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By the definition of the integral we have:

If $f$ is a function defined for $a \leq x \leq b$, we divide the interval $[a,b]$ into $n$ subintervals of equal width ${\Delta}x = \frac{b - a}{n}$. We let $x_{0} (= a), x_{1}, x_{2}, {\dots}, x_{n} (= b)$ be the endpoints of these subintervals and we let $x_{1}^{*}, x_{2}^{*}, \dots, x_{n}^{*}$ be any sample points in these subintervals, so $x_{i}^{*}$ lies in the $i$th subinterval $[x_{i - 1}, x_{i}]$. Then the definite integral of $f$ from $a$ to $b$ is

\begin{equation} \int_{a}^{b} f(x) \;\mathrm{d}x = \lim_{n \to \infty} \sum_{i = 1}^{n} f(x_{i}^{*}){\Delta}x \end{equation}

provided that the limit exists and gives the same value for all possible choices of sample points. If it does exist, we say that $f$ is integrable on $[a, b]$.

From this, if $f(x) = x^{x}$, $f$ is defined on $[a, b]$ if $a = 0$ and $b = \infty$ so $f$ should be integrable.

\begin{align} \int_{0}^{1} x^{x} \;\mathrm{d}x &= \lim_{n \to \infty} \sum_{i = 1}^{n} {x_{i}^{*}}^{x_{i}^{*}}{\Delta}x \end{align} with ${\Delta}x = \frac{b - a}{n} = \frac{1 - 0}{n} = \frac{1}{n}$

and $x_{i}^{*} = a + i{\Delta}x = 0 + i \times \frac{1}{n} = \frac{i}{n}$ \begin{align} \int_{0}^{1} x^{x} \;\mathrm{d}x &= \lim_{n \to \infty} \sum_{i = 1}^{n} \left(\frac{i}{n}\right)^{\left(\frac{i}{n}\right)}\left(\frac{1}{n}\right)\\ &= \lim_{n \to \infty} \left(\frac{1}{n}\right) \sum_{i = 1}^{n} \left(\frac{i}{n}\right)^{\left(\frac{i}{n}\right)} \end{align}

The issue now is that I'm not sure how to go about evaluating this Riemann Sum to determine whether the limit exists or not. I know that it's difficult (if not impossible) to find the indefinite integral of $x^{x}$ as there is no antiderivative using elementary functions. No function exists which when differentiated yields $x^{x}$. But this shouldn't have any effect on the integrability of the function since by the theorem stated above the function is defined on the interval therefore it should be integrable.

What information could you give me to point me in the right direction?

Kalcifer
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  • There is a theorem that says a continuous function on an open interval has an antiderivative there. Are you looking for something as abstract as that? – hardmath Dec 25 '20 at 06:06
  • @hardmath So a function is only integrable if it has anti derivative which exists on an open interval? The theorem I see in my textbook is as follows: If $F$ is an antiderivative $f$ on an interval $I$, then the most general antiderivative of $f$ on $I$ is $F(x) + C$ where C is an arbitrary constant. – Kalcifer Dec 25 '20 at 06:10
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    $x^{x} \to 1$ as $ x\to 0$. So you are looking at Riemann integral of a continuous function. – Kavi Rama Murthy Dec 25 '20 at 06:14
  • @Kalcifer: If a function has an antiderivative on an open interval, then it is Riemann integrable there. The theorem you found describes how the antiderivative (if it exists) is not unique (because of the famous "+C" relating any one antiderivative to another). However for the proposition that a continuous function on an open interval has an antiderivative, see Do all continuous functions have antiderivatives?. – hardmath Dec 25 '20 at 06:30
  • $$\int_a^b x^xdx=\text{Sphd}(1;b)-\text{Sphd}(1;a)$$ Sphd$(\alpha;x)$ is the so called Sophomores Dream function. See the properties in : https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function . – JJacquelin Dec 25 '20 at 11:07

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AFAIK, there is no closed form of this, but I have a uselful result.
This integral is known as sophomore's dream.
We have $$\int_{0}^{1}x^xdx=\int_{0}^{1}e^{x\log x}dx=\sum_{k=0}^{\infty}\int_{0}^{1}\frac{x^k\log ^k x}{k!}dx$$ I will cheat a bit by using mathematica, which gives

$$\int_{0}^{1}x^k\log^kxdx=(-1)^k(1+k)^{-(1+k)}k!$$ So $$\int_{0}^{\infty}x^xdx=\sum_{k=0}^{\infty}\frac{(-1)^k}{(1+k)^{1+k}}=-\sum_{k=1}^{\infty}\frac{(-1)^k}{k^k}\label{a}\tag{1}$$ This is a simple and nice sum representation of that integral.
You can also see this, this and this.