Deriving a differential equation for the Lambert W function.

Question

I would like to know if I am correct in the following:

Let $f(x) = e^{x}x$, and $g(x) = w(x)$, where $w(x)$ is the Lambert W function.

By the rule that inverse function integral relation, which states that:

$$\int_{a}^{b} f(x) \,dx = bf(b) - \int_{f(a)}^{f(b)} g(x) \,dx$$

$$\int_{a}^{b} e^{x}x \,dx = e^{b}b - \int_{f(a)}^{f(b)} W(x) \,dx$$

One can show that:

$$\int e^{x}x \,dx = e^{x}(x - 1) + C$$

Let $W(x)$ be the antiderivative of the Lambert W function, $w(x)$, then:

$$W(be^{b}) - W(ae^{a}) = e^{a}(a - 1) + e^{b}$$

$$\int_{ae^{a}}^{be^{b}} w(x) \,dx = e^{a}(a - 1) + e^{b}$$

Let $a = 0$, then:

$$\int_{0}^{be^{b}} w(x) \,dx = e^{b} - 1$$

Therefore:

$$\int_{0}^{x} w(t) \,dt = \frac{x}{W(x)} - 1$$

$$w(x) = \frac{w(x) - xw'(x)}{w^{2}(x)}$$

$$w^{3}(x) = w(x) - xw'(x)$$

$$w^{3}(x) - w(x) + xw'(x) = 0$$

I am unsure as to how to progress, or if these steps are even correct.

I would appreciate any help.

Thank you.

Answer 1

Here is what I tried: The function $x\mapsto w(x)$ in question satisfies $$x=w(x)e^{w(x)},\qquad w(0)=0\ .\tag{1}$$ Differentiating with respect to $x$ we obtain $$1=w'(x)e^{w(x)}\bigl(1+w(x)\bigr)\ .$$ We now substitute $e^{w(x)}={x\over w(x)}$ from $(1)$ and get the followsing differential equation: $$w'(x)={w(x)\over x\bigl(1+w(x)\bigr)}\ .$$

Answer 2

I think there are some errors. The theorem you cite says that holds $$ \int_{a}^{b}f\left(x\right)dx=bf\left(b\right)-af\left(a\right)-\int_{f\left(a\right)}^{f\left(b\right)}f^{-1}\left(x\right)dx $$ then if we put $f\left(x\right)=xe^{x} $ and $f^{-1}\left(x\right)=w\left(x\right) $ we have $$\int_{a}^{b}xe^{x}dx=b^{2}e^{b}-a^{2}e^{a}-\int_{ae^{a}}^{be^{b}}w\left(x\right)dx $$ then $$\int_{ae^{a}}^{be^{b}}w\left(x\right)dx=b^{2}e^{b}-a^{2}e^{a}-e^{b}\left(b-1\right)+e^{a}\left(a-1\right) $$ now if we take $a=0 $ we get $$\int_{0}^{be^{b}}w\left(x\right)dx=b^{2}e^{b}-e^{b}\left(b-1\right)-1 $$ so if we put $z=be^{b}=w\left(z\right)e^{w\left(z\right)} $ we have $$W\left(z\right)=\int_{0}^{z}w\left(x\right)dx=w^{2}\left(z\right)e^{w\left(z\right)}-e^{w\left(z\right)}\left(w\left(z\right)-1\right)-1 $$ now if you take the derivative, you get a differential equation.