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# Freshman's Dream!

$\displaystyle \int_0^1 {x^{x^{x^{.^{.^.}}}} \ dx} = \sum_{k=0}^{\infty}{{(-1)}^k {(k+1)}^{k-1}}$

Proof:

Since generalized exponential series -

$\displaystyle \mathcal{E}_t(x) = \sum_{k = 0}^{\infty}{{(tk+1)}^{k-1} \frac{x^k}{k!}}$

follow

$\displaystyle \mathcal{E}_t(x) = \exp(x\mathcal{E}_t(x)^t)$

Therefore, for $$t= 1$$,

$\displaystyle \mathcal{E}(x) = \exp(x\mathcal{E}(x))$

$\displaystyle \mathcal{E}(\ln(x)) = \exp(\mathcal{E}(\ln(x)) \ln(x)) = x^{\mathcal{E}(\ln(x))}$

That's our tetration.

$\displaystyle \mathcal{E}(\ln(x)) = x^{x^{x^{.^{.^.}}}}$

Also,

$\displaystyle \mathcal{E}(\ln(x)) = \sum_{k = 0}^{\infty}{{(k+1)}^{k-1} \frac{{\ln(x)}^k}{k!}}$

$\displaystyle \int_0^1{x^{x^{x^{.^{.^.}}}}\ dx} = \sum_{k = 0}^{\infty}{\frac{{(k+1)}^{k-1}}{k!} \int_0^1{{(\ln(x))}^k \ dx} }$

Let's take this integral separately.

$I = \int_0^1 {{(\ln(x)}^k\ dx}$

Consider

$I(n) = \int_0^1 {x^n \ dx} = \frac{1}{n+1}$

$I^{(k)}(0) = \int_0^1 {{(\ln(x)}^k\ dx} = {(-1)}^k k!$

Therefore,

$\displaystyle \int_0^1{x^{x^{x^{.^{.^.}}}}\ dx} = \sum_{k = 0}^{\infty}{\frac{{(k+1)}^{k-1}}{k!}{(-1)}^k k! }$

$\displaystyle \int_0^1 {x^{x^{x^{.^{.^.}}}} \ dx} = \sum_{k=0}^{\infty}{{(-1)}^k {(k+1)}^{k-1}}$

/*

Now this "oscillating divergence" is really not what I liked. Is it correct, anyways?

I am not really freshman yet. Will be in a month or so.

*/

Note by Kartik Sharma
2 months, 3 weeks ago