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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.

C.F.G
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Mário Marques
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2 Answers2

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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 ${e^{\ln x^x}}$

$$\int{x^xdx} = \int{e^{\ln x^x}dx} = \int{\sum_{k=1}^{\infty}\frac{x^k\ln^k x}{k!}}dx$$

Programmer
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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.

Claude Leibovici
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  • 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
  • I meant something like https://math.stackexchange.com/questions/83721/when-can-a-sum-and-integral-be-interchanged, https://math.stackexchange.com/questions/1334907/reversing-the-order-of-integration-and-summation and http://planetmath.org/criterionforinterchangingsummationandintegration. – Royi Sep 26 '17 at 13:26
  • @Royi If you view all of those links, there are 'bounds' on the integrals, none of the answers/last link go over indefinite integrals (simply because they don't make much sense IMO) – Simply Beautiful Art Sep 26 '17 at 13:28
  • It seems I'm missing something in your intention. For instance this http://planetmath.org/criterionforinterchangingsummationandintegration never mention any boundary of the integral. I guess it could be with boundaries or just calculation of the indefinite integral (Anti Derivative). By the way, this https://math.stackexchange.com/questions/1334907/reversing-the-order-of-integration-and-summation also never mention any boundary of the integral. – Royi Sep 26 '17 at 13:40
  • @Royi No, the integrals in that link are integrals over measure spaces. They aren't indefinite integrals at all. – Simply Beautiful Art Sep 26 '17 at 13:42
  • I see. I wonder what happens in cases of indefinite integral or integral on all $ R $. – Royi Sep 26 '17 at 13:54