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The factorial $n!$ has a nice representation as definite integral: $ n!=\Gamma(n+1)=\int_0^\infty t^{n} e^{-t}\, \mathrm{d}t. \! $ Is it possible to write down such an integral for $n^n$ as well?

I tried to come up with an integral that reproduces a $n$ factor, $n$-times, but without success. I don't see a way to stop the partial integration process like in the $n!$ case. So this might not work here and I currently can't think of another way. If it helps to restrict $n$, feel free to do so.

The only thing a found online so far is the Lambert's $W$ function, which is involved when solving $x^x=z$, but I'm not sure if this helps.

EDIT: Answers with integrals of the form $\displaystyle n^n=\int_0^\infty \cdots dt$ are preferred.

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    Asymptotics like $\displaystyle n^n\approx \int_0^\infty (2\pi n)^{-1}(et)^{n} e^{-t}\, \mathrm{d}t $ don't count.2012-07-26

4 Answers 4

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The correct question (which someone has just asked me) may be as follows:

Find $f(t)$ (positive and independent of $n$) such that

$n^n = \int_0^\infty t^n f(t) \,dt,\quad \forall\,n$

That is, find $f(t)$ such that $\{n^n\}$ is the sequence of moments of f(t)dt. This is a particular instance of the Stieltjes moment problem.

There is criterion (due to Carleman) which guarantees that such a solution is unique, but I don't know an explicit form for $f(t)$.

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    thanks for the background infos...2013-03-05
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This may not be the line of thinking you are after, but it does give the integral you desire.

$n^n=\int_0^n nx^{n-1}\ dx$

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    Thanks, no it's not. I thought about an infinite upper limit. But despite that it looks very interesting. +12012-07-26
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$\frac{\Gamma (\alpha)}{s^{\alpha}}=\int_{0}^{\infty }t^{\alpha-1}e^{-st}dt\tag{1}$

$\frac{\Gamma (s)}{s^{s}}=\int_{0}^{\infty }t^{s-1}e^{-st}dt\tag{2}$

$s^{-s}=\frac{1}{\Gamma (s)}\int_{0}^{\infty }t^{s-1}e^{-st}dt\tag{3}$

$s^{s}=\frac{\Gamma (s)}{\int_{0}^{\infty }t^{s-1}e^{-st}dt}\tag{4}$

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    Hmm, not bad +1. invert it and we are even closer $\displaystyle s^{s}=\frac{\Gamma (s)}{\int_{0}^{\infty }t^{s-1}e^{-st}dt}$, at least on the lhs...2012-07-26
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No need for the Lambert W function , respecting your limits and using $\Gamma$ we get that

$ n^{n}=\frac{1}{\Gamma (n+1)}\int_{0}^{\infty }e^{-\frac{t^{ \frac{1}{n}}}{n}} dt $

Update:

Lookup the Exponential Integral and its relationship with the Incomplete gamma function

$E_{n}(x)=x^{n-1}\Gamma(1-n,x)\tag{1}$

for $n=1-n$ and $x=\frac{1}{n}$ we get that :

$E_{1-n}(\frac{1}{n})=\int_{1}^{\infty}\frac{e^{-\frac{t}{n}}}{t^{1-n}}dt = n^{n}\Gamma (n,\frac{1}{n})\tag{2}$

Changing the integration limits to $[0,1]$ :

$\int_{0}^{1}\frac{e^{-\frac{t}{n}}}{t^{1-n}}dt = n^{n}(\Gamma(n)-\Gamma (n,\frac{1}{n}))\tag{3}$

Combining 2+3 we get the solution:

$n^{n}=\frac{1}{\Gamma(n)}\int_{0}^{\infty}\frac{e^{-\frac{t}{n}}}{t^{1-n}}dt$

The top result is due to the relationship :

$n E_{1-n}(\frac{1}{n})= \int_{1}^{\infty} e^{-\frac{t^{\frac{1}{n}}}{n}} dt\tag{4}$

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    Thanks again, great answer, vote up people$\color{green}{.}\color{goldenrod}{.}\color{red}{.}$2012-07-30