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Although one cannot find an elementary antiderivative of $f(x)=x^x$, we can still give a series representation for $\int_0^1 x^x dx$, namely:

$$I_1=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^n}=0.78343\ldots$$

One can even find an expression for the complete antiderivative in terms of infinite sums and the incomplete gamma function $\Gamma(a,x)$:

$$\int x^x dx =\sum_{n=1}^\infty \left(\frac{(-1)^{n+1}\Gamma(-n\ln(x),n)}{n^n \Gamma(n)}\right)+C$$

Considering special, non-elementary function, series, infinite products, etc. , is this also possible for $\int_0^1 x^{x^x} dx$?

Thank you in advance!

PinkyWay
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user765118
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1 Answers1

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

$$\exp(\ln(x)e^{x\ln(x)})= \sum^{\infty}_{n=0} \frac{\big(\ln(x)e^{x\ln(x)}\big)^n}{n!}$$

$$\int_0^1 x^{x^x}dx=\int_0^1\sum^{\infty}_{n=0} \frac{\big(\ln(x)e^{x\ln(x)}\big)^n}{n!}dx$$ Now using Fubini's theorem we swap the integral and sum

$$=\sum^{\infty}_{n=0} \frac{1}{n!} \int_0^1\big(\ln(x)e^{x\ln(x)}\big)^ndx$$ Now we evaluate the integral

$$\int_0^1\big(\ln(x)e^{x\ln(x)}\big)^ndx=\int_0^1 \ln^n(x)e^{nx\ln(x)}dx=\int_0^1 \sum^\infty_{n=0} \frac{\ln^n(x)(nx\ln(x))^n}{n!}dx$$ $$= \sum^\infty_{n=0} \int_0^1 \frac{\ln^n(x)(nx\ln(x))^n}{n!}dx=\sum^\infty_{n=0}\frac{n^n}{n!} \int_0^1 x^n \ln^{2n}(x)dx$$ Evaluating the integral using Bernoulli's proof

$$\int_0^1 x^n \ln^{2n}(x)dx= \Bigg[\frac{x^{n+1}}{n+1}\sum_{i=0}^{2n} (-1)^i \frac{(2n)_i}{(n+1)_i} (\ln(x))^{2n-1}\Bigg]_0^1$$

Therefore $$\int_0^1 x^{x^x}dx=\sum^{\infty}_{n=0} \frac{1}{n!} \int_0^1\big(\ln(x)e^{x\ln(x)}\big)^ndx= \sum^{\infty}_{n=0} \frac{1}{n!}\sum^\infty_{n=0}\frac{n^n}{n!} \int_0^1 x^n \ln^{2n}(x)dx$$

$$=\sum^{\infty}_{n=0} \frac{1}{n!}\sum^\infty_{n=0}\frac{n^n}{n!} \Bigg[\frac{x^{n+1}}{n+1}\sum_{i=0}^{2n} (-1)^i \frac{(2n)_i}{(n+1)_i} (\ln(x))^{2n-1}\Bigg]_0^1 $$ I don't know how to do the closed form for this integral but I hope this helps

hwood87
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  • @TymaGaidash I believe one would have to use limits. I just used the Wikipedia page linked and substituted $m=n$ and $n=2n$ to the equation – hwood87 Sep 07 '21 at 00:10
  • $\int_0^1(\ldots)^n,dx=\sum_{\color{red}{n}=0}^\infty\ldots$ is a mistake that ruins the rest of the answer. You should use a different index. – metamorphy Sep 07 '21 at 06:23
  • @metamorphy What would the rest of the answer be if a different index was used? – Тyma Gaidash Sep 07 '21 at 18:05
  • @TymaGaidash: Obviously, this would result in a double summation. Less useful, but at least correct. – metamorphy Sep 07 '21 at 19:01