I remember, that Carl had some code at his Yahoo group

It goes a bit a different route, but essentially it should work (not with Maple alone
and, of course, there is the original).

Let be s1, s2, etc the Fourier transforms of sinc(a(k)), those rectangulars are given
explicitely, F denotes the Fourier transform ( F^2 = id up to scaling, suppressed for
simplicity) and let be # = convolution.

Then the product n=4 is Int(F(s1)*F(s2)*F(s3) * f) = Int(F(s1#s2#s3) * f) [we do not
need to care for brackets by associativity; now by Plancherel + Convolution theorem:]
= Int( F(F(s1#s2#s3)) * F(f) ) ) = Int( s1#s2#s3 * F(f) )

Now take f = sinc(a(k)), k < 4 (because we are on k=1,2,3) and one has to verify:

  The 'smoothed trapezoid' s1#s2#s3 is constant on the support of F(f) = rectangle. 

This is due to the 'nested' behaviour of the transforms s_k, as we shrink by 2^k.

That means: the latter integrand equals (s1#s2#s3)(0) * F(f).

I have not cared for the exact constants, but wanted to give a kind of induction
step showing the idea/reason: 

It simplifies by a) nested functions and b) that lives on rectangular functions
allowing 'easy' reduction to the highest order.

The converse (decreasing) would be better, but I have no idea how to start it.

It goes a bit a different route, but essentially it should work (not with Maple alone
and, of course, there is the original).

Let be s1, s2, etc the Fourier transforms of sinc(a(k)), those rectangulars are given
explicitely, F denotes the Fourier transform ( F^2 = id up to scaling, suppressed for
simplicity) and let be # = convolution.

Then the product n=4 is Int(F(s1)*F(s2)*F(s3) * f) = Int(F(s1#s2#s3) * f) [we do not
need to care for brackets by associativity; now by Plancherel + Convolution theorem:]
= Int( F(F(s1#s2#s3)) * F(f) ) ) = Int( s1#s2#s3 * F(f) )

Now take f = sinc(a(k)), k < 4 (because we are on k=1,2,3) and one has to verify:

  The 'smoothed trapezoid' s1#s2#s3 is constant on the support of F(f) = rectangle. 

This is due to the 'nested' behaviour of the transforms s_k, as we shrink by 2^k.

That means: the latter integrand equals (s1#s2#s3)(0) * F(f).

I have not cared for the exact constants, but wanted to give a kind of induction
step showing the idea/reason: 

It simplifies by a) nested functions and b) that lives on rectangular functions
allowing 'easy' reduction to the highest order.

The converse (decreasing) would be better, but I have no idea how to start it.

It is named Plancherel formula (and usually used for 'theory', extending the function space and talking about isometry), in the special case of the task the conjugation drops off, since everything is real.

But that 'isometry' view explains, why I can not find something towards products to be cited ...

It is named Plancherel formula (and usually used for 'theory', extending the function space and talking about isometry), in the special case of the task the conjugation drops off, since everything is real.

But that 'isometry' view explains, why I can not find something towards products to be cited ...

By symmetry one can look at the integrals over the whole Reals and instead using the
Cosinus transform one can take the Fourier transform. Similar to the FCT there is a
formula relating products, after integrating over the Reals the values are the same.

  Int(              f(x)*g(x),              x=-infinity .. infinity) =
  Int( fourier(f(x),x,t)*fourier(g(x),x,t), t=-infinity .. infinity) / (2*Pi); 

The factor 2*Pi is needed, since Maple uses the non-normalized Fourier transform
(while for the discrete case, FFT, it uses a normalization).

I do not have a proof for this (but citations from old Wikipedias), the analog for
Sinus and Cosinus is given in the old book of Titchmarsh.

One needs to extend this to products of more than 2 functions. For which I do not
have a formula (may be in some lectures or books? however 'Parseval's theorem' is
almost always given in terms of norms, sadly). Anyway ...

Now compute the FT of sinc(x*a(k)), notations as in the old thread, and denote it
by sigma(k,t). Maple finds it, it is a rectangular function (as known to Physics),

  sigma := (k, t) -> piecewise(t < -2^(-k),0,t < 2^(-k),2^k*Pi,2^(-k) 
Products of such can be simplify very quickly by Maple.


After experimenting I came up with the following formula (*):

  Int(Product('sinc'(x*a(k)), k = 0 .. n), x = -infinity .. infinity)/Pi = 
  Int(Product('sigma'(k,t),   k = 0 .. n), t = -infinity .. infinity)*
    1/Pi^(n)/2^n /Pi/2^((n-2)*(n-1)/2);

Note that Product(1/'a'(k), k = 0 .. n) = Product(2^k,k = 0 .. n) = 2^(1/2*n*(n+1))

and (a bit of experimenting) 

  Int(Product('sigma'(k,t), k = 0 .. n), t = -infinity .. infinity) = 
    2^(1/2*(n-1)*n+1)*Pi^(n+1);


That is not a proof at all of course, just a sketch how to use Fourier transforms
similar to the proof sketched in the Borwein paper.

Using n=123 for the rhs of (*) quickly evaluates to 1, as expected and hoped.

The reason for the quick evaluation: the rectangles are nested, so only that of the
highest order remains and just the maxima are multiplied, so one gets

  Product('sigma'(k,t),   k = 0 .. n)/Pi^(n+1)  
  = Product(2^k*Pi for abs(t) L.T. 2^(-k),  k = 0 .. n)/Pi^(n+1) and zero else  
  = Product(2^k,  k = 0 .. n)
  = 2^(1/2*n*(n+1)) for abs(t) L.T. 2^(-n) and zero else

By symmetry one can look at the integrals over the whole Reals and instead using the
Cosinus transform one can take the Fourier transform. Similar to the FCT there is a
formula relating products, after integrating over the Reals the values are the same.

