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The formal definition of a Riemann Integral is written such that you can have uneven subintervals and it still works. Why do we need to generalize to the case of uneven subintervals? Why not insist our subintervals are always of equal length, and $\Delta x$ is the same for all of them?

Here's the full definition from Wiki:

The Riemann Integral of $f$ equals $s$ if: For all $\epsilon > 0$, there exists a $\delta > 0$ such that for any tagged partition $x_0 \dots x_n$ and $t_0 \dots t_{n-1}$ whose mesh is less than $\delta$ we have $$\left| \left(\sum_{i=0}^{n} f(t_i) (x_{i+1}-x_{i})\right) - s \right| < \epsilon$$

Notice that they have to reference the "mesh" in this definition, i.e. the length of the longest subinterval. Wouldn't we get a simpler definition by just requiring equal subintervals? E.g [my version]

The Riemann Integral of $f$ equals $s$ if: For all $\epsilon > 0$, there exists a $\delta > 0$ such that for any tagged partition $x_0 \dots x_n$ and $t_0 \dots t_{n-1}$ [with equal subintervals] less than $\delta$ we have $$\left| \left(\sum_{i=0}^{n} f(x_i)\Delta x_i \right) - s \right| < \epsilon$$

Or is there a use for having different subinterval lengths?

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    $\begingroup$ It amounts to the same thing. The standard definition is just more flexible, allowing for a wider range of proofs. $\endgroup$
    – TonyK
    Feb 1, 2022 at 21:20
  • $\begingroup$ What about the case where $ f(x)=0 $ if $ x\in\Bbb Q$ and $1$ elsewhere. $\endgroup$ Feb 1, 2022 at 21:21
  • $\begingroup$ when as a student can I first expect to see something where I need variable size subintervals? @hamam_Abdallah could you elaborate? couldn't we take even-sized subintervals where its irrational? There would be gaps between the subintervals but they would be the same size? $\endgroup$
    – Ben G
    Feb 1, 2022 at 21:37
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    $\begingroup$ Fermat used unequal interval lengths to aid in computations well before Cauchy and Riemann and others laid down formal definitions of integration -- definitions designed to capture what people had been doing all along before this. $\endgroup$ Feb 2, 2022 at 8:55
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    $\begingroup$ For discussions and references about variations one obtains when restricting to left (or right) endpoint selections and/or equal-length subintervals, see these sci.math posts: 10 December 2002 and 10 December 2002 and 7 November 2007. $\endgroup$ Feb 2, 2022 at 9:03

1 Answer 1

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First example

Suppose we wish to calculate $\int_0^1 \sqrt{x} dx$. This is trivial if we already have the fundamental theorem of calculus in our disposal, but let's see what happens if we try to calculate this integral directly from the definition.

If we take a partition with equal intervals, that is, for every $n\in\mathbb{N}$ we take $x_k = \frac{k}{n}$ for $k\in \{0,1,\ldots,n\}$, we get the Riemann sum $$\sum_{k=1}^n \Delta x_k f(x_k) = \sum_{k=1}^n \frac{1}{n}\sqrt{\frac{k}{n}}=\frac{1}{n\sqrt{n}}\sum_{k=1}^n \sqrt{k}$$ and since there's no formula for the sum of square roots it appears we are stuck.

Since we're bothered by the the square roots, it would be a nice idea to try a partition with unequal intervals given by $x_k = \frac{k^2}{n^2}$, because now the Riemann sum becomes

$$\sum_{k=1}^n \Delta x_k f(x_k) = \sum_{k=1}^n (x_k - {x_{k-1}})\sqrt{\frac{k^2}{n^2}}=\sum_{k=1}^n \left(\frac{k^2}{n^2}-\frac{(k-1)^2}{n^2}\right)\frac{k}{n}=\frac{1}{n^3}\sum_{k=1}^n 2k^2 - k$$ so we got rid of the square roots. By the well known formulas for $\sum k$ and $\sum k^2$ (see Faulhaber's formula) we get that the Riemann sum equals $$\frac{1}{n^3}\left(2\cdot\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}\right)$$ so that taking the limit as $n\to\infty$ we get $\int_0^1 \sqrt{x} dx=\frac{2}{3}$.


