Thank you :-)

Mine is Win 7 Professional (oops) SP 1 64 Bit, preconfigured on some Acer.

I am AVt (as admin - ok, not the best idea) on my system, the user directory has AVt and public only
and I it installed as AVt, access is it as AVt (and Maple works, except 'compile').

It does not even go to show up the directory where it fails.

I guess, it is an administration problem: where to write with what rights.

Edited: very strange

The *.mw works painless.

Inserting and executing the simple code in a classical sheet executes and returns a
correct result. But without the output of "Compiler:-Compile(tmp)". Then it crashes.

Sigh.

I want to use it in classical sheets, as before. Of course.

Thank you :-)

Mine is Win 7 Professional (oops) SP 1 64 Bit, preconfigured on some Acer.

I am AVt (as admin - ok, not the best idea) on my system, the user directory has AVt and public only
and I it installed as AVt, access is it as AVt (and Maple works, except 'compile').

It does not even go to show up the directory where it fails.

I guess, it is an administration problem: where to write with what rights.

Edited: very strange

The *.mw works painless.

Inserting and executing the simple code in a classical sheet executes and returns a
correct result. But without the output of "Compiler:-Compile(tmp)". Then it crashes.

Sigh.

I want to use it in classical sheets, as before. Of course.

I am not sure, whether it is a general and 'stable' recipe for those tasks to decompose
into positive and negative part.

Up to some scalings it is Int( fct * indicator, variables in a cube ).

Where the indicator function here is quite complicated and expressed in terms of Heaviside
(signum would do as well), selecting a set, where a certain trig expression is in a range.

That very set is semi-analytic (i.e. where the inner expression is positive), of dim=3 
(I think) and already plotting the boundary makes me unsure if one could describe it as
surface in a 'reasonable' way.

But that means, that the task actually is: Int( fct, variables in that 3-dim-set ).
A kind of 'abuse' ...


For Preben Alsholm's approach using tanh to approximate the indicator: it is very rough
and an (adaptive) routine will do better - if it works.

The technical problem with fct * indicator = integrand is: most integrators assume, that
the integrand is C^2 (and thus will refine beyond any fixed tanh-approximation). But it
is not 'true' for the task.

There are some (symbolic) ways for 1-dim integrals, but I guess the curse of dimensions
will kill them here in the current implementation by time+memory.


But that are just my side remarks.

fct = (sin(2*a)*cos(b)*cos(w)-2/5)*sin(a)*cos(b)^3*sin(w)*(1+1/4*cos(2*b))
set = {(a,b,w) | 0 <= abs(sin(2*a)*cos(b)*cos(w)-sin(a)^2*sin(b)*sin(w))-2/5}

@Markiyan Hirnyk 

Please find details appended for the aproach.

MP_some3dim_int2.mws

@Markiyan Hirnyk 

Please find details appended for the aproach.

MP_some3dim_int2.mws

I looked at that symmetry in a=Pi and then once more. Find a sheet appended.

MP_some3dim_int.mws

I looked at that symmetry in a=Pi and then once more. Find a sheet appended.

MP_some3dim_int.mws

I would expect to have included the full versions (not sure for Matlab).
And of course Mathematica is missing. If comparisions make sense.

Thank you, your task and critics motivated me.

Here is my update including the Laplace method and relating it to the numerical test (quite similar to what you did in your posted solution in "simple answer").

MP_asympt_integra.zip

Thank you, your task and critics motivated me.

Here is my update including the Laplace method and relating it to the numerical test (quite similar to what you did in your posted solution in "simple answer").

MP_asympt_integra.zip

Find it appended (as *.mws and *.pdf).

If needed then I can provide the 'Laplace method' for that asymptotics as well,
that should avoid to compute the integral.

MP_asympt_int_2F.zip

Find it appended (as *.mws and *.pdf).

If needed then I can provide the 'Laplace method' for that asymptotics as well,
that should avoid to compute the integral.

MP_asympt_int_2F.zip

@Markiyan Hirnyk 

Then I do not understand your question.


I think Maple has not implemented such asymptotics or limits for integrals.

And computing the integral results in hypergeometrics, where the limits have
to be taken w.r.t. parameters, which seems not be implemented as well (and
as Temme's article shows, that it would be lengthy already for 2F1).


Are you missing a reference for the formula I used?

Edited (some hours later): 

Ok, I see. You are right, that reasoning is incomplete and I have to think about it.

In a more simple form your task is: 

Limit( Int((n+1)*t^n/(1-z*t)^(1/2),t = 0 .. 1), n=infinity) =  1/sqrt(1-z), abs(z) < 1

@Markiyan Hirnyk 

Then I do not understand your question.


I think Maple has not implemented such asymptotics or limits for integrals.

And computing the integral results in hypergeometrics, where the limits have
to be taken w.r.t. parameters, which seems not be implemented as well (and
as Temme's article shows, that it would be lengthy already for 2F1).


Are you missing a reference for the formula I used?

Edited (some hours later): 

Ok, I see. You are right, that reasoning is incomplete and I have to think about it.

In a more simple form your task is: 

Limit( Int((n+1)*t^n/(1-z*t)^(1/2),t = 0 .. 1), n=infinity) =  1/sqrt(1-z), abs(z) < 1

In a first way I computed the integral giving two 2F1 hypergeometric function terms in
the unit circle hypergeom([1/2, n+1],[2+n],r) which I reduced with some handwaving for
the series to hypergeom([1/2, large],[large],r), which is 1/sqrt(r), giving the 26/3. (*)

But was warned by an article of Temme "Large parameter cases of the Gauss hypergeometric
function" (2002), http://arxiv.org/abs/math/0205065v1 about sloppy reasoning.

Hence looked into filed lectures about asymptotics. And after verifying the above is
correct. But one apply that directly, without computing the integral:

  Int((1-(3/26)*x-(37/49)*x^2)^n, x = 0 .. 1) = Int(f(t)*exp(n*phi(t)),t = 0 .. 1)


Now from asymptotic expansions for such an integral over an interval t = a ... b it is
know, that it can be approximated as follows (Laplace case):

  eval(%, f=1);
  eval(%, phi = 't -> ln(1-3/26*t-37/49*t^2)');
  simplify(%, exp) assuming 0
  1/n*f(b)/D(phi)(b)*exp(n*phi(b)):
  subs(b=a, %):
  J(n) = %% - %;

                   f(b) exp(n phi(b))   f(a) exp(n phi(a))
            J(n) = ------------------ - ------------------
                      n D(phi)(b)          n D(phi)(a)

And one continues by using that

  eval(%, f = 1):
  eval(%, phi = 't -> ln(1-3/26*t-37/49*t^2)');
  subs(a=0, b=1, %);
  simplify(%, exp);
  
  %*n;
  limit(%, n=infinity);


            limit(n*J(n),n = infinity) = 26/3

(*) Correction: 1/sqrt(1-r), use 'simplify' for that.