# Area under the infinite tetration curve

What is the area under the curve where the infinite power tower converges?

$$\lim_{y \to \infty} = {}^y x.$$

The formula for this curve is given by various sources as:

$$\frac{\mathrm{W}(-\ln x)}{-\ln x}.$$

And the limits are from $\mathrm{e}^{-\mathrm{e}}$ to $\mathrm{e}^{\frac1{\mathrm{e}}}$. So we have:

$$\int_{\mathrm{e}^{-\mathrm{e}}}^{\mathrm{e}^{\frac1{\mathrm{e}}}} \frac{\mathrm{W}(-\ln x)}{-\ln x} \,\mathrm{d}x.$$

Numerically this value is approximately $1.244131300633398$.

Is there an exact value for this integral known?

• My motivation is simply idle curiosity. Mar 10, 2018 at 19:05
• Can you add a few links: one for Lambert W and one for ${}^yx$? Would be nice of you :) Mar 10, 2018 at 19:09
• As requested, I have added a link to the Wikipedia pages for these functions to the bottom of the question. Mar 10, 2018 at 19:12
• By the inverse integral theorem this is equivalent to computing $\int_{1/e}^e t^{1/t} \, dt$. Which smells non elementary to me (e.g., it's reminiscent of the sophomore's dream integral). Mar 10, 2018 at 19:27
• Right, and $2.65+1.24\approx e \cdot e^{1/e} - \frac{1}{e} \cdot e^{-e}$. (If the initial integral computes the area below the curve, its inverse integral computes the area to the left of the curve...) Mar 10, 2018 at 19:34

From Micah:

Since we know that

$$y=x^y\implies x=y^{1/y}$$

the inverse of the function is given by $x^{1/x}$, hence we have

$$\int_{\exp(-e)}^{\exp(1/e)}x^{x^{x^{\dots}}}~dx=\exp\left(1+\frac1e\right)-\exp(-1-e)-\int_{1/e}^ex^{1/x}~dx$$

Note that there is no known closed form for the integral on the right, even though it is in a much more manageable form. Numerically, the integral is given by WolframAlpha as $\approx2.658607452339$, so

$$\int_{\exp(-e)}^{\exp(1/e)}x^{x^{x^{\dots}}}~dx\approx1.2441313006$$