I have evaluated $\frac{d}{dx} x^x$ to be $x^x \left( log(x) + 1 \right)$ is that correct? Furthermore, what is $\int x^x dx$? Is that even possible? If not, why?
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1Why don't you search for the integral (as I did), and check your derivative with WolframAlpha (as I did)? Yes, it's correct. – Oct 15 '15 at 02:39
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Ok, so I see that the derivative is correct, but Wolfram gives a very complicated solution to the antiderivative. How is that found? – TheLabyrinthMaker Oct 15 '15 at 02:42
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WolframAlpha states: "no result found in terms of standard mathematical functions". It then gives a series expansion which is an approximation to the answer. – Ian Miller Oct 15 '15 at 02:45
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You're correct, I didn't see that it was an approximation. Thank you. – TheLabyrinthMaker Oct 15 '15 at 02:49
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There is no closed form for the antiderivative, but for a definite integral there is the sophomore's dream:
$$ \int_0^1 x^x\; dx = \sum_{n=1}^\infty (-1)^{n+1} n^{-n} $$
Robert Israel
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@NormalHuman Why did someone have the urge to down vote a perfectly good answer? It is not the respondent's responsibility to comb through the archaic categorization of answers before proceeding to help. What is this perverse delight that some take for posting a comment such as yours? +1 to neutralize the absurdity! – Mark Viola Oct 15 '15 at 03:22
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@Rbert Israel. Proving that formula for $\int^1_0 x^x $ was a Putnam Competition problem, one of the easier ones. – DanielWainfleet Oct 15 '15 at 03:48
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Hi Robert Israel ! Look at this : https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function – JJacquelin Nov 30 '15 at 18:01