If $\mathrm{d}x$ is treated as a hyperreal infinitesimal we can easily do derivations.

How do we interpret and perform integrals using infinitesimals?

What is the $\mathrm{d}x$ in $\int x^3\,\mathrm{d}x$ ?

  • $\begingroup$ I think doing derivatives with infinitesimals is no easier (and if you have the experience, no harder) than with the usual methods. $\endgroup$ – GEdgar Jul 9 '16 at 13:32
  • $\begingroup$ You don't. Integration is, courtesy of FTC, purely algorithmic and shouldn't involve infinitesimals. The usual $dx$ before the equals on the LHS is therefore a bit misleading. In this answer here a $dx$ naturally appears but the technique is not as good as the alternative - where is doesn't. NB I'm really talking about nilsquare infinitesimals, not hyperreal ones. $\endgroup$ – user301988 Jul 12 '16 at 1:29
  • $\begingroup$ In what I've seen, $\mathrm{d}x$ holds the same meaning as it does in the standard sense; it's just a formal thing, or a differential form if you know about those. That said, I wouldn't be surprised if there was another formulation of differential geometry where $\mathrm{d}x$ was genuinely an infinitesimal-valued thing. $\endgroup$ – user14972 Jul 16 '16 at 7:55

$\DeclareMathOperator{\st}{st}$I will explain one way to treat your definite integral in the Internal Set Theory (IST) approach to Nonstandard Analysis (NSA). Before beginning, please be aware this NSA approach is entirely analogous to a classical integral with Riemann sums (whereas NSA topics like continuity are a bit more novel). However, it does allow us to interpret the integral as an actual sum using infinitesimals. I will try to give a basic idea how this works in IST.

Very Basic Overview of IST Terminology

The basic idea of the IST approach is you add 3 axioms and the unary predicate standard to ZFC to form a conservative extension of ZFC that lets you use infinitesimals and "infinite" numbers in e.g. $\mathbb{R}$ with its usual definition. Every set is either standard or not standard. If you can uniquely define a set with ZFC alone, you can be assured it is standard. Nonzero infinitesimals are, e.g., not standard.

In this approach, it's a conflation of terminology to refer to the "infinite" elements of $\mathbb{R}$ as infinite since they are in fact finite in the set-size sense of the term. Instead, I will use the terms limited and unlimited.

Finally, you should have a basic idea what the standard part function does. If $x\in\mathbb{R}$ is limited, the standard part function, $\st(x)$, is well defined and returns the unique standard number in $\mathbb{R}$ that differs from $x$ by no more than an infinitesimal.

We will now use our terminology to interpret the following standard integral (in particular, $t^3$ is a standard function and $x$ is a fixed standard number).

$$\int_0^x t^3\,\mathrm{d}t$$

You may have seen a definite integral defined as the limit of a Riemann sum as the number of rectangles goes to infinity where $x/n$ is the width of the rectangle and $n$ is the number of rectangles:

$$\int_0^x t^3\,\mathrm{d}t=\lim_{n\to\infty}\sum_{k=1}^n \left(k\frac{x}{n}\right)^3\frac{x}{n}$$

If we wrote this with the width of the interval, $x/n$, as $h$, we could instead write

$$\int_0^x t^3\,\mathrm{d}t=\lim_{h\to 0^+}\sum_{0\leq kh<x} (kh)^3h$$

However, in NSA, we can just choose a fixed infinitesimal $h>0$ that you might be excused to identify with $\mathrm{d}t$. Provided the integral exists, the result will be no more than an infinitesimal away from the integral, so we can take the standard part to evaluate it.

$$\int_0^x t^3\,\mathrm{d}t=\st\sum_{0\leq kh<x} (kh)^3h$$

Instead of picking an infinitesimal $h$, we could just as easily have picked an unlimited $n$ in the first Riemann sum formula and taken the standard part just the same (this is identifying the arbitrary infinitesimal $h$ you choose with $x/n$ for an unlimited $n$). Let's do this. Fix an unlimited $n$:

\begin{align*} \int_0^x t^3\,\mathrm{d}t&= \st \sum_{k=1}^n \left(k\frac{x}{n}\right)^3\frac{x}{n} \\ &=\st\bigg(\frac{x^4}{4}+\underbrace{\frac{x^4}{2n}+\frac{x^4}{4n^2}}_{\substack{\text{infinitesimal}\\ \text{($x$ is standard} \\ \text{$\implies x$ limited)}}}\bigg) \\ &=\frac{x^4}{4} \end{align*}

More generally, the axioms of IST let us prove that for an infinitesimal $h>0$ and standard integral, the following holds:

$$\int_a^b f(x)\,\mathrm{d}x=\st\sum_{a\leq kh<b} f(kh)\,h$$

If you wish to motivate the usual integral notation (though I prefer the above form for lack of ambiguity), we can define $x_k=kh$ and label our infinitesimal $\mathrm{d}x=h$ to get:

$$\int_a^b f(x)\,\mathrm{d}x=\st\sum_{a\leq x_k<b} f(x_k)\,\mathrm{d}x$$


IST works with real numbers (modulo certain modifications of the background set theory) rather than the hyperreal numbers mentioned in the question. In the hyperreal setting, one chooses an infinitesimal $dx$ for example as $\frac{b-a}{H}$ where $[a,b]$ is the interval of integration and $H$ is a fixed infinite hypernatural number: $H\in {}^\ast\mathbb N \setminus \mathbb N$. Then one partitions the interval $[a,b]$ into $H$ subintervals with partition points $x_i$ as $i$ runs from $0$ to $H$. One forms an infinite Riemann sum by summing the infinitely thin rectangles of width $dx$ and height $f(x_i)$. This is not quite the integral. To obtain the integral one applies the standard part function to the infinite Riemann sum. This is similar to the procedure for differentiation mentioned in the question: the last step is an application of standard part.


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