The question is self-explanatory, why does the integral of $$x^x$$does not exist? Can someone give a proof or reason?
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Related question: https://math.stackexchange.com/questions/141347/finding-int-xxdx – Robert Z Aug 28 '19 at 18:00
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9You are confusing integrability and having a primitive function expressed in terms of elementary functions. As said, continuous functions are integrable. – I was suspended for talking Aug 28 '19 at 18:01
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The question you probably meant to ask is, "Why does $x^x$ not have an elementary antiderivative?" The answer is, "Why should it?" Contrary to the impression many people receive in calculus courses, it is actually quite rare for elementary functions to have an elementary antiderivative. Of course certain very important classes of functions do, and those are the ones you typically see in examples and homework. But look at something like $f(x)^{g(x)}$, or even $f(x) g(x)$ where $f$ and $g$ are randomly-chosen functions, and it's quite likely that it won't have an elementary antiderivative.
Robert Israel
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$x^x$ is continuous on $(0,\infty)$, thus it is integrable with antiderivative $F: (0,\infty), x \mapsto \int_0^x f(t)\text{ dt}$ with $f(x) = x^x$ for $x > 0$ and $f(0)=\lim_{x\rightarrow 0} x^x = 1$.
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I assume that you meant $x\to0^+$ because typically we have a domain of $\mathbb{R}_{\gt0}$ for the function $x\mapsto x^x$. – Peter Foreman Aug 28 '19 at 18:26