Indefinite integral involving the Lambert W or product log function.

Question

How to calculate the integral

$\displaystyle \int \frac{1-x}{x W\left(\frac{1-x}{x}\right)} \, dx$ ?

I tried making a substitution but it doesn't seem to work.

Does this integral have a symbolic solution or a series expansion?

Answer 1

If you change variable $$\frac{1-x}x=t\implies x=\frac{1}{t+1}\implies dx=-\frac{dt}{(t+1)^2}$$ you end with $$I=\int \frac{1-x}{x W\left(\frac{1-x}{x}\right)} \, dx=-\int \frac {t}{(1+t^2) \, W(t)} \,dt$$ You can expand the integrand around $t=0$ using composition of Taylor series. This would give $$ \frac {t}{(1+t^2) \, W(t)}=1-t+\frac{1}{2}t^2+\frac{2 }{3}t^3-\frac{71 }{24}t^4+\frac{443 }{60}t^5-\frac{11627 }{720}t^6+\frac{86111 }{2520}t^7+O\left(t^8\right)$$

Edit

Thinking more about it, using Taylor around $x=1$, you have $$\frac{1-x}{x W\left(\frac{1-x}{x}\right)} =1-(x-1)+\frac{1}{2} (x-1)^2-\frac{2}{3} (x-1)^3+\frac{3}{8} (x-1)^4-\frac{19}{30} (x-1)^5+\frac{35}{144} (x-1)^6-\frac{601}{840} (x-1)^7+O\left((x-1)^8\right)$$

Integrating between $\frac 12$ and $\frac 32$,the above series would lead to $\frac{48241}{46080}\approx 1.0468967$ while the numerical integration would give $1.0467994$.

Answer 2

Hint:

Let $u=\dfrac{1-x}{x}$ ,

Then $x=\dfrac{1}{u+1}$

$dx=-\dfrac{du}{(u+1)^2}$

$\therefore\int\dfrac{1-x}{xW\left(\frac{1-x}{x}\right)}~dx=-\int\dfrac{u}{(u+1)^2W(u)}~du$

Let $v=W(u)$ ,

Then $u=ve^v$

$du=(v+1)e^v~dv$

$\therefore-\int\dfrac{u}{(u+1)^2W(u)}~du=-\int\dfrac{(v+1)e^{2v}}{(ve^v+1)^2}~dv$