# Proof by enumeration/proof by exhaustion of integral expression

I am looking to prove my conjecture that $$$$\int\cdots\int f(x_1,\dots,x_n) \,dx_1\cdots\,dx_n=I_n.$$$$ over the domain $$\{(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.

• Sounds like this would be a good case for a proof by Mathematical Induction. What is $f(x_1, \ldots, x_n)$? Jun 21, 2021 at 15:54
• 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. Jun 21, 2021 at 16:30

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.
• 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? Jun 21, 2021 at 18:55
• @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. Jun 22, 2021 at 0:57
• 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 $I_1,I_2,I_3$. Jun 22, 2021 at 11:15