Maple's help about SphericalY gives 3 references and talks about sign conventions:

  "Attention should be paid to the normalization conventions adopted. The requirement
   that the double integral mentioned is equal to one does not fix a phase, which can
   then be chosen in different ways ..."

and says they used a norming. But it really should be clear (I find it a bit messy).

Let be A:= (your first expression SP1) using Maple's notation, B:= (using your Ylm),
so both are the same (3,-3),(0,0) case. Then

  A/B; 
  eval(%, theta=arccos(x));     # 0 < theta < Pi/2 
  simplify(%) assuming 0
    I

Which "explains" interchanging real and imaginary part.  


That one of this is zero (depending on conventions) follows by looking at symmetries
of the integrand: plot3d indicates that w.r.t. phi it is 3 times the same (periodics),
but with range 0 ... 2*Pi / 3. And then a symmetry in phi = Pi/3 +- t, but changing
the sign (so modify it). That can be shown by Maple.

Now plotting again for the smaller range indicates a symmetry in theta = Pi/2 +- t,
which I can not prove, but plotting the difference is of error ~ 1E-13 (with 15 Digits)
where the actual magnitude is ~ 1E2. So at least it is "numerically true".

The range now is [theta=0..Pi/2, phi=0..2*Pi/3/2] times 12 gives the integral.

Using evalc *now* it should show the vanishing and the integral reduces. 

On that one can re-write it as algebraic task using arcsin, but it looks very ugly
and it is unclear, whether that helps.

Let me know whether I should clean up my sheet to load it up (but I will not care for
the messy problem of sign conventions).


Correction: Maple's help states the convention for integers at least and it is

SphericalY(a,b,theta,phi) = (-1)^b/2/sqrt(Pi)*
  sqrt((2*a+1)/(a+b)!*(a-b)!) * exp(phi*b*I) * LegendreP(a,b,cos(theta))

for integers a,b sucht that abs(b) <= a

See other conventions at http://en.wikipedia.org/wiki/Spherical_harmonics#Conventions
or that of Mathematica, http://mathworld.wolfram.com/SphericalHarmonic.html

                                                      b
But be aware of notations of upper and lower index  P   (x) = LegendreP(a,b, x),
                                                      a

en.wikipedia.org/wiki/Associated_Legendre_polynomials#The_first_few_associated_Legendre_functions

You SP1 has 0,0 not 3,3 in the 2nd sperical ?

edited: file attached, method = cuhre works with ~ 8 digits,
confirming Maple's result in 3 digits: real part "is" zero.

MP_spherical.mws

Edited: one can even prove symbolically, that the real part
is zero (using Maple's notations). Might it be a matter of
conventions used for the task?

I guess you mean 

Int(Int(f(x,y)*exp(I*w*g(x,y)),y=0..1),x=0..1);

                  1    1
                 /    /
                |    |
                |    |   f(x, y) exp(w g(x, y) I) dy dx
                |    |
               /    /
                 0    0
What is f (or at least its main properties) and
what is the magnitude of w?

Thank you, but I am not satisfied:

That f(x)/sqrt(1-x^2) 'cries' for Chebychev, but I think, that Maple converts to JacobiP
(the generalization) and then answers by MeijerG, using Prudnikov & Marichev.

There is a package by Moiseev to compute orthogonal series expansions covering Chebychev
(as special case of Jacobi), but I have not tried [I find it unhandy, being not a usual
Maple sheet classic style). May be it helps to develop those Int(2F1/sqrt) and have some
handy sum.


The actual mathematical reasoning is D. Cross, Linking Integral Projection at archiv.org,
http://arxiv.org/abs/0907.3446v1

In Prop 1. he reduces those linking numbers to winding numbers using homtopy invariance
and a geometric argumentation (for which my imagination is not enough).

The other interpretation is as degree of the Gauss map as integral (may be, but for me
that always is the number of generic points of a generic fibre).

For bothe one would need a reasonable presentation of the Hopf link ( = 2 rings in IR^3)
mentioned in your reference to math.stackexchange.com

For a true solution I have to give up :-(

Thank you, but I am not satisfied:

That f(x)/sqrt(1-x^2) 'cries' for Chebychev, but I think, that Maple converts to JacobiP
(the generalization) and then answers by MeijerG, using Prudnikov & Marichev.

There is a package by Moiseev to compute orthogonal series expansions covering Chebychev
(as special case of Jacobi), but I have not tried [I find it unhandy, being not a usual
Maple sheet classic style). May be it helps to develop those Int(2F1/sqrt) and have some
handy sum.


