Triple integral associated with sum of three cubes probability

Question

Main reference is an answer to the following question:

Probability that a natural number is a sum of two squares?

According to a comment by the same author, the method in this answer can be employed as well for solving the following problem:

What is the probability that a natural number is a sum of three cubes?

The reason is that $1/n = (1/n^{1/3})^3$, so an analogous method should work for three cubes.
We define an initial segment of the naturals with length $N$ and count all sums of three cubes in that segment. The idea is that if $N\to\infty$, then we have the desired probability. (Oh yeah, it assumed that zero is part of the naturals as well) The accompanying program is in Pascal:

program kubussen;

procedure test(N : integer; show : boolean);
var
  i,j,k,t,L : integer;
  memo : array of boolean;
begin
  t := 0;
  i := 0;
  SetLength(memo,N);
  while true do
  begin
    if sqr(i)*i > N-1 then Break;
    j := i;
    while true do
    begin
      if sqr(i)*i+sqr(j)*j > N-1 then Break;
      k := j;
      while true do
      begin
        L := sqr(i)*i+sqr(j)*j+sqr(k)*k;
        if L > N-1 then Break;
        if show then
        begin
          Writeln(L:3,' = ',i,'^3 + ',j,'^3 + ',k,'^3');
          if memo[L] then Writeln('duplicate');
        end;
        memo[L] := true;
        k := k + 1;
        t := t + 1;
      end;
      j := j + 1;
    end;
    i := i + 1;
  end;
  Writeln(t,'/',N);
end;

begin
  test(10,FALSE);
  test(100,TRUE);
  test(1000,FALSE);
  test(100000,FALSE);
  test(1000000,FALSE);
  test(10000000,FALSE);
  test(100000000,FALSE);
  test(1000000000,FALSE);
end.

Sample output, details (show=TRUE) for $N=100$ :

6/10
  0 = 0^3 + 0^3 + 0^3
  1 = 0^3 + 0^3 + 1^3
  8 = 0^3 + 0^3 + 2^3
 27 = 0^3 + 0^3 + 3^3
 64 = 0^3 + 0^3 + 4^3
  2 = 0^3 + 1^3 + 1^3
  9 = 0^3 + 1^3 + 2^3
 28 = 0^3 + 1^3 + 3^3
 65 = 0^3 + 1^3 + 4^3
 16 = 0^3 + 2^3 + 2^3
 35 = 0^3 + 2^3 + 3^3
 72 = 0^3 + 2^3 + 4^3
 54 = 0^3 + 3^3 + 3^3
 91 = 0^3 + 3^3 + 4^3
  3 = 1^3 + 1^3 + 1^3
 10 = 1^3 + 1^3 + 2^3
 29 = 1^3 + 1^3 + 3^3
 66 = 1^3 + 1^3 + 4^3
 17 = 1^3 + 2^3 + 2^3
 36 = 1^3 + 2^3 + 3^3
 73 = 1^3 + 2^3 + 4^3
 55 = 1^3 + 3^3 + 3^3
 92 = 1^3 + 3^3 + 4^3
 24 = 2^3 + 2^3 + 2^3
 43 = 2^3 + 2^3 + 3^3
 80 = 2^3 + 2^3 + 4^3
 62 = 2^3 + 3^3 + 3^3
 99 = 2^3 + 3^3 + 4^3
 81 = 3^3 + 3^3 + 3^3
29/100
178/1000
13142/100000
124398/1000000
1213524/10000000
11991346/100000000
119249735/1000000000

Trivial duplicates are: $$ i^3+j^3+k^3 = i^3+k^3+j^3 = k^3+i^3+j^3 = j^3+i^3+k^3 = j^3+k^3+i^3 = k^3+j^3+i^3 $$ They are eliminated in the program by starting loops accordingly : $i\ge 0,j\ge i,k\ge j$ .
If these duplicates are not eliminated, then the outcome must be divided by six.
Non-trivial duplicates are retained, though. For example: $$ 855 = 1^3 + 5^3 + 9^3 = 0^3 + 7^3 + 8^3 \\ 729 = 1^3 + 6^3 + 8^3 = 0^3 + 0^3 + 9^3 \\ 251 = 2^3 + 3^3 + 6^3 = 1^3 + 5^3 + 5^3 \\ 216 = 3^3 + 4^3 + 5^3 = 0^3 + 0^3 + 6^3 \\ 344 = 4^3 + 4^3 + 6^3 = 0^3 + 1^3 + 7^3 $$ With help of the above, it is conjectured that, in the limit for $N\to\infty$, the desired probability is equal to the following integral, over the domain $x\ge 0,y\ge 0,z\ge 0, x^3+y^3+z^3 \le 1$ : $$ \frac{1}{6} \iiint dx\,dy\,dz $$ But here I'm stuck. My knowledge about multiple integrals has become rusty and I have no idea how to calculate other than numerically (and the latter is rather prohibitive as well). It is suggested by the sample output that the outcome must be somewhat close to $0.119249735$ , but how good has it converged? So the question is: does there exist an exact solution for the integral?

Where does the conjecture come from? Integrating $1$ gives the expected number of solutions. But why should integrating $x^3 + y^3 + z^3$ give the probability? Does not seem natural to me. — zhoraster, Sep 10 '15 at 14:31
Are you sure you entered that integral correctly? The reason I ask is that it seems that a numerical value of that integral (which, it seems, can be expressed in a hypergeometric function) is $0.0593394$. I have not, in contrary to @zhoraster, read all the text, but rather just looked at the integral. — mickep, Sep 10 '15 at 14:34
But then this is not the probability, but rather the expected number. — zhoraster, Sep 10 '15 at 17:10
@zhoraster: if it is expected that it rains tomorrow, then isn't that the same as saying that there is a probability that it rains tomorrow? — Han de Bruijn, Sep 11 '15 at 13:10
Yes, it is the same. However, this is different from saying that the expected precipitation amount is 30mm. You measure the expected precipitation amount, not the probability, this is my point. — zhoraster, Sep 11 '15 at 13:22

score 1 · Accepted Answer · answered Sep 11 '15 at 13:07

The integral is a little bit simpler than I first thought: $$ \frac{1}{6} \iiint dx\,dy\,dz = \int_{x=0}^{x=1} \left[ \int_{y=0}^{y=(1-x^3)^{1/3}} (1-x^3-y^3)^{1/3}\, dy \right] dx $$ This is what MAPLE says about the (indefinite) inner integral:

> int((1-x^3-y^3)^(1/3),y);
                                                          3
             3 (1/3)                                     y
       (1 - x )      y hypergeom([-1/3, 1/3], [4/3], - -------)
                                                             3
                                                       -1 + x

The hypergeom is abacedabra for me. And if not: we still have the outer integral, which seemingly cannot be evaluated exactly. So let's evaluate the integral numerically, with help of MAPLE again:

> evalf(Int(Int((1-x^3-y^3)^(1/3),y=0..(1-x^3)^(1/3)),x=0..1))/6;

                             0.1186788238

Which I find satisfactory close to the last iterand ($0.119249735$) of the program.

Triple integral associated with sum of three cubes probability

1 Answers1