@spalinowy From a quick scan with the "naked eye" (i.e., no computation), I see that you can eliminate some subset of {Z1(s), Z2(s), Phi1(s), Psi2(s)}, but certainly not Psi1(s). If you want to proceed along those lines, the command is eliminate.

@mew26 Ah, I see the problem then. This is a bit more complicated than what subs can do. There is a related command subsindets that can handle this. But before I can solve it fully, I need to know whether in each target integral, the only exponent used on x is n for some specific value of n. (I don't consider the lone x as having an exponent, although, of course, in some contexts we would say that its exponent is 1.)

Would you please enter here in MaplePrimes the exact subs command that you're trying to use? Please type it here directly rather than copy-and-paste. Or you can upload a worksheet using the green uparrow on the toolbar in the MaplePrimes editor.

There's no reason why one shouldn't be able to substitute one unassigned indexed name for another with the same index. For example:

J:= Int(a[n]*x^n*u(x), x= -infinity..infinity):
subs(a[n]= b[n], J);

However, note that in a name such as a_n or a__n (rather than a[n]), one should consider the "index" strictly as the literal letter n rather than as an index that can take numeric values. In these cases, it'd be better to refer to n as a subscript rather than as an index.

@Muhammad Usman If you enter it into Maple, you get a result (in terms of Elliptic functions). Is that result from Maple not useful to you?

@maple2015 Your 6-term model is worthless. That's obvious from the p-values (the last column of the summary table). Once you see that, there's no point in pursuing any further analysis. Don't be fooled by the relatively high R², 86%. Adding terms will always increase R², regardless of whether the terms are significant.

Why are you ignoring the model that I proposed? I don't claim to have any great insight that led me to it. But, having stumbled upon it, I see that its p-values are really, really, good---so good that it looks like a textbook problem that was designed to lead you to that model.

@Muhammad Usman In that case, the answer is as I said. Just "freeze" u(eta) to a simple name such as u__eta, and enter the integral into Maple. The results do not depend on the discriminant. Also, without loss of generality you can make a[3]=1 and d[2]=0 because this can be accomplished with a trivial linear substitution.

@acer I hadn't found an exact simplification of the residuals to 0 at the time that I posted. My first attempt tried evala, but it balked, I think because of the abs. Since it was well past when I wanted to go to sleep, I was for the moment satisfied with a floating point verification, which I did but didn't post because it's trivial. I totally agree with you about the mathematical relevance of an exact simplification, and I'm impressed with the one that you found.

@Kitonum Sorry, you're right. I misinterpretted your assuming as an attempt to tell solve that the variables were real. And I interpretted it that way because that would be a natural and useful thing to do if it were possible.

This concept can be easily generalized to rtables of any number of dimensions, and it can be done in such a way that all previously covered cases are handled with the same syntax.

#The optional fourth parameter d is the dimension along which
#to perform the operation.
InsertSlice:= proc(A::rtable, V, p::posint, d::posint:= 1)
local D:= [rtable_dims(A)], n:= nops(D), L:= (..)$d-1, R:= (..)$n-d;
   ArrayTools:-Concatenate(
      d, 
      A[L, ..p-1, R], 
      `if`(
         V::rtable and rtable_num_dims(V) < n,
         ArrayTools:-Alias(V, 0, [D[..d-1][], 1, D[d+1..][]]),
         V
      ),
      A[L, p.., R]
   )
end proc:
   
A:= Array((1..3)$3, rand(-9..9)):
seq(A[..,k,..], k= 1..3);
                   [-1  -4  -8]  [-8  -3  -2]  [-7  7  -9]
                   [          ]  [          ]  [         ]
                   [ 4   1  -4], [-1  -1  -7], [-2  1   4]
                   [          ]  [          ]  [         ]
                   [ 9   5   0]  [-1   8   5]  [ 8  2   2]
M:= Matrix(3$2, rand(-9..9));
                                   [ 9  -1  -5]
                                   [          ]
                              M := [ 3  -9   8]
                                   [          ]
                                   [-1  -1  -6]
B:= InsertSlice(A, M, 2, 2):
seq(B[.., k, ..], k= 1..4);
            [-1  -4  -8]  [ 9  -1  -5]  [-8  -3  -2]  [-7  7  -9]
            [          ]  [          ]  [          ]  [         ]
            [ 4   1  -4], [ 3  -9   8], [-1  -1  -7], [-2  1   4]
            [          ]  [          ]  [          ]  [         ]
            [ 9   5   0]  [-1  -1  -6]  [-1   8   5]  [ 8  2   2]
#An example with a "thick" slice:
M1:= Array(1..3, 3..4, 1..3, rand(-9..9)):
seq(M1[..,k,..], k= 3..4);
                         [ 8   2   6]  [-3   4  -6]
                         [          ]  [          ]
                         [-8  -1  -5], [-1   8  -9]
                         [          ]  [          ]
                         [ 5  -4   7]  [-2  -2  -4]
B1:= InsertSlice(A, M1, 2, 2):
seq(B1[.., k, ..], k= 1..5);
     [-1  -4  -8]  [ 8   2   6]  [-3   4  -6]  [-8  -3  -2]  [-7  7  -9]
     [          ]  [          ]  [          ]  [          ]  [         ]
     [ 4   1  -4], [-8  -1  -5], [-1   8  -9], [-1  -1  -7], [-2  1   4]
     [          ]  [          ]  [          ]  [          ]  [         ]
     [ 9   5   0]  [ 5  -4   7]  [-2  -2  -4]  [-1   8   5]  [ 8  2   2]

@Kitonum It's mentioned in ?solve,details that solve ignores assuming real even if you use the option useassumptions. Thus the solutions need to be verified, which, in this case, is very messy if you want to do it exactly, but not bad if you're okay with a decimal approximation. Your three solutions are indeed correct; I'm just saying that they need to be treated with some suspicion at first.

@tomleslie Sadly, the ~ operator was added in Maple 12 or Maple 13. But that's easy to work around by using map or zip.

@das1404 David, I added a significant amount of new material to my Reply "eval" above, so you should probably read that again.

@tomleslie I suspect that the most-significant performance penalty of MutableSets will be making equality comparisons.

@JoshuaLeiter Thanks. My only regrets are that your one line of code turned into my 11 lines, plus comments, and that this happens so often in 3-D plots. There are so many possible anomalies. Yet it's just a pair of algebraic functions. It took me so many tries, and I've been doing this for many years. There should be some more fundamental ways to correct these. What I've learned today: Strive to use exactly one expression (or one parametric list of three expressions) for each connected component of the surface.

@9009134 I didn't ignore your problem. I worked on it for about two hours without making progress that I thought significant enough enough to report. I was neither able to do it nor prove that it couldn't be done. I suspect that some other correspondents here had the same experience. Perhaps if we detail our experiences, some collaboration will lead to progress.

I recall being able to algebraically eliminate one of the three dependent variables. I thought that that was odd, but not necessarily a bad sign.

I'll see if I can find the notes that I made about it. Recovering unsaved worksheets after a Maple GUI crash is iffy.