This integral is based off of two types of integrals shown off by Maths505, those being of the form$$\int f(x)^{dx}$$and$$\int f(dx)$$Now for those who don't know$$\int_a^bf(x)dx=\lim_{n\to\infty}\sum_{n\ge k\ge1}f(x_k)\Delta x$$where $\Delta x=\frac{b-a}n$. A product integral is defined as$$\int_a^bf(x)^{dx}=\lim_{n\to\infty}\prod_{n\ge k\ge1}f(x_k)\Delta x$$and the other integral type is defined as$$\int f(x)g(dx)=\lim_{n\to\infty}\sum_{n\ge k\ge 1}f(x_k)g(\Delta x)$$Now consider the integral$$I=\int x^{\ln(dx)}$$I know I can rewrite this as$$I=e^{\int\ln(x)\ln(dx)}$$but I'm unsure if this integral is evaluable from here (if it even makes sense), and if it is, how would I go about evaluating it?
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1See this . I have answered this hypothetical – Ninad Munshi Jan 05 '24 at 02:49
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@NinadMunshi How would this help me evaluate $\int\ln(x)\ln(dx)$ though? – CrSb0001 Jan 05 '24 at 02:55
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2It cannot be defined. $\ln(0)$ is not defined. – Ninad Munshi Jan 05 '24 at 02:59
1 Answers
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 https://youtu.be/VVF2nZ0WOY4?si=G_Cx6WSFs138sAR1 . 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 1 for the value of $C=e$ , therefore we will only look for solutions of this integral on the interval $x∈ [e,∞)$ and for the fact that I think this integral has no solution for values less than $e$. Therefore the integral collapses to $\int dx$ for $x=e$ or $\int0 dx $ for $x∈(e,∞)$ . Therefore out solution will be $x$ for $x=e$ and $c$ for $x∈ (e,∞)$ . This solution can defined as $(u(-x+e))x+c$ s.t. $x∈[e,∞)$
(the unit step function or u(x) can be seen from https://www.youtube.com/watch?v=QmptnAjoCi)
Also if you have any clarifying questions please ask or if you see an error in my solution or working please let me know.

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