Show that $\int\limits_0^1 \left(x^{x}\right)^{\left(x^{x}\right)^{\left(x^{x}\right)^{\left(x^{x}\right)^{⋰}}}}\ \mathrm{d}x=\frac{\pi^2}{12}$.

Question

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.

Answer 1

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

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

Answer 2

We generalize the integral as follows:

$$ \int_0^1 \left(x^{-\alpha x}\right)^{\left(x^{-\alpha x}\right)^{\left(x^{-\alpha x}\right)^{\cdots}}} \, dx = \frac{\mathrm{Li}_2(\alpha)}{\alpha} $$

This yields the following elegant evaluations:

$$ \int_0^1 \left(x^{x}\right)^{\left(x^{x}\right)^{\left(x^{x}\right)^{\cdots}}} \, dx = \frac{\pi^2}{12} $$

$$ \int_0^1 \left(x^{-x}\right)^{\left(x^{-x}\right)^{\left(x^{-x}\right)^{\cdots}}} \, dx = \frac{\pi^2}{6} $$

$$ \int_0^1 \left(x^{-x/2}\right)^{\left(x^{-x/2}\right)^{\left(x^{-x/2}\right)^{\cdots}}} \, dx = \frac{\pi^2}{6} - \ln^2(2) $$

Proof of the Generalization It's similar to the answer provided above by Mariusz Iwaniuk

We evaluate the general integral:

$$ \int_0^1 \left(x^{-\alpha x}\right)^{\left(x^{-\alpha x}\right)^{\left(x^{-\alpha x}\right)^{\cdots}}} \, dx $$

Let

$$ y = \left(x^{-\alpha x}\right)^{\left(x^{-\alpha x}\right)^{\left(x^{-\alpha x}\right)^{\cdots}}} $$

Then it can be deduced that:

$$ y = \frac{W(\alpha x \ln x)}{\alpha x \ln x} $$

where ( W(x) ) is the Lambert W function (see http://mathworld.wolfram.com/LambertW-Function.html ):

$$ W(x) = \sum_{n=1}^{\infty} \frac{(-n)^{n-1} x^n}{n!} $$

Using this, the integral becomes:

$$ \int_0^1 \frac{W(\alpha x \ln x)}{\alpha x \ln x} \, dx $$

Expanding $ W(\alpha x \ln x) $ into a power series:

\begin{align*} \int_0^1 \frac{W(\alpha x \ln x)}{\alpha x \ln x} \, dx &= \int_0^1 \left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \alpha^{n-1} n^{n-1} x^{n-1} \ln^{n-1}(x)}{n!} \right) dx \\ &= \sum_{n=1}^{\infty} \int_0^1 \frac{(-1)^{n-1} \alpha^{n-1} n^{n-1} x^{n-1} \ln^{n-1}(x)}{n!} \, dx \end{align*}

Now we use the integral identity:

$$ \int_0^1 x^{n-1} \ln^{n-1}(x) \, dx = (-1)^{n-1} \left( \frac{1}{n} \right)^n \Gamma(n) $$ Proof- $$ \int_0^1 x^{n-1} \ln^{m-1}(x) \, dx = \frac{d^{m-1}}{dn^{m-1}}\int_0^1 x^{n-1} \, dx=\frac{d^{m-1}}{dn^{m-1}}\frac{1}{n}=\frac{(-1)^{m-1}\Gamma(m)}{n^m} $$ All that remains is taking limit m tending to n.

And thus, $$ \int_0^1 x^{n-1} \ln^{n-1}(x) \, dx = (-1)^{n-1} \left( \frac{1}{n} \right)^n \Gamma(n) $$ Therefore:

\begin{align*} \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \alpha^{n-1} n^{n-1}}{n!} \cdot (-1)^{n-1} \left( \frac{1}{n} \right)^n \Gamma(n) &= \sum_{n=1}^{\infty} \frac{\alpha^{n-1}}{n^2} \\ &= \frac{\mathrm{Li}_2(\alpha)}{\alpha} \end{align*}

Hence proved. $\blacksquare$