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:

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


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 2


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.

  • $\begingroup$ 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). $\endgroup$
    – Royi
    Sep 26, 2017 at 11:05
  • $\begingroup$ @Royi Its kind of difficult getting a notion of rigorously interchanging the sum and integral when no bounds are given. $\endgroup$ Sep 26, 2017 at 13:13
  • $\begingroup$ @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. $\endgroup$
    – Royi
    Sep 26, 2017 at 13:17
  • $\begingroup$ @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. $\endgroup$ Sep 26, 2017 at 13:18

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