Integral of $x^{\frac{1}{x}}$

Question

I was just wondering if there was a way to find the integral of $x^{\frac{1}{x}}$, either as a function or a value for between $0$ and $\infty$, although I think this is unbounded, as graphically the function seems to asymptote to $y = 1$. I am guessing that this is probably non-elementary, but if anyone knows something about this, (or can even prove that it asymptotes), I would be interested to find out. Thanks.

(I have included a tag on $e$, because $x = e$ is the maximum value of the function.)

According to wolfram alpha, there is no indefinite integral: https://www.wolframalpha.com/input?i=integrate+x%5E%281%2Fx%29 — 1123581321, Aug 14 '23 at 07:36
Your proof that this integral equals $+\infty$ is of course correct. Your question is unclear. What else (if any) do you want, what for, and what did you try? — Anne Bauval, Aug 14 '23 at 10:25
Of course the indefinite integral $\int x^x,dx$ "exists". But it is not an "elementary function". — GEdgar, Aug 14 '23 at 12:19
Does this answer your question? How to integrate the nth root of n — Тyma Gaidash, Aug 14 '23 at 12:53

score 2 · Answer 1 · answered Aug 14 '23 at 09:44

Although it seems there is not a formula for the indefinite integral of such a function, you can say for sure that

$$\int_1^{\infty} x^{\frac{1}{x}} = +\infty.$$

Indeed, $\lim_{x \to \infty} x^{\frac{1}{x}} = 1$ implies so. See Divergence of improper integral if f has limit $l>0$ and Limit at infinity of x^(1/x) for reference.

Integral of $x^{\frac{1}{x}}$

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