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I am working on a problem in which there is an integral containing $x^\frac1x$. I have looked at this question for some help: Behavior of the Nth root of N?

I have changed $x^\frac1x$ to $e^{ln(x)/x}$ but I still have not found a way to integrate it. How can I integrate either equation? Thanks. If you vote down, please tell me why so I can improve this question.

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  • $\begingroup$ Wolfram alpha say "no result found in terms of standard mathematical functions", so there is probably no easy way to do this. What's your problem exactly, is your integral definite or indefinite? $\endgroup$ – Tryss Feb 24 '15 at 14:26
  • $\begingroup$ @Tryss The integral is definite(but inside a limit). I have already tried many of the online function integrators out there and none have given me an answer. $\endgroup$ – Progo Feb 24 '15 at 14:29
  • $\begingroup$ You may have more luck in posting the whole expression to calculate. Maybe only an asymptotic growth or an approximation is necessary for your problem $\endgroup$ – Tryss Feb 24 '15 at 14:31
  • $\begingroup$ As said, it would be better to know what is the integral. Please, post it. $\endgroup$ – Claude Leibovici Feb 24 '15 at 14:37
  • $\begingroup$ $\int x^{1/x}\,dx$ is not an elementary function. Some (indeed most) integrals are like that. $\endgroup$ – GEdgar Feb 24 '15 at 14:39
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Define $$ F(x) = \int_{1}^{x} s^{1/s}\;ds,\qquad x>0 $$ Then $F'(x) = x^{1/x}$ so $F$ is an antiderivative. There is no simpler way to write this.

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  • $\begingroup$ So, x^(1/x) is the anti-derivitive of x^(1/x)? $\endgroup$ – Progo Mar 7 '15 at 15:03
  • $\begingroup$ No that function $F$ is the antiderivative of $x^{1/x}$. $\endgroup$ – GEdgar Mar 7 '15 at 15:05
  • $\begingroup$ Ok, thanks! This is very helpful! $\endgroup$ – Progo Mar 7 '15 at 15:06

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