I saw a YouTube video that messed up all my intuition. They were calculating the integral of something like -1+x^dx which in the end he solved by taking the limit as dx->0 and he got the solution. But everyone in the comments was saying complex stuff like like “well to be rigorous x^dx can be thought of an element of C*(R,R), the exterior algebra on R”. Huh. When did dx get any different from lim x->0.
In: 2
dx can be interpreted in a couple of different ways. They all produce the same (or nearly the same, up to some technicalities about domain) results, so these are at the end of the day just different ways to formalize the same ideas.
—–
One way to think of dx is, as you roughly say, limit of delta-x as delta-x -> 0. This intuition can get you surprisingly far with two rules:
* lim delta-x -> 0 is dx
* dx^2 is “0” (or more properly, infinitesimally smaller than dx)
So for example, suppose we want to compute the area of a polar figure. To do that, we need to take two close-by points on the polar curve and compute the area of the wedge between them. This ends up being the combination of a circular wedge of area (delta-theta/2pi) * pi * r^(2), plus an approximately triangular shape of area (1/2) (r * delta theta) (delta r).
The first term gets us (1/2) r^2 delta-theta, and the second term gets us 1/2 * r * delta-theta * delta-r. If we then think about this in the limit, the first term is (1/2) r^2 d(theta), and the second term ends up with a d(theta) times a dr -> two d’s multiplied -> effectively zero.
So we get, correctly, the area element (1/2) r^2 d(theta).
In this approach, you’re thinking of your integrals as the Riemann integrals you probably learned first: you have a bunch of bars, you make them infinitely thin, and you add them up.
—–
But there’s another, deeper, way to think about what dx represents, and it turns out to be a way that generalizes better to more complex math.
That way is to think about dx as assigning a “weight” to each segment of the thing you’re integrating over. So for example, dx assigns a weight of (b-a) to every interval [a,b] of the real line. And then your integral over an interval is simply the “height” of the function f(x) times the weight of each tiny interval dx. Similarly, you write dA, and what you’re really doing is assigning a weight (b-a)(d-c) to each rectangle in R^(2).
Up until now, we’ve thought of integrals as having two basic inputs: the set you’re integrating over (typically an interval) and the function you’re integrating. But with this formulation, an integral has *three* basic inputs: the set you’re integrating over, the function you’re integrating, **and the weight function dx you’re using**.
Formally speaking, we think of such integrals as regular integrals of some special product between the function and the weight function, more technically called a *distribution* or a *measure* (depending on exactly what you’re doing; the concepts are equivalent-ish). And because functions form a vector space, where you can add functions or multiply them by constants, and because the product of a function and a distribution or measure produces a real number, the measure is (by defintion) a function from vectors to scalars, which is what we call a *functional*.
The reason that this is useful is that it’s more generalizable. For example, if you’ve had some differential equations, you may have encountered the Dirac delta function, a function with the property that:
* delta(x) = 0 everywhere except x = 0 and
* integral(a,b) delta(x) dx = 1 if the interval contains 0, and 0 otherwise.
No actual function has this property. So the Dirac delta is not, strictly speaking, a function. However, it *is* a *distribution*, and what we more properly mean by this definition is that:
* integral(a,b) f(x) delta(x) = f(x) if the interval contains 0 and integral(a,b) f(x) delta(x) = 0 otherwise
And *this* idea is well-defined.
With me so far?
The most important thing to take away is that the integral of x^d^x is not a defined expression and is fundamentally meaningless.
The dx in the context of classic integrals and derivatives is purely symbolic and does not have a definition but *represents* the width of an infinitely small interval. While it can sometimes be convenient to treat the dx like a factor or divisor, that is technically not rigorous and can lead to errors.
Therefore, the question “what is the integral of x^d^x – 1” does not have a proper answer because in the context of calculus exponentiating to dx is not a thing. However, you can still think about what this expression *could* represent *if* you ignore that it is not rigorous, in the same way that I could ask you what the color purple should taste like, even though there is no rigorous answer to that either. [Here](https://math.stackexchange.com/questions/3201425/integrating-a-function-with-mathrm-dx-as-an-exponent) are some interesting answers that one might get using different approaches.
The specific terminology in the comment you mentioned is referencing a concept called “differential forms”.
The exact theoretical explanation is kind of tricky, but the following should give you some idea what this is about: On the number line, a “differential 1-form” like 3x dx basically gives you a “scale” at every point x, kind of like on a measuring tape. For example at x=2, the form 3x dx gives you the scale 6 dx that’s six times as dense as the standard scale; at x=1 the form gives you the scale 3 dx that is only 3 times as dense as the standard scale; at x=-1 the scale -3 dx you get is thrice the usual density, and also increases in the minus direction instead of the plus direction. From this point of view, integrating 3x dx from 1 to 3 is understood as follows: you break up [1,3] into (infinitesimally) tiny segments, check the length of each segment against your scale at the corresponding point, and then add the results up.
For basic integrals over a 1D interval, this sort of conceptual abstraction is kind of overkill. But as one goes further in math, in particular into multivariable calculus, these things turn out to eventually build a beautiful theory that makes some patterns behind the Calculus results we use way more obvious. This approach to differential calculus is less commonly taught at lower levels, since even though it does reveal some patterns, it does have the downside of having to wrap your head around these differential forms that are somehow less intuitive than functions or vector fields. So since it’s not strictly needed, it is left for later, mostly for people that specialize in something that actually really needs this. The typical point where these would be encountered is if one starts working with curved spaces, where the standard approach to calculus starts becoming too cumbersome not to introduce this point of view.
Latest Answers