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How can it be shown that $$\lim_{p\to\infty}I(p)= \lim_{p \to \infty}\int^{1}_0 (x^x)^{\scriptscriptstyle {(x^x)^{(x^x)^{(x^x)^{(x^x)^{(x^x)...(p \; times)}}}}}} dx= \frac{\pi^2}{12}$$

$I(1)=\int^{1}_0 x^x dx=\sum_{n=0} ^{\infty} \frac{1}{n!}\int^{1}_0 \ln(x)^nx^n dx$

also $\int^{1}_0 \ln(x)^nx^n dx=(-1)^n (1+n)^{-1-n}n!$ .

I don't know how to calculate $I(p)$ beyond $p=1.$

Note: This integral was proposed on Romanian Mathematical Magazine and two solutions can be found here.

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    $\begingroup$ I guess you should use Lambert W function $\endgroup$
    – user715522
    Feb 8, 2020 at 12:43
  • $\begingroup$ I've numerically verified it and found it's equal to $-\int_{0}^{1}\frac{W_{0}\left(-x\ln x\right)}{x\ln x}\,\mathrm{d}x$ with the principal branch of the Lambert W but can't figure out how best to proceed. $\endgroup$
    – Jam
    Feb 8, 2020 at 14:09
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    $\begingroup$ Part of the solution may be showing that this infinite exponent converges for $0 < x < 1$. It is known that the infinite exponential $x^{x^{x^\cdots}}$ diverges by oscillation for $x$ near $0$. $\endgroup$
    – GEdgar
    Feb 8, 2020 at 14:45
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    $\begingroup$ @GEdgar true. But this is straightforward as $x^x$ has a minimum of $e^{-1/e}$ at $x=1/e$ which is well above the point where $x^{x^{...}}$ starts oscillating (i.e. $x < e^{-e}$) $\endgroup$ Feb 8, 2020 at 16:03
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    $\begingroup$ It is interesting to note that the related integral below has much slower convergence. $$\lim_{p\to\infty}I(p)= \lim_{p \to \infty}\int^{1}_0 (x^{-x})^{\scriptscriptstyle {(x^{-x})^{(x^{-x})^{(x^{-x})^{(x^{-x})^{(x^{-x})...(p \; times)}}}}}} dx=\frac{\pi^2}{6}$$ $\endgroup$ Feb 8, 2020 at 18:30

1 Answer 1

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From here: $$W(x)=\sum _{n=1}^{\infty } \frac{(-n)^{n-1} x^n}{n!}$$

and with CAS help: $$\begin{align} \int_0^1 -\frac{W(-x \ln (x))}{x \ln (x)} \, dx &=\int_0^1 \left(\sum _{n=1}^{\infty } \frac{(-1)^{2 n} n^{-1+n} x^{-1+n} \ln ^{-1+n}(x)}{n!}\right) \, dx \\ &=\sum _{n=1}^{\infty } \int_0^1 \frac{(-1)^{2 n} n^{-1+n} x^{-1+n} \ln ^{-1+n}(x)}{n!} \, dx\\ &=\sum _{n=1}^{\infty } \frac{\left((-1)^{2 n} n^{-1+n}\right) \int_0^1 x^{-1+n} \ln ^{-1+n}(x) \, dx}{n!}\\ &=\sum _{n=1}^{\infty } \frac{\left((-1)^{2 n} n^{-1+n}\right) \left(-\left(-\frac{1}{n}\right)^n \Gamma (n)\right)}{n!}\\ &=\sum _{n=1}^{\infty } \frac{(-1)^{1+3 n}}{n^2}\\ &=\frac{\pi ^2}{12}. \end{align}$$

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    $\begingroup$ Where does the closed form for $\displaystyle\int_0^1 (x\ln x)^m\,\mathrm{d}x$ in the last step come from? I could only solve it in terms of the incomplete gamma function. $\endgroup$
    – Jam
    Feb 8, 2020 at 14:36
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    $\begingroup$ @Jam You can do repeated integration by parts and the boundary terms will vanish $\endgroup$ Feb 8, 2020 at 14:46
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    $\begingroup$ @Jam. $\int_0^1 (x \ln (x))^m \, dx=-\left(-\frac{1}{1+m}\right)^{1+m} \Gamma (1+m)$ $\endgroup$ Feb 8, 2020 at 14:47
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    $\begingroup$ @Jam Consider $\displaystyle f(m) = \int_0^1 x^m \,{dx} = \frac{1}{1+m}$. Then $\displaystyle f^{(m)}(m) = \int_0^1 (x \ln{x})^m \,{dx} = \frac{(-1)^m m! }{(1+m)^{n+1}}$ $\endgroup$
    – NoName
    Feb 8, 2020 at 14:53

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