularity (more exactly there is a sequence of poles at x_n, x_n < 0,  x_n --> 0) 
  So, the limit at 0 does not exist. The same (of course) for its derivative.
• The series given by MultiSeries is for x>0 (recall that it uses directional limits) and it is a "generalized series", not a Taylor one.
  (Note that the standard :-series refuses to give the expansion and invites to use asympt).

@Christopher2222 

No need to check, the expression is symmetric in x and y.

@Christopher2222 

A double integral must be computed:

with(Statistics):
X := RandomVariable(Normal(0, 1)): Y := RandomVariable(Uniform(-2, 2)):
#Probability(X*Y < 0);
f:=t->PDF(X,t);  g:=t->PDF(Y,t);
int(Heaviside(-s*t)*f(s)*g(t), s=-infinity..infinity, t=-infinity..infinity);
      1/2

@epostma 

But wouldn't this be just an ad hoc patch?
I think it would be interesting to investigate why those undefined and Dirac have appeared under int.
(And also why int is sometimes unable to compute an integral containing Heaviside without a convert/piecewise.)
 

Yes, I did not noticed that X3,X4 are uniform.
Then the CDF of Z is of course
CDF(X1, t)^2*CDF(X3, t)^2;

So, with assumptions Maple is correct.
Strange bug anyway.

@Christopher2222 

Unfortunately, all are wrong.
The correct CDF is:
CDF(X1,t)^4;
==>  (1/2+(1/2)*erf((1/2)*t*sqrt(2)))^4
(so, Mathematica's answer is ok).
Note that Maple also fails for  CDF(max(X1,X2,X3), t)
for which it results a limit at infinity < 0.6 (!)
but strangely, CDF(max(X1,X2), t) is correct.

@mmcdara 

- When seen as a distribution, the value of H at 0 does not matter.
- Before trying to compute HD, this entity must be defined as a distribution. How do you define it?

@mmcdara 

fsolve(CDF(X,x)=0.2, x);

@Markiyan Hirnyk 

My comment was towards the rest of the Maple community because I have anticipated the (usual) title of your reply.
The past demonstrates that our maths, Maple and communication manner are too different.