Timeline for Is $\int_0^\infty{dx\over x^{x^{x^x}}}<\int_0^\infty{dx\over x^{x^{x^{x^{x^x}}}}}<\int_0^\infty{dx\over x^{x^{x^{x^{x^{x^{x^x}}}}}}}<\cdots$ true?

Current License: CC BY-SA 4.0

11 events

when toggle format	what		by	license	comment
Jan 16, 2023 at 1:16	history	edited	Eric Naslund	CC BY-SA 4.0	edited body
Jan 14, 2023 at 22:49	history	edited	Eric Naslund	CC BY-SA 4.0	added 1 character in body
Jan 13, 2023 at 21:52	comment	added	Fred Hucht		The $u$ in the third integral after Prop. 2 is still missing, should be $e^{e^{-u}u}$. Furthermore, this integral can be simplified by a partial integration to simply give $$ e^{1/e-1} - e^{1-e} + \int_{1/e}^e\mathrm dx \, x^{-x}. $$
Jan 13, 2023 at 19:20	history	edited	Eric Naslund	CC BY-SA 4.0	Updated the answer based on Fred Hucht's comment
Jan 13, 2023 at 19:08	comment	added	Eric Naslund		@FredHucht: That's great, thank you!
Jan 13, 2023 at 18:47	comment	added	Fred Hucht		First note some typos: You mixed up $L(x)$ and $1/L(x)$ a bit in your answer, and a $u$ is missing in the third integral after Prop. 2. Now my point: By integrating the inverse function of $1/L$ we find \begin{align}\int_{0}^{e^{-e}}\frac 1 {L(x)} \,\mathrm dx = e^{1-e}-\int_{1}^{e} e^{u\, W_{-1}[-\log(u)/u]} \, \mathrm du,\end{align} with $W_{k}(z) = \mathrm{ProductLog}[k,z]$. The limiting value from Prop. 1 now is $$1.91873106231887960413475821627\ldots\,.$$
Jan 13, 2023 at 17:28	history	edited	Eric Naslund	CC BY-SA 4.0	added 123 characters in body
Jan 13, 2023 at 5:50	history	edited	Eric Naslund	CC BY-SA 4.0	added 111 characters in body
Jan 13, 2023 at 4:35	history	edited	Eric Naslund	CC BY-SA 4.0	added 10 characters in body
Jan 13, 2023 at 4:29	history	edited	Eric Naslund	CC BY-SA 4.0	added 67 characters in body
Jan 13, 2023 at 4:23	history	answered	Eric Naslund	CC BY-SA 4.0