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I am looking to prove my conjecture that \begin{equation} \int\cdots\int f(x_1,\dots,x_n) \,dx_1\cdots\,dx_n=I_n. \end{equation} over the domain $\{(x_1,\dots,x_n):0< x_1\leq\cdots\leq x_n<\infty\}$ where $I_n$ is some expression.

Given that it is very difficult to solve for the $n$ integrals in my case, my approach was to instead solve the single, double and tripple integrals to obtain $I_1$, $I_2$ and $I_3$ which is doable and from there deduce the general expression $I_n$ by identifying the patterns of $I_1$, $I_2$ and $I_3$.

Would this be a correct way of proving the top expression? I have read about proof by enumeration/proof by exhaustion but am not sure whether this falls under that category.

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  • $\begingroup$ Sounds like this would be a good case for a proof by Mathematical Induction. What is $f(x_1, \ldots, x_n)$? $\endgroup$
    – gt6989b
    Jun 21, 2021 at 15:54
  • $\begingroup$ What you mention would be a good way to start. But it would not be a proof. After you get the formula, then try proof by induction. $\endgroup$
    – GEdgar
    Jun 21, 2021 at 16:30

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In order to prove this conclusively, you would need to use proof by induction. Enumeration and exhaustion only work when the set of $n$ is finite, but it seems like you want to prove that works for all $n\in \mathbb N$.

That is, letting $S(n)$ be the statement that your equation is true for that value of $n$, you would need to show $S(1)$ is true, and to prove that $S(n)$ implies $S(n+1)$ for all $n\in \mathbb N$. So, you would not need to reduce the integral entirely, but show how the integral for $n+1$ can rearranged so that the integral for $n$ appears inside of it. Without more details, I cannot give any more specific advice for what that would look like.

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  • $\begingroup$ This was actually something I was thinking whether I would need to resort to induction or not. Will do that then! $\endgroup$
    – index
    Jun 21, 2021 at 16:08
  • $\begingroup$ After reading your answer again, I realize that in my case $n$ will actually be finite. Then, would the proof described in my post suffice? Does it fall then under the category proof by enumeration? $\endgroup$
    – index
    Jun 21, 2021 at 18:55
  • $\begingroup$ @index I suppose that if you want to prove that your expression is correct for $n=1,\dots,10$, e.g, then you could call that proof by exhaustion. But if you are still using the basic strategy of an inductive proof (relating case $n$ to case $n+1$), I would still call it an inductive proof. Why, may I ask, do you care about the terminology so much? None of my answer or this comment should give you any help solving your particular problem. $\endgroup$
    – Mike Earnest
    Jun 22, 2021 at 0:57
  • $\begingroup$ In my case $I_1,\dots,I_n$ are expressions but not in closed form. Therefore, I find it quite difficult to make an inductive proof. Hence, I am constrained to find the expression for $I_n, n<\infty$, by identifying the pattern for $I_1,I_2,I_3$. $\endgroup$
    – index
    Jun 22, 2021 at 11:15

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