Integrating $x^{x^x}$ Although one cannot find an elementary antiderivative of $f(x)=x^x$, we can still give a series representation for $\int_0^1 x^x dx$, namely: 

$$I_1=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^n}=0.78343\ldots$$ 

One can even find an expression for the complete antiderivative in terms of infinite sums and the incomplete gamma function $\Gamma(a,x)$:

$$\int x^x dx =\sum_{n=1}^\infty \left(\frac{(-1)^{n+1}\Gamma(-n\ln(x),n)}{n^n \Gamma(n)}\right)+C$$ 

Considering special, non-elementary function, series, infinite products, etc. , is this also possible for $\int_0^1 x^{x^x} dx$?
Thank you in advance!
 A: $$x^{x^x}=\exp(x^x\ln(x))=\exp(\ln(x)e^{x\ln(x)})$$
$$\exp(\ln(x)e^{x\ln(x)})= \sum^{\infty}_{n=0} \frac{\big(\ln(x)e^{x\ln(x)}\big)^n}{n!}$$
$$\int_0^1 x^{x^x}dx=\int_0^1\sum^{\infty}_{n=0} \frac{\big(\ln(x)e^{x\ln(x)}\big)^n}{n!}dx$$
Now using Fubini's theorem we swap the integral and sum
$$=\sum^{\infty}_{n=0} \frac{1}{n!} \int_0^1\big(\ln(x)e^{x\ln(x)}\big)^ndx$$
Now we evaluate the integral
$$\int_0^1\big(\ln(x)e^{x\ln(x)}\big)^ndx=\int_0^1 \ln^n(x)e^{nx\ln(x)}dx=\int_0^1 \sum^\infty_{n=0} \frac{\ln^n(x)(nx\ln(x))^n}{n!}dx$$
$$= \sum^\infty_{n=0} \int_0^1 \frac{\ln^n(x)(nx\ln(x))^n}{n!}dx=\sum^\infty_{n=0}\frac{n^n}{n!} \int_0^1 x^n \ln^{2n}(x)dx$$
Evaluating the integral using Bernoulli's proof 
$$\int_0^1 x^n \ln^{2n}(x)dx= \Bigg[\frac{x^{n+1}}{n+1}\sum_{i=0}^{2n} (-1)^i \frac{(2n)_i}{(n+1)_i} (\ln(x))^{2n-1}\Bigg]_0^1$$
Therefore
$$\int_0^1 x^{x^x}dx=\sum^{\infty}_{n=0} \frac{1}{n!} \int_0^1\big(\ln(x)e^{x\ln(x)}\big)^ndx=
\sum^{\infty}_{n=0} \frac{1}{n!}\sum^\infty_{n=0}\frac{n^n}{n!} \int_0^1 x^n \ln^{2n}(x)dx$$
$$=\sum^{\infty}_{n=0} \frac{1}{n!}\sum^\infty_{n=0}\frac{n^n}{n!} \Bigg[\frac{x^{n+1}}{n+1}\sum_{i=0}^{2n} (-1)^i \frac{(2n)_i}{(n+1)_i} (\ln(x))^{2n-1}\Bigg]_0^1
$$
I don't know how to do the closed form  for this integral but I hope this helps
A: Let A179230 be the Exponential Generating function for $(x+1)^{(x+1)^{(x+1)}}$, then the antiderivative is given by the Taylor series of $x^{x^{x}}$ at $x=1$:
$$\int x^{x^{x}} dx=\int\sum_{n=0}^\infty \frac{\left(\frac{d^n}{dx^n} x^{x^{x}}\right)_{x=1}\ (x-1)^n}{n!}dx= C+\sum_{n=0}^\infty \frac{\left(\frac{d^n}{dx^n} x^{x^{x}}\right)_{x=1}\ (x-1)^{n+1}}{(n+1)!}= C+\sum_{n=0}^\infty \frac{\text A179230(n)(x-1)^{n+1}}{(n+1)!}=  C+\sum_{n=1}^\infty \frac{\text A179230(n-1)(x-1)^{n}}{n!} = C+\sum_{n=1}^\infty \frac{\left(\frac{d^{n-1}}{dx^{n-1}} x^{x^{x}}\right)_{x=1}\ (x-1)^n}{n!} =C+(x-1)+\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}+\frac{3(x-1)^4}{8}+\frac{4(x-1)^5}{15}+\frac{(x-1)^6}{4}+\frac{53(x-1)^7}{280}+\frac{233(x-1)^8}{1440}+\frac{5627(x-1)^9}{45360}+\frac{2501(x-1)^{10}}{25200}+O\left(x^{11}\right)$$
The final result also uses Big-O notation which is just a way of saying “the rest of the terms”. Here is a graph of the integral function’s series. The Taylor series at other points is beautiful, but complicated. Please correct me and give me feedback!
