This Integral Breaks Math

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Recommends N Flinn · 2024-01 · Mathematics Stack Exchange · on “Is $\displaystyle\int x^{\ln(dx)}$ possible to integrate?”

This isn't a rigorous approach as it I treat differentials like a fraction but I think this should give you a valid solution. For the expression $f(x) = \int{x^{ln(dx)}}$ we will rewrite it as $f(x) = \int{\frac{x^{ln(dx)} }{dx}dx}$ since the 2 new dx will "cancel out". Now $\frac{x^{ln(dx)} }{dx}$ is approximately the $\lim_{\Delta x \to 0^{+}} \frac{x^{ln(\Delta x)} }{\Delta x}$ (this approach is from Brithemathguy from the video [this video] . Now since for this limit $\Delta x$ and $x$ are independent from each other $x$ will be considered a constant $C$ and now we have $\lim_{\Delta x \to 0^{+}} \frac{C^{ln(\Delta x)} }{\Delta x}$ . This limit only converges to $0$ for values of $C>e$ or to

Cites Mark S. · 2021-05 · Mathematics Stack Exchange · on “Can the integral $\int e^{dx}$ be solved?”

@David To be fair, sense can arguably be made of $\int x^{dx}-1$ (as seen in [this video] ) and a whole lot of mathematical discoveries have been made from unjustified formal manipulations, including the original solution to the Basel problem and solving the cubic, etc.

Cites User123 · 2021-05 · Mathematics Stack Exchange · on “Can the integral $\int e^{dx}$ be solved?”

Nice, the result. Can someone explain how "the video method" ( [this video] ) given by @MarkS in the comments doesn't have a limit (thus no sense), but this is valid (probably)?

Cites Jojo · 2021-12 · Mathematics Stack Exchange · on “Integral with $x^{dx}$”

I was just watching this video [this video] , where he calculates $$I := \int x^{dx} -1.$$ The reasoning is that $$I =\int\frac{x^{dx} -1}{dx}dx,$$ and then he uses $$\lim_{h\rightarrow0} \frac{x^{h} -1}{h} = \ln x$$ to give that $$I = \int\ln x dx.$$ The argument makes sense to me, at least informally. Is it standard to ascribe meaning to integrals of this form as the author has done in the video, or is it just a curio that he's come up with? Is there some way to extend his method to a wider class of integrals of the form $\int f(x, dx)$ ? Clearly the $f$ in his example has been chosen to give the required result. Are there applications where you would want to be able to calculate $\int f(x, dx)

Cites user905694 · 2022-03 · Mathematics Stack Exchange · on “Is $\int x^{dx}-1$ well defined?”

My question is precisely the question here [LINK] . A quick summary of the algebraic process to "solve" the integral follows (taken from miniprime1 [LINK] '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 inte

Critical Turquoise Tilt · 2023-12 · Mathematics Stack Exchange · on “Does $\int (x^{dx}-1)$ make sense?”

I came across the YouTube video "I Computed An Integral That Breaks Math" [this video] 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? Wha