Integration of Lambert W function

Question

I am interested in the integration of Lambert W function. Differentiation is ok but I am unable to integrate it. How to perform it?

Answer 1

Although this question is older, the question and result are vague. The following may be of assistance.

For the integral $$\int W(a x) \, dx$$ let $t = W(a x)$, or $x = \frac{1}{a} \, t \, e^{t}$ and $dx = \frac{1}{a} \, (t + 1) \, e^{t} \, dt$ to obtain \begin{align} \int W(a x) \, dx &= \frac{1}{a} \, \int t \, (t + 1) \, e^{t} \, dt \\ &= \frac{1}{a} \, \int e^{t} \, (t^2 + t) \, dt \\ &= \frac{1}{a} \, \left[ (t^2 - 2 t + 2) + (t -1) \right] \, e^{t} + c_{0} \\ &= \frac{1}{a} \, (t^2 - t + 1) \, e^{t} + c_{0} \\ &= t \, \frac{t e^{t}}{a} - \frac{t e^{t}}{a} + \frac{t e^{t}}{a \, t} + c_{0} \\ &= x \, (W(a x) -1) + \frac{x}{W(a x)} + c_{0}. \end{align} Notice that if $x = 0$ then $W(x)|_{x=0} = 0$ and the integral result will be invalid. To remove this difficulty it is required to use $$e^{W(x)} = \frac{x}{W(x)}.$$ The integral then becomes $$\int W(a x) \, dx = \begin{cases} x \, (W(a x) -1) + \frac{x}{W(a x)} + c_{0} & x \neq 0 \\ x \, (W(a x) - 1) + e^{W(a x)} + c_{0} & x \text{ is allowed to be } 0. \end{cases}$$

As an example is it now fairly easy to determine that $$\int_{0}^{1} W(a x) \, dx = \frac{(W(a) - 1)^{2}}{W(a)}.$$

Answer 2

Substitute the whole integrand, i.e. let $u=W(x)$ in $\int W(x)\mathrm dx$.

EDIT: To a bit more explicit: You'll find $\mathrm dx = \mathrm e^u(1+u)\mathrm du$ and ther rest is a simple computation...