3
$\begingroup$

How to prove that : $$ \int x^{n-1}W(x)dx = \frac {x^ne^{[-nW(x)]}[-nW(-x)]^{-n}[n\Gamma(n+1, -nW(x)- \Gamma(n+2, -nW(x))]} {n^2} $$ Where $W(x)$ is the Lambert-W function https://en.wikipedia.org/wiki/Lambert_W_function and $\Gamma$ is the incomplete gamma function.

I found this on wolfram alpha https://www.wolframalpha.com/input/?i=int+(x)%5E(n-1)W(x), but have no idea how it got this.

For those who are wondering why I want to know the proof of this integral it is because with this integral I can find the integral of the infinite titration of x as :

$ \int x^{x^{x^{.^{.......}}}} dx = \sum_{n=0}^{\infty}-\frac {(-1)^n( \ln x)^ne^{[-nW(-\ln x)]}[-nW(-\ln x)]^{-n}[n\Gamma(n+1, -nW(-\ln x)- \Gamma(n+2, -nW(-\ln x))]} {(n!)n^2} $

Lol the answer is so long that it doesnt fit on one line.

$\endgroup$
  • 2
    $\begingroup$ You probably want to make the substitution $ue^u = x$, $dx = (u+1)e^u du$ that would give you $\int u^{n-1} e^{(n-1)u} W(ue^u) (u+1)e^u du$. We know that $W(ue^u) = u$ Hence our integral is $\int u^{n} e^{(nu}(u+1) du$. What next? Well $\Gamma(x) = $ (tbc) $\endgroup$ – George Dewhirst Apr 3 at 22:47
  • 1
    $\begingroup$ If we try integration by parts: $$I=\int x^{n-1}W(x)dx$$ $u=W(x)$ then $u'=\frac{W(x)}{x[W(x)+1]}$ and $v'=x^{n-1}$ so $v=\frac{x^n}{n}$ so: $$I=\frac{x^n}{n}W(x)-\int\frac{x^{n-1}}{n}\frac{W(x)}{W(x)+1}dx$$ This may get somewhere $\endgroup$ – Henry Lee Apr 4 at 3:29
7
$\begingroup$

You probably want to make the substitution $ue^u = x$,

$dx = (u+1)e^u du$ that would give you $\int u^{n-1} e^{(n-1)u} W(ue^u) (u+1)e^u du$.

We know that $W(ue^u) = u$

Hence our integral is

$\int u^{n} e^{nu}(u+1) du$.

What next?

Well $\Gamma(x) = \int t^{x-1} e^{-t}dt$ So our integral looks kind of like a gamma integral but not quite.

We will separate it into

$\int u^{n+1} e^{nu} du (1) + \int u^{n} e^{nu} du (2)$.

$t = -nu$

$(1) = \int (-1/n)^{n+1} t^{n+1} e^{-t} dt/(-n) = (-1/n)^{n+2} \Gamma(n+2)$

$(2) = \int (-1/n)^{n} t^{n} e^{-t} dt/(-n) = (-1/n)^{n+1} \Gamma(n+1)$

Final answer $(-1/n)^{n+2} \Gamma(n+2)+(-1/n)^{n+1} \Gamma(n+1)$

(tbc)

$\endgroup$
  • 1
    $\begingroup$ Obviously think about what limits you'd like on the integrals as well, as the Gamma function is an integral over $[0,\infty)$ $\endgroup$ – George Dewhirst Apr 3 at 23:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.