Second example

Suppose we wish to find

$$\lim_{n\to\infty} \frac{1}{n^2}\sum_{k=1}^n (2k-1) \cos\left(\frac{\pi k^2}{2n^2}\right)$$

Noticing that $2k-1 = k^2 - (k-1)^2$, the limit equals

$$\lim_{n\to\infty} \sum_{k=1}^n \frac{k^2 - (k-1)^2}{n^2} \cos\left(\frac{\pi k^2}{2n^2}\right) = \frac{2}{\pi} \lim_{n\to\infty} \sum_{k=1}^n \left(\frac{\pi k^2}{2n^2} - \frac{\pi (k-1)^2}{2n^2}\right)\cos\left(\frac{\pi k^2}{2n^2}\right)$$ Denoting $x_k = \frac{\pi k^2}{2n^2}$ we see that this is $\frac{2}{\pi} \lim_{n\to\infty} \sum_{k=1}^n \left(x_k - x_{k-1}\right)\cos\left(x_k \right)$. Aside for $\frac{2}{\pi}$ this is the limit of a Riemann sum for $f(x)=\cos x$ on the interval $[0,\frac{\pi}{2}]$, so it equals $\int_0^{\frac{\pi}{2}} \cos x dx$ since we know $f(x)=\cos x$ is integrable. By the fundamental theorem of calculus $\int_0^{\frac{\pi}{2}} \cos(x)dx = 1$, so that the limit of the whole expression equals $\frac{2}{\pi}$.

Without knowing that Riemann sums of integrable functions converge to the same value (the integral) regardless of how we partition the domain of integration, this limit would probably be very hard to calculate.

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  • $\begingroup$ Assuming we did already have the fundamental theorem of calculus, would there still be cases where we'd want to utilize uneven intervals? $\endgroup$
    – Ben G
    Feb 2, 2022 at 2:52
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    $\begingroup$ @bgcode One example I can think of is that sometimes we are given some limit and in order to calculate it we interpret it as an appropriate Riemann sum of some integrable function. Are you familiar with this type of exercise? The idea is that after we interpret it as the limit of a Riemann sum then we note that the function is integrable (usually because it is continuous) and so we find the limit using the fundamental theorem of calculus. It is definitely conceivable that such a limit would match a Riemann sum with intervals of unequal lengths. $\endgroup$
    – Snaw
    Feb 2, 2022 at 3:08
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    $\begingroup$ Moreover, it seems that purely from a theoretical perspective it is definitely satisfying to know that if we've shown that a function is integrable (for instance because it is continuous) then no matter how we split its domain into subintervals (whether of equal lengths or not) the Riemann sum is going to converge to the same value. $\endgroup$
    – Snaw
    Feb 2, 2022 at 3:11
  • $\begingroup$ It seems to trivially follow from the equal interval case that you could mix it with bigger intervals and it would still work? Since you could always slice things into a common divisor, and create a new equal interval thing with smaller intervals that filled in both. $\endgroup$
    – Ben G
    Feb 2, 2022 at 3:13
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    $\begingroup$ @bgcode It's not as trivial as it seems, see en.wikipedia.org/wiki/Riemann_integral#Similar_concepts "Another popular restriction is the use of regular subdivisions of an interval .. alone this restriction does not impose a problem, but the reasoning required to see this fact is more difficult than in the case of left-hand and right-hand Riemann sums. However, combining these restrictions, so that one uses only left-hand or right-hand Riemann sums on regularly divided intervals, is dangerous .. under these conditions the indicator function $I_{\mathbb{Q}}$ will appear to be integrable" $\endgroup$
    – Snaw
    Feb 2, 2022 at 3:26

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