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A very simple question, but I cannot seem to find the answer. I know that $f(x)=x^x$ simply isn't integrable, but that doesn't mean the indefinite integral doesn't exist, right? Is there anything that you could put in place of the "$?$" that could satisfy the indefinite integral? $$\int x^x dx =?$$

Sam
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  • You contradict to yourself. How the indefinite integral can exist if underlying function is not integrable? – Kaster Aug 21 '15 at 07:39
  • What do you mean by "$f(x)=x^x$" simply isn't integrable? Of course it is integrable. – 5xum Aug 21 '15 at 07:39
  • Do you mean that you just want a function $f(x)$ such that $f'(x) = x^x$? – Sempliner Aug 21 '15 at 07:39
  • You can just say "Let $F:[0, \infty)\to \Bbb R$ be defined as $F(x) = \int_0^x t^t dt$", and you're good to go. – Arthur Aug 21 '15 at 07:39
  • @Sempliner I thought that was pretty obvious. – Sam Aug 21 '15 at 07:41
  • Then you just use the fundamental theorem of calculus... – Sempliner Aug 21 '15 at 07:42
  • Note that $|x|^x$ is a "differentiable" extension of $x^x$ to all of the reals. Well, not quite: at $x=0$, the limit of the derivative from either side is $-\infty$. But since the limit is the same from either side, the curve is smooth in a sense. – 2'5 9'2 Aug 21 '15 at 07:57

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So it seems you are very confused here. It is not the case that $x^x$ is not integrable, that would be absurd as it would have to have discontinuities of positive measure which it most certainly does not. What you actually have heard is that $x^x$ does not have an elementary antiderivative, i.e. there is not a well known function that is a polynomial in $e^x$, sin, cos, or $x$ that expresses the antiderivative of this function. However one can obviously define $\int_{0}^x t^t dt$ to be the antiderivative of this function.

Sempliner
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