Integral $\int_0^e \left(\operatorname{W}(x)^{2}x-\frac{6x}{8}-\frac{3\operatorname{W}(x)}{8}+\frac{3}{8}\right)\,dx=0$

Question

Hi I was playing with the Lambert function when I wondering myself about that :

Prove that : $$\int_0^e \left(\operatorname{W}(x)^2 x-\frac{6x}{8}-\frac{3\operatorname{W}(x)}{8}+\frac{3}{8} \right) \, dx=0$$

My try

It's straightforward if we have :

\begin{align} & \int \left(\operatorname{W}(x)^2 x-\frac{6x}{8}-\frac{3\operatorname{W}(x)}{8}+\frac{3}{8} \right) \, dx \\[8pt] = {} & \frac{(x (\operatorname{W}(x) - 1) (4 x \operatorname{W}(x)^3 - 3 \operatorname{W}(x)^2 + 3 (x + 1) \operatorname{W}(x) - 3 x))}{(8 \operatorname{W}(x)^2)} \\ & {} + \text{constant} \end{align}

And after using the fundamental theorem of calculus.

My question

How to prove it using others method?

Thanks in advance for your time.

Answer 1

By the definition of the Lambert $W$ function we have, taking $W(x)=u\Rightarrow x=ue^u$, that $$\int_0^e W(x)^2 x \, dx = \int_0^1 u^3 e^{2u} (u+1) \, du = \frac{3}{8} (e^2-1)$$ and $$-\frac{3}{8} \int_0^e W(x) \, dx = \int_0^1 ue^u (u+1) \, du = -\frac{3}{8} (e-1)$$ and trivially $$\int_0^e \left(\frac{3}{8}-\frac{6}{8}x\right) \, dx = -\frac{3}{8} (e-1)e.$$