  Int(              f(x)*g(x),              x=-infinity .. infinity) =
  Int( fourier(f(x),x,t)*fourier(g(x),x,t), t=-infinity .. infinity) / (2*Pi); 

The factor 2*Pi is needed, since Maple uses the non-normalized Fourier transform
(while for the discrete case, FFT, it uses a normalization).

I do not have a proof for this (but citations from old Wikipedias), the analog for
Sinus and Cosinus is given in the old book of Titchmarsh.

One needs to extend this to products of more than 2 functions. For which I do not
have a formula (may be in some lectures or books? however 'Parseval's theorem' is
almost always given in terms of norms, sadly). Anyway ...

Now compute the FT of sinc(x*a(k)), notations as in the old thread, and denote it
by sigma(k,t). Maple finds it, it is a rectangular function (as known to Physics),

  sigma := (k, t) -> piecewise(t < -2^(-k),0,t < 2^(-k),2^k*Pi,2^(-k) 
Products of such can be simplify very quickly by Maple.


After experimenting I came up with the following formula (*):

  Int(Product('sinc'(x*a(k)), k = 0 .. n), x = -infinity .. infinity)/Pi = 
  Int(Product('sigma'(k,t),   k = 0 .. n), t = -infinity .. infinity)*
    1/Pi^(n)/2^n /Pi/2^((n-2)*(n-1)/2);

Note that Product(1/'a'(k), k = 0 .. n) = Product(2^k,k = 0 .. n) = 2^(1/2*n*(n+1))

and (a bit of experimenting) 

  Int(Product('sigma'(k,t), k = 0 .. n), t = -infinity .. infinity) = 
    2^(1/2*(n-1)*n+1)*Pi^(n+1);


That is not a proof at all of course, just a sketch how to use Fourier transforms
similar to the proof sketched in the Borwein paper.

Using n=123 for the rhs of (*) quickly evaluates to 1, as expected and hoped.

The reason for the quick evaluation: the rectangles are nested, so only that of the
highest order remains and just the maxima are multiplied, so one gets

  Product('sigma'(k,t),   k = 0 .. n)/Pi^(n+1)  
  = Product(2^k*Pi for abs(t) L.T. 2^(-k),  k = 0 .. n)/Pi^(n+1) and zero else  
  = Product(2^k,  k = 0 .. n)
  = 2^(1/2*n*(n+1)) for abs(t) L.T. 2^(-n) and zero else

If you have a sequence of integers (in your case: 8), then there is
always a unique polynomial producing them. And it automatically
maps integers to integers.

Thus the expected answer makes no sense to me: if you do that,
then you actually demand a longer sequence ...

F := x -> -165+35663/84*x-74113/180*x^2+48101/240*x^3-
    1931/36*x^4+8*x^5-28/45*x^6+11/560*x^7

for which F(9) = 156

If you demand it to be 35, then the polynomial function is

F := x -> -286+632873/840*x-7723579/10080*x^2+193123/480*x^3-
    697007/5760*x^4+1729/80*x^5-6511/2880*x^6+143/1120*x^7-121/40320*x^8

Besides chain rule it is the Rodrigues formula

'HermiteH(n,x) = (-1)^n*exp(x^2)*diff(exp(-x^2),`$`(x,n)) ';

Diff(erf(x),`$`(x,n+1)) = 2*(-1)^(-n)/Pi^(1/2)*exp(-x^2)*HermiteH(n,x);

Besides chain rule it is the Rodrigues formula

'HermiteH(n,x) = (-1)^n*exp(x^2)*diff(exp(-x^2),`$`(x,n)) ';

Diff(erf(x),`$`(x,n+1)) = 2*(-1)^(-n)/Pi^(1/2)*exp(-x^2)*HermiteH(n,x);

It was just luck, that I vaguely rembered they did some crazy integrals over products,
which I have saved to those folders one always want to study, but never actually do.

You are right, it is the final part of Thm 2 (ii). But looking into the  proof they
do some iterative steps (suczessiv integration by parts) and use identities, which
will be difficult to reproduce with Maple (though they do not say "it is obvious,
that the following holds true"). Personally I would not try with Maple.

I think it may follow from a paper by Borwein+Borwein, Some remarkable properties of sinc and related integral (year?)

Int(Product(sin(x*a(k))/x/a(k), k = 0 .. n), x = 0 .. infinity),

a:= k -> 1/2^k, sinc = 't -> sin(t)/t'

I hate those notations from Engineering / Physic as given in the first picture

The integral is over ... w.r.t. d(beta*h)/(beta*h). Do I have to understand it as given
in your uploaded sheet?

I hate those notations from Engineering / Physic as given in the first picture

The integral is over ... w.r.t. d(beta*h)/(beta*h). Do I have to understand it as given
in your uploaded sheet?

At least the contour method does not work when it could:

p:=x^5-x+1 is not solvable by radicals, r:=RootOf(_Z^5-_Z+1,index = 3) is the only
real root, thus Int( (x-r)/p, x=-infinity .. infinity) can be computed through the
residues at poles in the upper halfplane, which are r1:= ...index=1, r2:= ... index=2

That gives an 'ugly' expression because of notations (command 'alias' helps).

However Maple returns unevaluated after trying that method. And failed before to
compute the limit of the antiderivative (which is not very explicit, of course).

MP_int_residues.mws