Integrate $W(e^u)$

Question

Please help to integrate

$$\int_{e+1}^{2+e^2} W(e^u) \, du,$$

where $W$ is the Lambert W function? Shall feyman technique of integration should be used here? I am not able to do it by integration by parts.

Answer 1

You can start by making the sub $W\left(e^u\right)=t$, $u=t+\ln \left(t\right)$, $du=1+\frac{1}{t}\:dt$

Then having $$\int _{e+1}^{e^2+2}W\left(e^u\right)\:du=\int _{W\left(e^{e+1}\right)}^{W\left(e^{e^2+2}\right)}t\:\left(1+\frac{1}{t}\right)dt=\int _{W\left(e^{e+1}\right)}^{W\left(e^{e^2+2}\right)}t\:dt+\int _{W\left(e^{e+1}\right)}^{W\left(e^{e^2+2}\right)}dt$$ By simply integrating we get, $$\boxed{\frac{W\left(e^{e^2+2}\right)^2}{2}+W\left(e^{e^2+2}\right)-\frac{W\left(e^{e+1}\right)^2}{2}-W\left(e^{e+1}\right)\approx28.2753212}$$ Which seems to agree with WolframAlpha.