Functional equation

I don't care if it is the Lambert W function .

$x=f(x)e^{f(x)} \implies f^{-1}x=xe^x$ .

$\displaystyle I=\int_0^e f(x)\mathrm dx$ .

Let $u=f(x)$ , $x=ue^u$ .

When $x=0$ , $u=0$ . When $x=e$ , $u=1$ .

$\displaystyle I=\int_0^1 u\mathrm dx=\int_0^1 u\mathrm d(ue^u)$ .

Integration by parts:

$I=[u^2e^u]_0^1-\int_0^1ue^u\mathrm du$

$I=[u^2e^u]_0^1-\int_0^1u\mathrm d(e^u)$

$I=[u^2e^u]_0^1-[ue^u]_0^1+\int_0^1e^u\mathrm du$

$I=[u^2e^u]_0^1-[ue^u]_0^1+[e^u]_0^1$

$I=(e-0)-(e-0)+(e-1)$

$I=e-1$ .

Let $I = \int^{e}_{0} f(x) dx$ See that $\frac{d}{dx} [f(x)e^{f(x)}]=1 \implies I=\int^{e}_{0} \frac{d}{dx}[f(x)e^{f(x)}]f(x)$
Then: $I = \bigg|^{e}_{0} f(x)^{2}e^{f(x)} - \int^{e}_{0} f(x) \frac{d}{dx} [e^{f(x)}] dx$ $\implies I = \bigg|^{e}_{0} e^{f(x)} (f(x)^{2} - f(x) +1 )$ Now we know that when $x=0$ , $f(x)e^{f(x)} =0$ Assuming that: $\lim_{x \rightarrow 0} f(x) \neq - \infty$ (or more broadly, that the function is well defined at $x= 0$ ) We also know that this implies $f(0) = 0$ . Similarly, when $x=e$ , we know: $f(e)e^{f(e)} = e \implies f(e) = e^{1-f(e)}$ This is easily solved to yield $f(e) = 1$ Plugging the numbers back into the formula for $I$ , we get : $I = e(1-1+1)-1(0+0+1) = e-1$

$f(x)$ , the inverse of $xe^{x}$ , is known as the the Lambert W function .

It can be seen by integration by parts that

$\int_{0}^{e} f(x) dx = e \times 1 - \int_{0}^{1} xe^{x} dx$

$= e-|xe^x - \int e^{x} dx|_{0}^{1} = e-|e^{x}(x-1)|_{0}^{1}$

$= \boxed{e-1}$

I don't know the Lambert function, but here is my solution:

Take $f(x) =y$ :

We have the equation: $x=y\cdot e^y$

$\Rightarrow \frac{x}{y} = e^y \Rightarrow ln(x) - ln(y) = y$ [Taking log (base e) on both sides]

Now differentiating both sides w.r.t $x$ , we get:

$\frac{1}{x} - \frac{1}{y} \cdot \frac{dy}{dx} = \frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx} = \frac{y}{x(y+1)} \Rightarrow y\cdot dx = x(y+1) dy = ye^y(y+1) dy = e^y(y^2+y)dy$

Now, we need to find $\int_0^e y \, dx = \int_0^1 e^y(y^2+y) \, dy$

Applying Integration by Parts, we have:

$\int_0^1 e^y(y^2+y) \, dy = [e^y(y^2 -y+1)]_0^1 = e-1 = 1.7182$

\int { f(x) } =xf(x)+g(x)\\ where\quad g(x)\quad was\quad a\quad function\\ f(x)=f(x)+x\frac { d }{ dx } f(x)+\frac { dg(x) }{ dx } \\ g(x)=-\int { xdf(x) } \\ g(x)=-\int { f(x)e^{ f(x) }df(x) } \\ g(x)=-[f(x)e^{ f(x) }-e^{ f(x) }]\\ then\\ \int { f(x)dx } =xf(x)+e^{ f(x) }(1-f(x))\\ and\\ \int _{ 0 }^{ e }{ f(x)dx } =[ef(e)+e^{ f(e) }(1-f(e))]-[0f(0)+e^{ f(0) }(1-f(0))]\\ note\quad that\\ f(0)=0\quad and\quad f(e)=1\\ then\\ \int _{ 0 }^{ e }{ f(x)dx } =e-1\approx 1.71828183

