My question is precisely the question here.
A quick summary of the algebraic process to "solve" the integral follows (taken from miniprime1's answer):
$\int x^{dx}-1 = \int \frac{x^{dx}-1}{{dx}} {dx} = \int (\lim_{h \to 0}\frac{x^h-1}{h}) {dx} = \int \ln x {dx} = x\ln x-x+const.$
The reason I ask this question (again) is that to me, none of the answers to the linked question actually answer the question, particularly what the geometric meaning of these manipulations are. I can't wrap my head around thinking about how this integral relates to Riemann sums or Taylor Series. I have some further problems/questions as well:
- How well-defined is the operation of multiplying the integral with $dx/dx$ (I assume this should be a reasonable step)?
- Does taking the limit break an intrinsic connection between $dx$ and $x$?
- Different sources say different things – the linked question has answers that boldly assert (or seem to assert) that $x \ln(x) – x + C$ is a perfectly valid solution. A Quora answer by Alon Amit on a very similar question suggests that this is an abuse of notation that makes no sense.
- If this operation is well-defined, can it be extended to any $f(x)$ such as follows:
$$\int x \, (dx)^2 = \int \lim_{h \to 0} xh \, dx = \int 0 \, dx = C$$
There are many videos that talk about this problem as well such as Flammable Math's video and BriTheMathGuy's video but none of them seem to give me a good enough intuition to establish whether the integral actually makes sense.
So at the end of the day, is $\int x^{dx}-1$ well defined?
Note: If this post violates any Math StackExchange (duplication) policy, please leave a comment and I will edit or (if I have to) delete the question. A comment is appreciated much more than a downvote 🙂
Best Answer
If you think about what a Riemann integral is, it is essentially the limit of some structure of Riemann sums, where a Riemann sum is defined $$\sum_{m=0}^{n-1}f(t_m)\Delta{x}_m,$$ where $[x_0,x_1,...,x_{n-1},x_n]$ is a partition of the interval $[a,b],$ the domain of $f,$ and $[t_0,t_1,...,t_{n-1}]$ is a set of numbers satisfying $t_m\in[x_{m+1},x_m].$ The limit of the above Riemann sums as $$\mathrm{max}_{0\leq{m}\leq{n}}\Delta{x}_m\to0$$ is equal to the integral $$\int_a^bf(x)\,\mathrm{d}x.$$ This notation is chosen because the symbol $\int$ is roughly analogous to $\sum$ and the symbol $\mathrm{d}x$ is roughly analogous to $\Delta{x}_m.$ The way people often explain it on the Internet, in a very handwavy way, is that $\Delta{x}_m\to\,\mathrm{d}x.$
With this motivation in mind, if we want to give a definition to the symbol $$\int_a^b{x^{\mathrm{d}x}-1},$$ then that definition should be motivated by the notation chosen, such that the it corresponds to the definition of the Riemann integral in terms of Riemann sums. So we can consider the sums $$\sum_{m=0}^{n-1}t_m^{\Delta{x}_m}-1,$$ and if we get convergence of these sums in the same precise sense that the Riemann sums converge to the Riemann integral, then we have sufficient justification to consider the symbol $$\int_a^b{x^{\mathrm{d}x}-1}$$ well-defined.