The actual mathematical reasoning is D. Cross, Linking Integral Projection at archiv.org,
http://arxiv.org/abs/0907.3446v1

In Prop 1. he reduces those linking numbers to winding numbers using homtopy invariance
and a geometric argumentation (for which my imagination is not enough).

The other interpretation is as degree of the Gauss map as integral (may be, but for me
that always is the number of generic points of a generic fibre).

For bothe one would need a reasonable presentation of the Hopf link ( = 2 rings in IR^3)
mentioned in your reference to math.stackexchange.com

For a true solution I have to give up :-(

The base change cos(x) + 1 = 2* t^2 gives the integral with the two elliptic functions.

Sorry for the typo, it is - 1, yes. Please find a sheet attached.

MP_linkingintegra.mws
MP_linkingintegra.pdf

PS: cos(x) = t works as well for the shown approach - however I hoped
that having sqrt(1 -x^2) in the denominator would help to proceed ...

The base change cos(x) + 1 = 2* t^2 gives the integral with the two elliptic functions.

Sorry for the typo, it is - 1, yes. Please find a sheet attached.

MP_linkingintegra.mws
MP_linkingintegra.pdf

PS: cos(x) = t works as well for the shown approach - however I hoped
that having sqrt(1 -x^2) in the denominator would help to proceed ...

I agree with Markiyan and also suggest that you check, whether you really want it that way - or have a typo.

Intuitively using exp(sigma*U) etc in the exponent would decay to zero in such an extreme speed for the task that I can not understand what its practical use it would be.

But in the case you are sure: what is the context for the task?

I do not quite understand your graphics, but looking at it you have
x and y as angles. But in the picture they are not the same: else it
would be a trapezoid.
May be you can plot it again using h1 = 3 * h2/2 3 * h1/2

I do not quite understand your graphics, but looking at it you have
x and y as angles. But in the picture they are not the same: else it
would be a trapezoid.
May be you can plot it again using h1 = 3 * h2/2 3 * h1/2

I think that the sheet gives all solution and you get them by using the final values
given in 'My' to evalute S34, but only 2 are real:

[x = .406715103897382, y = .406715103897382], [x = 2.73487754969241, y = 2.73487754969241]

The first was already given and it implies s < 0, the second one is missed by careless
use of fsolve.

Note that Eq 3 and Eq 4 already enforce, that x=y (since inverting cos is unique for
your desired real range 0 < x,y < Pi/2.

I agree, that it might be possible you have some error in the setting of your equations.

I think that the sheet gives all solution and you get them by using the final values
given in 'My' to evalute S34, but only 2 are real:

[x = .406715103897382, y = .406715103897382], [x = 2.73487754969241, y = 2.73487754969241]

The first was already given and it implies s < 0, the second one is missed by careless
use of fsolve.

Note that Eq 3 and Eq 4 already enforce, that x=y (since inverting cos is unique for
your desired real range 0 < x,y < Pi/2.

I agree, that it might be possible you have some error in the setting of your equations.

Markiyan, I am not used to that package and its notations. For me it is the free Z-module in Z^4 having basis [0, -1, 0, 1], [1, -2, 1, 0], [12, -12, 0, 0]. So 'abstract' it is Z^3

May be one could do two things: a) how does that package handle Z^3 (I guess that it tries to avoid embeddings) or b) how does the package allow to describe subgroups? This the Abelian case, of course.

Markiyan, I am not used to that package and its notations. For me it is the free Z-module in Z^4 having basis [0, -1, 0, 1], [1, -2, 1, 0], [12, -12, 0, 0]. So 'abstract' it is Z^3

May be one could do two things: a) how does that package handle Z^3 (I guess that it tries to avoid embeddings) or b) how does the package allow to describe subgroups? This the Abelian case, of course.

@Markiyan Hirnyk 

Thx, but my old sins have been in the field of algebraic/analytic geometry and algebra :-)

The symmetries allows to write it as 

  2*2* 1/(4*Pi)*Int((cos(t)-sin(s)-cos(t)*sin(s))/
    (3+2*(cos(t) - sin(s) - cos(t)*sin(s)))^(3/2),
    [t=0..Pi,s=Pi/2..Pi/2 + Pi]);

Now use Change(%, {t=y+Pi/2, s = x+Pi}) and call the integrand K0(x,y) = 
  -(sin(y)*sin(x)+sin(y)-sin(x))/(3-2*sin(y)+2*sin(x)-2*sin(y)*sin(x))^(3/2)

Then K0(x,y) = K0(-y,-x), but I forgot the trick to use it for that situation :-(

May be one should try polar coordinates, since it even looks symmetric to rotations,
  plot3d(K0(x,y), y=-Pi/2 .. Pi/2, x=-Pi/2 .. Pi/2, axes=boxed, orientation=[135,15])