Hi it's a problem of mine :
Prove that (without calculating the integral) :
$$\int_{0}^{e}\operatorname{W(x)^{\pi}}\mathrm{d}x>1$$
Where $\operatorname{W(x)}$ denotes the Lambert's function .
I have tested during 2 hours some methods but it fails always because the inequality is pretty sharp .Particulary I have used the power series of the Lambert's function but it's not convincing . So I really need help on this question .
If you have nice ideas…
…Thanks a lot.
Best Answer
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*}