# 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.

## 2 Answers

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.

• You're probably correct (I'm no expert on those things). But shouldn't you justify the switch of sum and the integral (Probably as the sum converge as this is why it was used to begin with). – Royi Sep 26 '17 at 11:05
• @Royi Its kind of difficult getting a notion of rigorously interchanging the sum and integral when no bounds are given. – Simply Beautiful Art Sep 26 '17 at 13:13
• @SimplyBeautifulArt, But there are known conditions when the two are interchangeable, right? If I remember correctly is has something to do with the convergence of the sum. – Royi Sep 26 '17 at 13:17
• @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
• – Royi Sep 26 '17 at 13:26