# Trying to solve this integral $\int x^xdx$ [duplicate]

I am trying to solve the following integral which supposedly cannot be solved by "normal" means.

$$\int x^xdx$$

Here's what I've come up with so far:

$$x^x=e^{x\ln(x)}$$

$$\frac{d(x^x)}{dx}=x^x\ln(x)+x^x$$

This must mean that:

$$\int \left(x^x\ln(x)+x^x\right)dx=x^x$$

this means: $$x^x-\int x^x\ln(x)dx=\int x^xdx$$

If I can solve $$\int x^x\ln(x)dx$$

I can solve $$\int x^xdx$$

Any ideas on how to solve the integral? Thanks in advance.

## marked as duplicate by Simply Beautiful Art, kingW3, José Carlos Santos, B. Goddard, Antonios-Alexandros RobotisSep 27 '17 at 0:02

You have to solve it as a series since there is no way to evaluate it as an 'exact' integral. Note, the series comes from the expansion of ${e^{\ln x^x}}$
$$\int{x^xdx} = \int{e^{\ln x^x}dx} = \int{\sum_{k=1}^{\infty}\frac{x^k\ln^k x}{k!}}dx$$
In exactly the same spirit as Programmer $$\int{x^xdx} = \int{\sum_{k=1}^{\infty}\frac{x^k\ln^k (x)}{k!}}\,dx=\sum_{k=1}^{\infty}\frac{1}{k!}\int{{x^k\ln^k(x)}}\,dx$$ and $$\int{{x^k\ln^k(x)}}\,dx=-\ln ^{k+1}(x)\, E_{-k}(-(k+1) \ln (x))$$ where appears the exponential integral function.
• @Royi Well, you still probably want bounds in order to get some meaning of it. But I do believe for any $0\le a\le x\le b$, the integral and sum may be rearranged. – Simply Beautiful Art Sep 26 '17 at 13:18