My brother gave me the following problem:
Let $f:[1;\infty)\to[1;\infty)$ be such that for $x≥1$ we have $f(x)=y$ where $y$ is the unique solution of $y^y=x$. Then calculate: $$ \int_0^e f(e^x)dx $$ I couldn't figure out something helpful, so any help is highly appreciated. I think he has the problem from some contest, but I don't know, from which.
On
First of all we have to solve $y^y=x$ for $y$. Using techniques from here, we get
$$f(x)=y=\frac{\ln x}{W(\ln x)},$$
where $W$ is the Lambert W function.
Now if $x>0$, then
$$f(e^x)=\frac{x}{W(x)}.$$
Form here we have to know, that
$$\int \frac{x}{W(x)} \, dx = \frac{x^2(2W(x)+1)}{4W^2(x)} + C.$$
This integral comes from a tricky substitution.
Between $0$ and $e$ integral limits it has the value
$$\int_0^e f(e^x) \, dx = \int_0^e \frac{x}{W(x)} \, dx = -\frac{1}{4}+\frac{3}{4}e^2.$$
Substitute $x=u\ln u$. Note that $1\ln 1 = 0$ and $e\ln e = e$. Thus $$ \int_0^e f(e^x) \,dx = \int_1^e f(e^{u\ln u}) (1+\ln u)\,du = \int_1^e u(1+\ln u)\,du \\ = \int_1^e (u + u\ln u)\,du = \left[\tfrac14 u^2 + \tfrac12 u^2\ln u\right]_1^e = \tfrac34 e^2 - \tfrac14 $$