Digits:=15;
exp(-.3872983346*r*(2.*r-2.))/(r*HeunB(0., -.8801117368*I, -.1936491673, 0., (.8801117368*I)*r)^2);
Int(Re(%), r = 0.6 .. 1, method = _d01ajc); 
evalf(%);

                          0.529829821983743

Indeed. Most weekends do not last 19 days. Ahem :-)

May be not enough Digits? Have you tried with Digits = 500 ?

Looking at it (interface(rtablesize=12): evalf[2](PP);) it seems to be quite extreme, between 1 and 10^460.

Another way would be A:=convert(PP, rational): evalf(Determinant(A));

That way I get ~ -.23E639 for Digits = 50

I do not know what they are using as source for tracking (though I get different answers through my posted links, depending on my access / proxy, not all 'know' my country)

PS: MaplePrimes is the only site which does not work properly with my security settings ...

Preben, sometimes I miss to see displays for results of your fine answers (ok, I hate to use the board functionality as well)

Please provide a valid Matrix. It does not make sense to consider one with 'undefined' entries.

You may wish to provide the formal ranges for variables. And say why you think your guess is true (beyond the experiments)

Modify Kitonum's code to t = 0 .. 300

Markiyan, I read that Wiki as "especially for ..." and have not looked in the various definitions (you may check the German version as well).

Are you sure what happens if you ever use or copy some code using j as an integer, say counting or looping?

with(inttrans);

hilbert(f(t), t, s); convert(%, Int);

    1       f(t)
   ---- Int(-----, t = -infinity .. infinity, CauchyPrincipalValue)
    Pi      t - s

H(j*w)=1/(1+tau*j*w); subs(j=I, %); # and w, not omega
subs(w=-I*t, %);

                                      1
                           H(t) = ---------
                                  t tau + 1

hilbert(f(t), t, s);
eval(%, f = 't -> 1/(t*tau+1)'):
simplify(%);

                         hilbert(f(t), t, s)

           {                                       1
           {         0                 tau = --------------
           {                                 RootOf(Im(_Z))
           {
           { ln(-tau) - ln(tau)
           { ------------------             otherwise
           {   (s tau + 1) Pi

So it depends on Im(tau) = 0 or not, i.e. tau purely real or not. ?

Working on Win in case of needs I export as gif or jpeg and everything is fine, using OpenOffice or LibreOffice.

Exporting as *.eps (looking ugly) it can be read by gsview, which converts to pdf (if the kids do not like cmd line).

A very easy way is: rename *.eps ---> *.ps and just use SumatraPdf (it understands a whole bunch of formats and avoids some risks of Acrobat)

For your very example I would use impliciteplot3d

Though I very roughly can imagine what you mean it would be not bad to give some details. Do you need "sternförmig"for the regions (sorry, do not know the English word ~ star shaped ?). And in higher dimensions that may be laborious (not astonishing though)

Ioannis, I once used that wayback machine for your (joined) work on Kepler's equation - but they saved only parts (especially I wanted to look at the references in your Russian article, how to use and understand your usage and notations of the function "HW", including Maple code). Maybe you want to create an extra topic on that?

strange ... I used compile, it is slow. But yours works. Sigh.