6

I came across the YouTube video "I Computed An Integral That Breaks Math" by BriTheMathGuy, where the problem is computing

$$\label{eq_1}\tag{1}\int (x^{dx}-1)$$

And basically to solve this integral we use a little trick writing the argument of $\eqref{eq_1}$ like $$\label{eq_2}\tag{2}\color{red}{\frac{x^{dx}-1}{dx}}\cdot dx$$ The red part of $\eqref{eq_2}$ could be interpreted as $$\label{eq_3}\tag{3}\lim_{h\to 0} \frac{x^h-1}{h} = \log (x)$$ And then we can compute the integral of $$\eqref{eq_1} = \int \log(x)dx = x\log(x) - x + c$$

Is this possible, does $\eqref{eq_1}$ even makes sense to write an integral with $dx$ in that unusual position, or is a glamour social media trick? What disturbs me about $\eqref{eq_2}$ is that even tough we can think of $dx$ as an infinitesimal quantity, deeply it isn't, it is a differential. Has someone some thoughts on this or is just a waste of time? Thank you so much.

Blue
  • 75,673
Turquoise Tilt
  • 1,192
  • 8
    I would have demanded to have a definition of $x^{dx}$ given before I would continue any further. But that’s probably why people say I’m grumpy. – Arkady Dec 14 '23 at 23:17
  • @Arkady you're right... I do not think about that. It does exist? – Turquoise Tilt Dec 14 '23 at 23:21
  • Yes, as someone comments on the site, there are formal manipulations in advanced mathematics that seem like this (Chern character, etc.), but they are written in a way that it's clear that one integrates the $n$-form part of a Taylor expansion over an $n$-manifold. And they are written with even-degree differential forms (like curvature) so that wedge product is commutative. I realize this is all too fancy for this post ... – Ted Shifrin Dec 14 '23 at 23:27
  • @TedShifrin would you like to share some links to the topic you are suggesting? Or is hardcore math that is best left to sleep... – Turquoise Tilt Dec 14 '23 at 23:33
  • 2
    I don't know any instances of this notation/concept that are not pretty advanced. For starters, one has to know about integration of differential forms. However, you might get away with interpreting this using "infinitesimals" in non-standard analysis (although I don't remember seeing such expression there). The standard text on that is by Robinson; there's also a book by Henle and Kleinberg. You might also look at this survey article by Henle. – Ted Shifrin Dec 14 '23 at 23:41
  • 2
    Although I wouldn't say it is impossible to make the expression $\int(x^{\mathrm{d}x}-1)$ sensible, I don't see a good reason to do this. The beauty of differentials lies in their power of linearizing complicated objects. So, when a mathematician want to think of a quantity like $x^{\mathrm{d}x}-1$, he/she knows that it is equivalent to $\log x,\mathrm{d}x$ and will use it in the first place. Another reason is that, when we begin to regard $\mathrm{d}x$ as a standalone object, like a differential form or measure, it becomes even harder to justify operations like this. – Sangchul Lee Dec 15 '23 at 00:22
  • 1
    If you really want to give a sense to an expression like $\int x^{\mathrm{d}x}$, it can be interpreted as a product integral (see https://math.stackexchange.com/questions/4605603/does-the-formula-lim-delta-x-rightarrow-0-sqrt-delta-x-fracfx-delta/4605755#4605755 or https://en.wikipedia.org/wiki/Product_integral), which can be seen as the inverse operation of the geometric derivative, which itself measures how a function varies "multiplicatively". But the manipulations in the video you linked are nonsensical abuses of notation. – Abezhiko Dec 15 '23 at 16:18

1 Answers1

2

You could interpret $\int_{a}^{b}(x^{dx} - 1)$ naturally as a limit of sums $\sum_{i = 1}^{N - 1}(\xi_i^{x_{i + 1} - x_i} - 1)$ where $a = x_1 < \dots < x_N = b$ is a partition of $[a, b]$, $\xi_i \in [x_i, x_{i + 1}]$, and the limit is taken as $\max_{1 \leq i \leq N - 1}(x_{i + 1} - x_i) \to 0$. You could probably recover the result you found rigorously. A similar notation arises in survival analysis for the relation of the survival function to the hazard function.

Kakashi
  • 1,381
  • @TedShifrin Yeah I should just call them sums; they definitely aren't Riemann sums. – Kakashi Dec 14 '23 at 23:46
  • OK, this seems like a good start. Is there a reasonable way to proceed to evaluate that limit? – Ted Shifrin Dec 14 '23 at 23:48
  • 1
    @TedShifrin Probably write it as $\sum_{i = 1}^{N - 1}\frac{\xi_i^{\Delta x_i} - 1}{\Delta x_i}\Delta x_i$ and estimate $$|\int_{a}^{b}\log(x),dx - \sum_{i = 1}^{N - 1}\frac{\xi_i^{\Delta x_i} - 1}{\Delta x_i}\Delta x_i|.$$ – Kakashi Dec 15 '23 at 01:55
  • OK, you obviously need to invoke the MVT. (Oh, your sums need to start at $i=0$.) – Ted Shifrin Dec 15 '23 at 02:08
  • @Kakashi sensei, The absolute value you've written above should be numerically approximated or we can use some convergence results in your opinion? – Turquoise Tilt Dec 15 '23 at 11:14
  • 1
    @TurquoiseTilt My opinion is that this definition combined with your proof can easily be turned into a rigorous proof that $\int_{a}^{b} (x^{dx} - 1) = \int_{a}^{b}\log(x),dx$ for $b > a > 0$: Using the MVT as Ted Shifrin suggests turns the sum into a Riemann sum for $\log(x)$, which we already know converges to the integral of $\log(x)$. – Kakashi Dec 15 '23 at 18:08