Nasty summation

$\Large f(x) = x^{x^{x^{.^{.^{.}}}}}$ is the infinite power tower (tetration), which is continuous and differentiable in $[e^{-e},e^{1/e}]$ . The integral

$\displaystyle \int_a^b \cdots \int_a^b \prod_{i=1}^n f(x_i) \ dx_1 \cdots dx_n = \prod_{i=1}^n \int_a^b f(x_i) \ dx_i = \left( \int_a^b f(x) \ dx \right)^n$ for $a$ and $b$ in the domain of $f$ .

Let $\displaystyle I = \int_a^b f(x) \ dx \\ \displaystyle \implies G(n) = I^n \longrightarrow \sum_{k=1}^{n} I^k \ = \ \dfrac{I^{n+1}-I}{I-1}. \quad \left( \text{Recalling from algebra that } 1+z+z^2+\cdots+z^{n-1} = \dfrac{z^n-1}{z-1} \right)$

Now we need only worry about evaluating $I$ . Recall that $\displaystyle \int_a^b f(x) \ dx + \int_{f(a)}^{f(b)} f^{-1}(x) \ dx = bf(b) - af(a)$ . For $a = e^{-e}$ and $b = e^{1/e}$ , $f(a) = \frac{1}{e}$ and $f(b) = e$ .

$\displaystyle \int_{1/e}^e x^{1/x} \ dx \approx 2.658607452339433 ... \ \implies \left( e^{1/e} \right) e - \left( e^{-e} \right) \frac{1}{e} - 2.658607452339433 ... \approx I$

So since $\displaystyle I \approx 1.2441313006514 \cdots \ \implies \ \sum_{n=1}^{50} G(n) \approx 282192.6725612...$ .

Hence, applying the greatest integer function yields $\boxed{282192}$ .