Understanding the $\mathrm{d}x$ in integrals

Question

The integral is usually introduced as an extension of the Riemann sum, i.e.

$$ \lim_{n\to\infty}\sum_{i=0}^n f(x_i)\,\Delta x_i = \int_a^b f(x)\,\mathrm{d}x $$

with the usual explanation that $\sum \to \int$ (finite sum to infinite sum) and $\Delta x \to \mathrm{d}x$ (finite element to infitesimal element). This is a nice intuitive explanation, but it doesn't explain what $\mathrm{d}x$ is. Attempting to read the definition of a differential gives a very broad and general definition that does not seem to be immediately applicable to integrals.

However, these $\mathrm{d}$-things appear in different settings as well, such as probability theory $$ \int_{\mathbb{R}} \mathrm{d}F(x) $$ and differential equations (and variable substitution) $$ \frac{\mathrm{d}y}{\mathrm{d}x} = x \iff \mathrm{d}y = x\,\mathrm{d}x \iff \int\mathrm{d}y = \int x\,\mathrm{d}x $$ with no apparent connection to the original $\Delta x$. This may be confusing, and does only appear to complicate things unnecessarily. It would also seem possible to define all these concepts without this $\mathrm{d}$-thing: $$ \lim_{n\to\infty}\sum_{i=0}^n f(x_i)\,\Delta x_i = {\Large\mathcal{I}}_a^b f(x) \\ {\Large\mathcal{I}}_a^b F'(x) \\ \frac{\mathrm{d}y}{\mathrm{d}x} = x \iff {\Large\mathcal{I}}y' = {\Large\mathcal{I}}x $$

So what is the purpose of the $\mathrm{d}$-things? Is it just a historical artefact, or is it possible to assign some intuition and purposeful meaning to $\mathrm{d}x$ without appealing to too abstract concepts in differential geometry? Maybe a simplified version that's easily explained and applicable to integrals.

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It seems clear that in some cases this cannot be remedied by making an alternative integration function ${\Large\mathcal{I}}$, as there are fields where $\mathrm{d}x$ does appear to have some meaning: $$ \int_{\partial\Omega}\omega = \int_\Omega \mathrm{d}\omega $$ (unless this is just deceivingly similar notation to normal integration.)

So how can one motivate the usage of $\mathrm{d}$ that is accessible when one first encounters integrals, but does not cause confusion when encountering it later? I.e. are there any intuitions besides $\Delta x\to \mathrm{d}x$ that gives insight into the interpretation of $\mathrm{d}x$?

Answer 1

In most cases, we may say $\mathrm{d}$ is simply notation used to give context. I argue there is no good "universal definition" of $\mathrm{d}$, at least one that isn't overly complex for those beginning to learn about calculus. However, there are shared themes found among the uses of $\mathrm{d}$.

In each case, one must explain the notation(s) behind a certain idea. In doing this, we somewhat divorce the notation from the rigorous meaning, getting rid of ideas such as "$\mathrm{d}x$ is a number" or "$\mathrm{d}y$ gets divided by $\mathrm{d}x$".

So, I will try to highlight several cases where $\mathrm{d}$ pops up, and attempt to motivate it.

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The definition of the derivative is given as

$$\frac{\mathrm{d}f}{\mathrm{d}x} = \lim_{h \to 0 } \frac{\Delta f}{\Delta x} = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$

Both "$\mathrm{d}$"s informally capture the fact that the derivative is a limiting ratio. Moreover, this notation carries the intuition in our brains about "$\Delta$"s to "$\mathrm{d}$"s. Both are simply notation, yet, they lead to actual mathematical facts as the notation has been connected to intuition. Note:

$$\frac{f(g(x+h_1))-f(g(x))}{g(x+h_1)-g(x)} = \frac{f(g(x)+h_2)-f(g(x))}{h_2} = \frac{\Delta (f \circ g)}{\Delta g}$$

and

$$\frac{\Delta g}{\Delta x} = \frac{g(x+h_1)-g(x)}{h_1}$$

so, (the "$\Delta$"s correspond to different changes, $h_1$ and $h_2$)

$$\frac{\Delta (f \circ g)}{\Delta x} = \frac{\Delta (f \circ g)}{\Delta g} \frac{\Delta g}{\Delta x}$$

And when we make this rigorous($h_1, h_2 \to 0$), it turns out that

$$\frac{\mathrm{d}(f \circ g)}{\mathrm{d}x} =\frac{\mathrm{d}(f \circ g)}{\mathrm{d}g} \frac{\mathrm{d}g}{\mathrm{d}x}$$

Or if $f$ is a function of $g$ and $g$ is a function of $x$, notationally,

$$\frac{\mathrm{d}f}{\mathrm{d}x} =\frac{\mathrm{d}f}{\mathrm{d}g} \frac{\mathrm{d}g}{\mathrm{d}x}$$

