Integrate $\int x^x dx$

Question

I have proceeded as follows: $$I = \int x^x dx = \int \sum \frac {(x \log x)^k}{k!} dx = \sum \frac {\Gamma [k+1, -(k+1) \log x]}{(-1)^k(k+1)^{k+1}k!} + C$$ But I am unable to go further to get rid of the series form and have a closed form expression. Please don't get confused with another post on the same topic . I am just trying to go a little deeper. Approximate closed expressions are well invited.

Answer 1

You arrived at $$\int { { x }^{ x } } dx=\sum _{ n=0 }^{ \infty }{ \int { \frac { { { x }^{ n }\left( \ln {x} \right) }^{ n } }{ n! } } dx}$$ which is equal to $$=\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } \int { { x }^{ n } } { \left( \ln { x } \right) }^{ n } dx }$$ Consider $u = {\left(\ln {x} \right)}^{n} $, $dv = {x}^{n} dx $.

Now, you should use the integration by parts to complete the exercise.

Answer 2

A closed form for this integral was proposed in the document : https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function $$\text{Sphd}(\alpha\:;\:x)=\int_0^x t^{\alpha t}dt$$ Thus $$\int_0^x t^t dt=\text{Sphd}(1\:;\:x)$$ $$\int x^x dx=\text{Sphd}(1\:;\:x)+\text{constant}$$ The document referenced above provides a lot of relationships, properties and related infinite series.

See also : https://en.wikipedia.org/wiki/Sophomore%27s_dream