@candy898 For positive a, the integral doesn't converge because of the singularity at x=a. You can enter the integral into Maple as

int(ln(x)/(x-a)/(x-1), x= 0..infinity);

Note that it is ln(x), not In(x).

I don't know exactly what Mathematica is doing here. I think it's called Wynn extrapolation or Wynn's epsilon method. Note that Mathematica does express skepticism about its own answers. Also note that the book that you referenced merely explains why Mathematica gives numeric answers for such integrals; it does not claim that those answers are correct (as far as I can tell). I think that the author is more-or-less apologizing for Mathematica's behaviour.

@adel-00 I don't really understand your comment. Are you saying that the program doesn't work for values of theta and phi other than zero? What exactly happens?

@Kitonum Actually, there is no missing parenthesis. The sqrt definitely encompasses the whole expression.

@Kitonum There is a missing right parenthesis in the original. My answer assumed that it closed the whole expression. Yours assumes that it closes the sqrt.

@Adri van der Meer The dynamic Vector approach is significantly faster for sequential output. My earlier Reply was wrong about that.

@Hermitage What tools would you have available on a test? Perhaps a graphing calculator? Maple? Although the problem is somewhat tedious to do analytically, the most basic plot shows that the range is [0, +infinity).

@Carl Love Being able to upload worksheets for display in a post is of fundamental importance for MaplePrimes, IMO. Are there any progress reports on returning this functionality? Are you planning to return this functionality?

@adel-00 Remove the line tau:= Pi from the top of the program. That's all that you need to do. I said that before.

The table in your question is not a Cayley table. There are no repeated elements in any row or column of a Cayley table.

@candy898 The natural logarithm of x is written ln(x), not In(x): letter L for Logarithm.

So, did you mean for (x-1) to be in the denominator?

@candy898 What does In(x) mean? There is no standard function with that name. Also, it is not clear if you mean for (x-1) to be in the numerator or denominator. If you don't know how to make the symbols, then type it out in English words.

Yeah, I know what an integral is, and I know that you mean that.

tau:= Pi:
P1:= plot(Spec, -10..10, axes=boxed, title=tit, color=black, font=[2,3,18],
            thickness=2, tickmarks=[3,3], titlefont=[SYMBOL,14], font=[1,1,18],
            linestyle=1
       ):

Normalize:= proc(P::specfunc(anything, PLOT))
local A,Smax1;
     A:= op([1,1], P);
     Smax1:= max(A[..,2]);
     if A::list then  A:= Matrix(A)  end if;
     A[..,2]:= A[..,2]/Smax1;
     subsop([1,1]= A, P)
end proc:

P1:= Normalize(P1):
for k from 2 to 5 do
     tau:= k*Pi;
     P||k:= plot(Spec, -10..10);
     P||k:= plottools:-translate(Normalize(P||k), 0, k-1)
od:
display([P||(1..5)]);

@JohnS What you want to do is essentially step through the variables in the basis of the linear solution, setting each to 1 and the rest to zero. I've written some code to do this; it's just a few lines. If you provide a complete example of the equations to solve and the Z1, ..., Zn equations, then I'll post my code.

This would be so much easier if numerical summation had a digits option akin to the option for numerical integration. The reason that the sum doesn't evaluate numerically for all x is that for some x it can't get Digits accuracy. It would be easy for digits < < Digits.

The numeric summation of this series can be accelerated using a technique based on the integral test, as the summand is eventually decreasing and has a symbolic antiderivative. I've written a procedure to do this numeric summation and it has an equivalent of the digits option. But it is much slower than Kitonum's solution, so I'll save it for later. However, it will numerically sum some series that evalf(Sum(...)) won't touch for any x.