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The second derivative is simply the derivative of the derivative. However, we can view it as such with "$\Delta$"s:

$$\frac{\Delta(\frac{\Delta f}{\Delta x})}{\Delta x} = \frac{f(x+2h)-2f(x+h)+f(x)}{h^2} = \frac{\Delta^2 f}{(\Delta x)^2}$$

where $\Delta^2$ was just defined to be this "double difference." One can clearly see how the notation

$$f''(x) \equiv \frac{\mathrm{d}^2 f}{\mathrm{d}x^2}$$

parallels this fact. Moreover, from this intuition about differences, we can begin to investigate and find:

$$\Delta^n f = \sum_{k=0}^{n} (-1)^{n-k}\binom{n}{k}f(x+kh)$$

So the exponent-like notation fits very well, as $\mathrm{d}x^n$ corresponds to $(\Delta x)^n$ notationally, and $\mathrm{d}^n f$ corresponds to $\Delta ^n f$, which almost takes the form of the polynomial

$$(x-1)^n = \sum_{k=0}^{n} (-1)^{n-k}\binom{n}{k}x^k$$

Again, $\mathrm{d}$ serves to connect these seemingly disparate ideas in our brains through notation.

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A separable differential equation can be defined as an equation of the form

$$\frac{\mathrm{d}y}{\mathrm{d}x} = f(x)g(y)$$

Now, consider a point $(x_0, y(x_0))$ for some solution $y$. Using the defintion of the derivative, we see that

$$\lim_{h \to 0} \frac{y(x_0 + h) - y(x_0)}{h} = f(x_0)g(y(x_0))$$

Or in terms of "$\Delta$"s,

$$\frac{\Delta y}{\Delta x} = f(x_0)g(y(x_0)) + \epsilon \iff \frac{1}{g(y(x_0))} \Delta y = f(x_0) \Delta x + \frac{\epsilon \Delta x}{g(y(x_0))}$$

Now, under certain conditions($f$ can be integrated, some condition on the zero sets of $g$, etc.), we can pick an interval $[a,b]$ such that the sums

$$\sum \frac{1}{g(y(x_0))} \Delta y \to \int_{x=a}^{x=b} \frac{1}{g(y)} \mathrm{d}y$$ $$\sum f(x_0) \left(\Delta x + \frac{\epsilon \Delta x}{g(y(x_0))}\right) \to \int_{x=a}^{x=b} f(x) \mathrm{d}x$$

Notationally, we have

$$\frac{1}{g(y)} \mathrm{d}y = f(x) \mathrm{d}x$$

which is just notation that reflects what was shown above.

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In Stokes',

$$\int_{\partial \Omega} \omega = \int_{\Omega} \mathrm{d}\omega$$

the $\Delta x$ intuition simply fails; it is too one-dimensional. We can, however, reconcile the formalization of $\mathrm{d}$ with the intuition that is common to almost all uses of $\mathrm{d}$.

The sum of "infinitesimal" changes in the interior is equal to the total change as viewed on the boundary.

Ultimately, this use of $\mathrm{d}$ captures the notion of "the infinitesimal" quite well, and not only in an analytic sense but in a geometric fashion. I mean, just look at this image!

Answer 2

In physics it makes sense to include the d-thing in the notation for the sake of unit consistency. An example: $$s = \int v\,\mathrm{d}t$$ Here we integrate the velocity (unit m/s) over time (unit s) to get the distance (unit m). Looking at this notation it makes sense that multiplication of the unit m/s with the unit s yields the unit m. This kind of intuition would be lost if the d-thing were hidden within the definition of the integral. This, along with the $\Delta x\to \mathrm{d}x$ intuition, makes the notation very sensible in my opinion.

Answer 3

I should also add (would comment if had the reputation) to the answers above that these "differentials", when used in volume integration for example, are useful as good notation for remembering the correct order in which the variables should be integrated, but it does not make much sense to me saying that their "multiplication" yields a "volume element" (not forgetting the Jacobian!); however, there is an interesting way to connect this notion with the abstract definition of differentials. See, for instance, O'Neill's book, Chap. 4.: https://www.google.com.br/search?q=barrett+o%27neill+elementary+differential+geometry&oq=barrett+o%27neill+elementary+differential+geometry&aqs=chrome..69i57j0l5.9502j0j7&sourceid=chrome&ie=UTF-8

Answer 4

Besides and in support to the explicative "dimensional" justification in Ovaflo's answer, the $dx$ is the differential of $x$ (or $f(x)=x$).
And it is definitely necessary and must be present to tell, e.g., that $$ \int_{}^{} {f(x^{\,2} )dx} \ne \int_{}^{} {f(x^{\,2} )d\left( {x^{\,2} } \right)} $$ and $$ \int_{}^{} {f(x/t)dx} \ne \int_{}^{} {f(x/t)dt} $$ and be able to change the integration variable, integrate by parts, etc. etc.