Freshman's Dream!

01xxx... dx=k=0(1)k(k+1)k1\displaystyle \int_0^1 {x^{x^{x^{.^{.^.}}}} \ dx} = \sum_{k=0}^{\infty}{{(-1)}^k {(k+1)}^{k-1}}

Proof:

Since generalized exponential series -

Et(x)=k=0(tk+1)k1xkk!\displaystyle \mathcal{E}_t(x) = \sum_{k = 0}^{\infty}{{(tk+1)}^{k-1} \frac{x^k}{k!}}

follow

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

Therefore, for t=1t= 1,

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

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

That's our tetration.

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

Also,

E(ln(x))=k=0(k+1)k1ln(x)kk!\displaystyle \mathcal{E}(\ln(x)) = \sum_{k = 0}^{\infty}{{(k+1)}^{k-1} \frac{{\ln(x)}^k}{k!}}

01xxx... dx=k=0(k+1)k1k!01(ln(x))k dx\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=01(ln(x)k dxI = \int_0^1 {{(\ln(x)}^k\ dx}

Consider

I(n)=01xn dx=1n+1I(n) = \int_0^1 {x^n \ dx} = \frac{1}{n+1}

I(k)(0)=01(ln(x)k dx=(1)kk!I^{(k)}(0) = \int_0^1 {{(\ln(x)}^k\ dx} = {(-1)}^k k!

Therefore,

01xxx... dx=k=0(k+1)k1k!(1)kk!\displaystyle \int_0^1{x^{x^{x^{.^{.^.}}}}\ dx} = \sum_{k = 0}^{\infty}{\frac{{(k+1)}^{k-1}}{k!}{(-1)}^k k! }

01xxx... dx=k=0(1)k(k+1)k1\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 years, 4 months ago

No vote yet
1 vote

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