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!
from Hot Weekly Questions - Mathematics Stack Exchange
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