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In my old writes I found next formula, where is ${_{}^2}x$ is tetration:

$\int_0^1 {_{}^2}x \ dx = \sum\limits_{i=1}^\infty \frac {(-1)^{i+1}} {{_{}^2}i} \approx 0.783430511\ldots$

And now I am interested in series of generalized case of tetration:

$\int_0^1 {_{}^n}x \ dx = ?$

Could anybody find out it with explanation?

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    I just proved a generalization of the Sophomore's Dream for [equation $(2)$](http://math.stackexchange.com/questions/205860/the-sum-of-a-series/205896#205896). Sophomore's Dream, as cited above, uses $q=-1$.2012-10-02

1 Answers 1

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Let $n \in \mathbb{Z}^+$, $x>0$ and

$a_{n,k}= \begin{cases} 1 & \quad \text{if $k=0$}\\ \dfrac{1}{k!} & \quad \text{if $n=1$}\\ \displaystyle \frac{1}{k}\sum_{j=1}^k ja_{n,k-j}a_{n-1,j-1} & \quad \text{otherwise.}\\ \end{cases} $

Then

$ \int {}^n x\, dx= \sum_{k=0}^n \frac{(-1)^k (k+1)^{k-1}\Gamma(k+1, -\log x)}{k!} + \sum_{k=n+1}^\infty (-1)^k a_{n,k} \Gamma(k+1, -\log x) + C. $

Source: I.N. Galidakis, On an Application of Lambert’s W Function to Infinite Exponentials, Corollary 10.9.

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    Yes. Did you surprise because of this series expansion is trivial for you? :)2012-10-03