@Markiyan Hirnyk 

Agreed.

However showing z=1/2 is a discontinuity I failed (after looking closer at my reasoning),
so I only state a guess without proof:

In z=1/2 the integral equals 3/sqrt(5) * Pi and the limits from left and right jump by Pi,

['(3/sqrt(5)+1)*Pi','(3/sqrt(5)-1)*Pi']; evalf(%);

                  /       3   \     /   3       \
                [ |1 + -------| Pi, |------- - 1| Pi ]
                  \    sqrt(5)/     \sqrt(5)    /


                 [7.35648149221421, 1.07329618503463]

@Markiyan Hirnyk 

Agreed.

However showing z=1/2 is a discontinuity I failed (after looking closer at my reasoning),
so I only state a guess without proof:

In z=1/2 the integral equals 3/sqrt(5) * Pi and the limits from left and right jump by Pi,

['(3/sqrt(5)+1)*Pi','(3/sqrt(5)-1)*Pi']; evalf(%);

                  /       3   \     /   3       \
                [ |1 + -------| Pi, |------- - 1| Pi ]
                  \    sqrt(5)/     \sqrt(5)    /


                 [7.35648149221421, 1.07329618503463]

@pagan 

But that is correct, I think (say working with 10 Digits), since evalf will ignore
the last those tiny tails - as a float. If one wants more (relative precision ~
number of places) one has to increase Digits, no?

May be the improvemant would be evalf(s) and proceed from there.

@pagan 

But that is correct, I think (say working with 10 Digits), since evalf will ignore
the last those tiny tails - as a float. If one wants more (relative precision ~
number of places) one has to increase Digits, no?

May be the improvemant would be evalf(s) and proceed from there.

@Markiyan Hirnyk 

So you convinced yourself, that z = 1/2 is the only discontinuity?

@Markiyan Hirnyk 

So you convinced yourself, that z = 1/2 is the only discontinuity?

I do not quite understand your reasoning: do you mean the change
of variables by cos(t) = w (and symmetry w.r.t. zero)?

PS: the *.mw does not even export poperly as Maple input *.mpl, grrr.

I do not quite understand your reasoning: do you mean the change
of variables by cos(t) = w (and symmetry w.r.t. zero)?

PS: the *.mw does not even export poperly as Maple input *.mpl, grrr.

@Markiyan Hirnyk 

Markiyan, have nothing at hand for a general (non-degnerated) triangle in the plane.

The simple form above uses: 2 side are parallel to the axes (and one point = zero,
but that is a shift, say for point nearest to zero).

Up to that shift it means: map to vectors onto the axes. That is a base change in
IR^2. One can even do it to the unit vectors, without changing from interior of the
triangle (invariance of domain would be a sledge hammer as argument).

But even having not worked it out I would guess that up to a shift one can reduce
it like that (and transform back to get the indicator), it seems to 'obvious' by
sketching on paper with pencil.

Could you agree?

But it would need to replace simplex[convexhull] in your solution by (affine)
linear algebra in the plane.


Or somewhat shorter: 

there is an affine linear transform, which sends the 3 points to (0,0), (0,1)
and (1,0). The inverse transform maps that 'unit traingle' to the given one.

@Markiyan Hirnyk 

Markiyan, have nothing at hand for a general (non-degnerated) triangle in the plane.

The simple form above uses: 2 side are parallel to the axes (and one point = zero,
but that is a shift, say for point nearest to zero).

Up to that shift it means: map to vectors onto the axes. That is a base change in
IR^2. One can even do it to the unit vectors, without changing from interior of the
triangle (invariance of domain would be a sledge hammer as argument).

But even having not worked it out I would guess that up to a shift one can reduce
it like that (and transform back to get the indicator), it seems to 'obvious' by
sketching on paper with pencil.

Could you agree?

But it would need to replace simplex[convexhull] in your solution by (affine)
linear algebra in the plane.


Or somewhat shorter: 

there is an affine linear transform, which sends the 3 points to (0,0), (0,1)
and (1,0). The inverse transform maps that 'unit traingle' to the given one.

@Markiyan Hirnyk reading the points again it must be modified a bit,
but in this case it is easy I think

ind:= (x,y) -> `if`( 0<=x and 0<=y and y<=1 and x <= y,1,0);

f:= (x,y) -> `if`( 0<=x and 0<=y and y<=1 and x <= y, sin(x+y),0);
plot3d(f, -0.2 .. 1.2, -0.2 .. 1.2, axes=boxed, 
  labels=["x","y",""], orientation=[-105,70],
  lightmodel=light3);

@Markiyan Hirnyk reading the points again it must be modified a bit,
but in this case it is easy I think

ind:= (x,y) -> `if`( 0<=x and 0<=y and y<=1 and x <= y,1,0);

f:= (x,y) -> `if`( 0<=x and 0<=y and y<=1 and x <= y, sin(x+y),0);
plot3d(f, -0.2 .. 1.2, -0.2 .. 1.2, axes=boxed, 
  labels=["x","y",""], orientation=[-105,70],
  lightmodel=light3);

@Markiyan Hirnyk 

Do not have Maple here ... but meant f(x,y) = sin(x+y) * indicator(x,y)
indicator = 0 or 1 (here piecewise should do, see above) to avoid the
simplex commands

@Markiyan Hirnyk 

Do not have Maple here ... but meant f(x,y) = sin(x+y) * indicator(x,y)
indicator = 0 or 1 (here piecewise should do, see above) to avoid the
simplex commands

These are (x,y), both positive and y is below the falling diagnonal + 1, 
i.e. { (x,y)| 0 <= x, 0 <= y, y <= -x+1 } as indicator.