Thus, the question to answer now is: do we actually get that type of convergence? This is the question that requires an actual investigation via real analysis. Now, usually, in YouTube videos that discuss this, such as in Flammable Maths' videos, the idea that gets used is the idea of multiplying the numerator and denominator by $\mathrm{d}x.$ If we are to do this rigorously, then this step is analogous to saying $$\sum_{m=0}^{n-1}t_m^{\Delta{x}_m}-1=\sum_{m=0}^{n-1}\frac{t_m^{\Delta{x}_m}-1}{\Delta{x}_m}\,\Delta{x}_m.$$ So far, there are no issues, but the next step that these mathtubers employ is the one that seriously requires justification, and the justification is by no means trivial. The argument that is always made is that $$\lim_{\epsilon\to0}\frac{x^{\epsilon}-1}{\epsilon}=\ln(x),$$ so we should be able to say $$\int_a^b\frac{x^{\mathrm{d}x}-1}{\mathrm{d}x}\,\mathrm{d}x=\int_a^b\ln(x)\,\mathrm{d}x.$$ This is the step that makes the demonstration problematic for all of those mathematics videos. The argument presented is not a valid argument, because it assumes $$\lim_{\Delta{x}_m\to0}\sum_{m=0}^{n-1}\frac{t_m^{\Delta{x}_m}-1}{\Delta{x}_m}\,\Delta{x}_m=\lim_{\Delta{x}_m\to0}\sum_{m=0}^{n-1}\left[\lim_{\delta\to0}\frac{t_m^{\delta}-1}{\delta}\right]\Delta{x}_m,$$ and it is not at all clear that this statement should be true. Really, this statement is the entire bulk of their demonstration, and yet it is the statement that gets completely skipped by the videos.
But, this statement does turn out to be true. Proving it is very tedious, and it requires using the Moore-Osgood theorem at some point, which in turn, requires proving uniform convergence. You would need several pages of a written paper dedicated solely to proving this statement, but it is doable. So, yes, the statement does hold, and so you can indeed say $$\int_a^b\frac{x^{\mathrm{d}x}-1}{\mathrm{d}x}\,\mathrm{d}x=\int_a^b\ln(x)\,\mathrm{d}x.$$
Unfortunately, this is not rigorous enough. Everything I said above is a heuristic that I used based on drawing analogies with the notation. But I have not provided a definition, nor have I specified over what class of functions is this definition supposed to be applicable to. But the heuristic is useful in that it tell us what we should expect from a rigorous definition, and it gives us a hint as to what class of functions we should be focusing on.
So here is my proposal. Consider a function $f:[a,b]\times(-r,r)\to\mathbb{R}$ with $r\gt0,$ such that $$\frac{f(x,\delta)}{\delta}\to{h(x)}$$ as $\delta\to0$ uniformly, with $h:[a,b]\to\mathbb{R}$ being Riemann integrable, and such that every $g_{\delta}:[a,b]\to\mathbb{R}$ is Riemann integrable. Thus, I define $$\int_a^bf(x,\mathrm{d}x):=\int_a^bh(x)\,\mathrm{d}x.$$ With a proof analogous to the one briefly sketched a few paragraphs above, one can show that this is indeed well-defined.
With this definition, though, something like $$\int_a^bx\,\mathrm{d}x^2$$ is not well-defined. My definition does not accommodate for such things, so some extension of my definition would be required. An appropriate extension would be to consider defining $$\int_a^bf(x,\mathrm{d}x)\,\mathrm{d}x$$ instead of $$\int_a^bf(x,\mathrm{d}x)$$ as I did. This can be accomplished by keeping the original definition I proposed, but replacing the condition $$\frac{f(x,\delta)}{\delta}\to{h(x)}$$ with $$f(x,\delta)\to{h(x)}.$$ Then simply $$\int_a^bf(x,\mathrm{d}x)\,\mathrm{d}x=\int_a^bh(x)\,\mathrm{d}x,$$ and $$\int_a^bx\,\mathrm{d}x^2=0$$ by this definition. You can take these generalizations even further. If you want to define $$\int_a^bf(x,\mathrm{d}x)\,\mathrm{d}x^k,$$ then consider $$\int_a^bf(x,\mathrm{d}x)\mathrm{d}x^{k-1}\,\mathrm{d}x=\int_a^bh(x)\,\mathrm{d}x,$$ and to make a rigorous definition for this, simply replace $$\frac{f(x,\delta)}{\delta}\to{h(x)}$$ with $$\frac{f(x,\delta)}{\delta}\delta^k\to{h(x)},$$ and this can be made to work for arbitrary integers $k.$ The nice thing about this generalization is that if $f$ is defined by $f(x,\delta)=h(x)\delta,$ then the familiar, basic Riemann integral results naturally. So the Riemann integral is just a special case of this more general integral I have defined here.