$\frac { dx }{ dx } =\frac { d(f\left( x \right) { e }^{ f\left( x \right) }) }{ dx } ={ e }^{ f\left( x \right) }f^{ ' }\left( x \right) +{ e }^{ f\left( x \right) }f^{ ' }\left( x \right) f\left( x \right) =1$

Which implies that

${ \frac { 1 }{ f^{ 1 }\left( x \right) } =e }^{ f\left( x \right) }+{ e }^{ f\left( x \right) }f\left( x \right)$ (eq.1)

Now, looking at the integral:

$\int { f\left( x \right) dx= } \int { \frac { f\left( x \right) }{ f^{ ' }\left( x \right) } } df\left( x \right)$

Using eq.1

$\int { \frac { f\left( x \right) }{ f^{ ' }\left( x \right) } } df\left( x \right) =\int { { f\left( x \right) }^{ 2 } } { e }^{ f\left( x \right) }df\left( x \right) +\int { { f\left( x \right) } } { e }^{ f\left( x \right) }df\left( x \right)$

Integrating by parts

$\int { \frac { f\left( x \right) }{ f^{ ' }\left( x \right) } } df\left( x \right) =\int { { f\left( x \right) }^{ 2 } } { e }^{ f\left( x \right) }df\left( x \right) +\int { { f\left( x \right) } } { e }^{ f\left( x \right) }df\left( x \right)=$ $={ f\left( x \right) }^{ 2 }{ e }^{ f\left( x \right) }-2\int { { f\left( x \right) } } { e }^{ f\left( x \right) }df\left( x \right) +\int { { f\left( x \right) } } { e }^{ f\left( x \right) }df\left( x \right)$

Integrating by parts again

${ f\left( x \right) }^{ 2 }{ e }^{ f\left( x \right) }-\int { { f\left( x \right) } } { e }^{ f\left( x \right) }df\left( x \right) ={ f\left( x \right) }^{ 2 }{ e }^{ f\left( x \right) }+{ e }^{ f\left( x \right) }-f\left( x \right) { e }^{ f\left( x \right) }$

Therefore:

$\int _{ 0 }^{ e }{ f\left( x \right) dx=e[f(e)-1] } +{ e }^{ f\left( e \right) }-{ e }^{ f(0) }$ (eq.2)

Back to the integral:

$\int { f(x)dx=\int { ln(x)dx-\int { ln(f(x))dx } } }$

Using (eq.1) in the second term and using integral by parts in both terms, we find that:

$\int { f(x)dx=\quad x[ln(x)-1]+{ e }^{ f\left( x \right) } } [1-f(x)ln(f(x))]$

And therefore:

$\int _{ 0 }^{ e }{ f(x)dx } ={ e }^{ f(e) }-eln(f(e))-{ e }^{ f(0) }$ (eq.3)

Using eq.2 and eq.3, we find that

$f(e)-1=ln(f(e))\Rightarrow f(e)=1$

Since ${ e }^{ f\left( 0 \right) }=1$ , we conclude, by eq.2 or eq.3, that

$\int _{ 0 }^{ e }{ f(x)dx=e-1 }$

let y=f(x) therefore, y(e)^y=x, it is an invertible function. => ∫ydx is area of y=f(x) with x-axis from x=0 to x=e which is equal to [e*1-∫xdy] where y is from 0 to 1. =e-1

If $x=ye^y$ then $y=W(x)$ , where $W(x)$ is the Lambert-W function.

$\int_0^e W(x) dx = e-1 = \boxed{1.718}$