Integral inequality $\int_{0}^{e}\operatorname{W(x)^{\pi}}>1$

Another approach where we don't have to integrate the Lamberts function, but only one standard integral of the form $\int_0^1 x^n \exp(x) \text{d}x$ for $n \in \mathbb{N}$, which can be solved explicitly using partial integration.

Let $g(x) = W^{-1}(x) = x \exp(x)$, then your integral becomes after a change of coordinates \begin{align*} \int_0^e W(x)^\pi \text{d}x &= \int_0^1 W(g(x))^\pi g'(x) \text{d}x \\ &= \int_0^1 x^\pi\left(\exp(x) + x \exp(x) \right) \text{d}x \\ &= \int_0^1 x^{\pi} \exp(x) \text{d}x + \int_0^1 x^{\pi + 1} \exp(x) \text{d}x. \end{align*} We can use partial integration for the first integral to obtain \begin{align*} \int_0^e W(x)^\pi \text{d}x &= \frac{1}{\pi+1} x^{\pi+1} \exp(x)\bigg|_{x=0}^{x=1} - \frac{1}{\pi+1} \int_0^1 x^{\pi+1} \exp(x) \text{d}x + \int_0^1 x^{\pi + 1} \exp(x) \text{d}x \\ &= \frac{e}{\pi+1} + \frac{\pi}{\pi+1} \int_0^1 x^{\pi+1} \exp(x) \text{d}x. \end{align*} We apply another round of partial integration to find \begin{align*} \int_0^e W(x)^\pi \text{d}x &= \frac{e}{\pi+1} + \frac{e \pi}{(\pi+1)(\pi+2)} - \frac{\pi}{(\pi+1)(pi + 2)} \int_0^1 x^{\pi+2} \exp(x) \text{d}x. \end{align*} We can now bound the last integral from below by $- \int_0^1 x^5 \exp(x)\text{d}x$ as $x^5 > x^{\pi+2}$ on the interval $[0,1]$ to obtain $\int_0^e W(x)^\pi \text{d}x \ge 0.999$, hence we apply two extra rounds of partial integration to obtain \begin{align*} \int_0^e W(x)^\pi \text{d}x &= \frac{e}{\pi+1} + \frac{e \pi}{(\pi+1)(\pi+2)} - \frac{e\pi}{(\pi+1)(pi + 2)(\pi+3)} + \frac{e\pi}{((\pi+1)(pi + 2)(\pi+3)(\pi+4))} - \frac{\pi}{((\pi+1)(pi + 2)(\pi+3)(\pi+4))} \int_0^1 x^{\pi+4} \exp(x) \text{d}x. \end{align*} We now bound the integral from below by $- \int_0^1 x^7 \exp(x)\text{d}x$, to obtain $\int_0^e W(x)^\pi \text{d}x \ge 1.00018$.

Furthermore, we could keep applying integration by parts to find the integral of $\int_0^e W(x)^\pi \text{d}x$ as \begin{align*} \int_0^e W(x)^\pi \text{d}x &= \frac{e}{\pi+1} + e \sum_{n=2}^\infty (-1)^n \frac{\pi}{\prod_{m=1}^n (\pi + m)}. \end{align*}

Not very rigorous.

Use the Taylor series of $W(x)$ around $x=e$ up to order $O\left((x-e)^{n+1}\right)$, raise it to power $\pi$ to get things like $$1+\frac{\pi (x-e)}{2 e}+\frac{\pi (2 \pi -5) (x-e)^2}{16 e^2}+\frac{\pi \left(45-30 \pi +4 \pi ^2\right) (x-e)^3}{192 e^3}+\frac{\pi (-583+510 \pi -120 \pi ^2+8 \pi ^3) (x-e)^4}{3072 e^4}+O\left((x-e)^5\right)$$ Integrate termwise between the bounds to get $\big[e\times P_n(\pi)\big]$

Below are given the decimal representation of the results $$\left( \begin{array}{cc} n & \text{result} \\ 1 & 0.58334827279065 \\ 2 & 0.81164122032718 \\ 3 & 0.92027110969382 \\ 4 & 0.96636673732565 \\ 5 & 0.98665793725515 \\ 6 & 0.99583123255972 \\ 7 & 0.99996493384791 \\ 8 & 1.00173198491504 \\ 9 & 1.00236972369043 \\ 10 & 1.00247298096855 \\ 11 & 1.00233463559724 \\ 12 & 1.00210020862900 \\ 13 & 1.00184132446202 \\ 14 & 1.00159192031259 \\ 15 & 1.00136660705921 \\ 16 & 1.00117017261118 \\ 17 & 1.00100256311893 \\ 18 & 1.00086150355766 \\ 19 & 1.00074386588970 \\ 20 & 1.00064636610536 \end{array} \right)$$