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In the thumbnail of this April-fools YouTube video there's the following expression:

$$\int \left(1+dx\right)^{\frac{x}{dx}}dx\tag{1}$$

Is this well-formulated?

Knowing a certain formula for $e$, one might be tempted to treat the two $dx$s appearing inside the integrand as $\left(\lim_{\varepsilon\to 0^{+}}\varepsilon\right)$ like so: $$\lim_{\varepsilon\to 0^{+}}\int \left(1+\varepsilon\right)^{\frac{x}{\varepsilon}}dx=\int \left(\lim_{\varepsilon\to 0^{+}}\left(1+\varepsilon\right)^{\frac{1}{\varepsilon}}\right)^{x}dx=\int e^{x}dx=e^{x}+C$$ but the step from $\int (1+dx)^{\frac{x}{dx}}dx$ to $\lim_{\varepsilon \to 0^{+}}\int (1+\varepsilon)^{\frac{x}{\varepsilon}}dx$ looks very wrong for many reasons. So is $(1)$ just a joke?

Omri Shavit
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  • I gave an answer to these sorts of integrals here. The short answer is that the expression cannot be reasonably saved because $f(x,y) = (x+y)^{\frac{x}{y}}y$ is not continuous in $y$ at $y=0$. There is no situation where you can be allowed treat some of the $dx$'s as different from the other $dx$'s as your work does. – Ninad Munshi Apr 25 '24 at 08:14
  • I see. I was aware of the following formula: $\int_a^{b}f(x)dx=\lim_{n\to\infty}\sum_{k=1}^{\infty}f\left(a+k\frac{b-a}{n}\right)\frac{b-a}{k}$ but not of $\int_a^{b}f(x,dx)=\lim_{n\to\infty}\sum_{k=1}^{\infty}f\left(a+k\frac{b-a}{n},\frac{b-a}{k}\right)$ Thanks! – Omri Shavit Apr 25 '24 at 19:19
  • There was no formula to be aware of, I defined that. – Ninad Munshi Apr 25 '24 at 20:34

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