基于离散求和型积分公式证明主要积分性质的可行性问询

Hey there, great question! First off, let's clarify: the discrete-sum limit definitions you're referencing are just uniform-partition versions of the Riemann integral. Instead of using arbitrary partitions of [a,b] (the standard approach in most analysis textbooks), you're fixing equal-width subintervals of size d (or 1/d, depending on the notation). This is totally a valid way to define the Riemann integral for continuous functions (and even many discontinuous ones), so proving the core integral properties using this form is absolutely possible—it's just not as commonly shown because the general partition approach is more powerful (it works for a broader class of integrable functions).

Let's walk through how you'd prove some of those key properties with your definition:

1. $\int_a^b f(x)dx = -\int_b^a f(x)dx$ #

Take your first definition: $\int_a^b f(x)dx = \lim_{d\to0^+} \sum_{x=a/d}^{b/d} d f(dx)$. If we swap a and b, the sum becomes $\sum_{x=b/d}^{a/d} d f(dx)$—which is just the negative of the original sum, since reversing the order of summation flips the sign. Taking the limit as $d\to0^+$ preserves that sign flip, so the equality holds.

2. $\int_a^a f(x)dx = 0$ #

Again using the first definition: when a = b, the upper and lower bounds of the sum collapse to the same value as $d\to0^+$, meaning the sum either has zero terms or reduces to a single term $d f(a)$ which approaches 0 as d goes to 0. Either way, the limit is 0, matching the property.

3. $\int_a^b f(x)dx + \int_b^c f(x)dx = \int_a^c f(x)dx$ #

Pick a d small enough that all the sums align cleanly. The sum for $\int_a^b$ runs from $x=a/d$ to $x=b/d$, and the sum for $\int_b^c$ runs from $x=b/d$ to $x=c/d$. Adding them together gives a sum from $x=a/d$ to $x=c/d$, which is exactly the sum used in the definition of $\int_a^c f(x)dx$. Taking the limit as $d\to0^+$ keeps this equality intact.

4. Integration by Parts #

This one is a bit trickier, but still doable. Start with the discrete version of the product rule: $d [u(dx) v(dx)] = u(dx) d v(dx) + v(dx) d u(dx)$. Summing both sides from $x=a/d$ to $x=b/d$, you get:
$$\sum_{x=a/d}^{b/d} d [u(dx) v(dx)] = \sum_{x=a/d}^{b/d} u(dx) d v(dx) + \sum_{x=a/d}^{b/d} v(dx) d u(dx)$$
The left-hand side is a telescoping sum! As $d\to0^+$, it approaches $u(b)v(b) - u(a)v(a)$. Rearranging the terms gives the integration by parts formula:
$$\int_a^b u(x) dv(x) = u(b)v(b) - u(a)v(a) - \int_a^b v(x) du(x)$$

So why don't you see these proofs often? Mostly because:

• The uniform partition definition is a special case of the general Riemann integral, so textbooks focus on the more general case to cover all integrable functions (not just those where uniform partitions work, though for continuous functions they're equivalent).
• The general partition proofs are more concise once you've established the basics of Riemann sums, but the uniform version is totally valid and straightforward for most common functions.

It's absolutely not impossible—you just happened to miss these proofs because they're not the standard fare in most intro analysis courses. If you dig into numerical analysis textbooks, you might see more of this kind of reasoning, since numerical integration relies exactly on these discrete sums!

备注：内容来源于stack exchange，提问作者Sig Moid