For $y \ge 0$, we have $$f(y) e^{f(y)} = y$$
for some unique $f(y)$. Find $$\int_0^e f(x) \ dx.$$
I tried letting $g(x) = xe^x$, so then, $g(f(y)) = y$ and $f, g$ are inverse functions. But, I'm not sure how to continue... How can I apply this inverse property in my solution?
First, let me apologize for any mistakes in formatting since this is my first time answering a question on this website.
So, the problem is the same as finding the inverse function of $f(x)=xe^x$, which is called "Lambert W. Function" and is denoted by $W(x)$. As any inverse function $g(x)$, it satisfies the basic properties $f(g(x))=x$ and $g(f(x))=x$, therefore $W(y)e^{W(y)}=y$.
Regarding the domain of the function, it's defined for $x\ge-\frac{1}{e}$, so for our integral there will be no issues, as the bounds of integration are within the domain, we also don't have to worry about its second branch as we are working with values $x\ge 0$ (you can read more about this on the Wikipedia article if you are interested, I'm not an expert on this topic so I can't say much more than this).
Now that we found the unknown function, all that remains is evaluating the integral, which is relatively easy but can prove tricky if you aren't too familiar with the Lambert W.Function.
First, the identities $W(x)e^{W(x)}=x$ and $W(xe^x)=x$ will be needed for the evaluation of the integral, as well as integration by parts and u-substitution. We will begin by finding the derivative of $W(x)$ through the formula for the derivative of an inverse function, namely, if $g(x)$ is the inverse function of $f(x)$, then $$g'(x)=\frac{1}{f'(f^{-1}(x))}$$ This, when applied to $W(x)$, translates to: $$W'(x)=\frac{1}{e^{W(x)}(W(x)+1)}=\frac{1}{W(x)e^{W(x)}+e^{W(x)}}=\frac{1}{x+e^{W(x)}}$$ Now, getting back to the original integral, we can first find its antiderivative: apply integration by parts by differentiating $W(x)$ and integrating $1$ to obtain:
$$\int W(x)dx= xW(x)-\int \frac{x}{x+e^{W(x)}}dx$$ then consider the integral $$\int \frac{x}{x+e^{W(x)}}dx$$ and substitute $u=e^{W(x)}$, we can see that: $$du=\frac{e^{W(x)}}{x+e^{W(x)}}dx;\qquad \ln(u)=W(x)$$ therefore the integral equals to: $$\int \frac{x}{e^{W(x)}}du$$ Now to complete the change of variable we can multiply both the numerator and denominator by $W(x)$, yielding: $$\require{cancel} \int \frac{xW(x)}{W(x)e^{W(x)}}du=\int \frac{\cancel{x}W(x)}{\cancel{x}}du=\int W(x)du=\int \ln(u)du $$ This can now be easily solved by integration by parts, so the antiderivative of the initial integral is: $$\int W(x)dx \ = \ xW(x)-e^{W(x)}(W(x)-1) +C \ = \ x(W(x)-1)+e^{W(x)}+C$$ At this point, all that is left is evaluating the integral with bounds of integration $[0,e]$: $$\int_0^e W(x)dx \ = \ \left.[x(W(x)-1)+e^{W(x)}\right\rvert_0^e \ = \ e(W(e)-1)+e^{W(e)}-0(W(0)-1)-e^{W(0)}$$ Note that since $W(xe^x)=x$, it follows that $W(e)=W(1e^1)=1$ and similarly $W(0)=W(0e^0)=0$, then the above expression evaluates to: $$\ e(1-1)+e^{1}-0(0-1)-e^{0}=e-1$$ therefore: $$f(y)=W(y); \qquad \int_0^e f(x) dx=e-1$$ Let me know if there are any mistakes or if you find anything confusing, have a nice day!