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Yesterday, I was reading about the Fundamental Theorem of Calculus and I asked this question here: $FTC$ problem $\frac d{dx}\int_1^\sqrt x t^tdt$. Then I realised that I never thought about $\int x^xdx$ at all.

So I tried to see what the result is if I take the $x$-derivative of $x^x$, $$y=x^x$$ $$\ln y=x\ln x$$ $$\frac 1y \frac {dy}{dx} = \ln x + 1$$ $$\therefore \; dy = x^x\ln x + x^x dx$$ Since I can take the derivative of $x^x$ I figured I could try to use integration by parts on $\int x^xdx$.

I dont think trying $u=x$ to get $du=dx$ is allowed and it will fail immediately anyway since it would mean $dv=x^xdx$ which poses the same question as the original integral when trying to find $v=\int dv = \int x^xdx$.

Thus, knowing that $du = x^x\ln x + x^x dx$, I'm left with trying to let $u=x^x$ and see what happens.

It got me got nowhere the integral just gets more complex, $$\int x^xdx = x\left(x^x\right)-\int x\left(x^x\ln x+x^x\right)dx$$ and if I try $u=x$ and $dv=x^x\ln x+x^x dx$ I get $$\int x^xdx = x(x^x)-\left(x(x^x)-\int x^xdx\right)$$ which is useless.

So I asked Mathematica about $\int x^xdx$ and it just has no answer. I looked here :Finding $\int x^xdx$ but my question is a bit different. Hence, if it is possible to solve $\int x^xdx$ explicitly, how can this be done, or does $\int x^x dx$ even exist and if it does not, why is that so?

Calculus always blows my mind and I absolutely love this topic, I want to know what is really going on when I use this powerful tool before I even think about to try and learn Real Analysis in the future.

Thank you for the answers and all help is always vastly appreciated!

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    $\begingroup$ If you assume that $$\int x^x\ dx=f(x)x^x$$ and differentiate both sides, you get a differential equation for $f$, which can be seen to be unsolvable in terms of elementary functions. $\endgroup$ – Simply Beautiful Art Feb 28 '17 at 18:41
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    $\begingroup$ Yes, an antiderivative exists: $\int_1^x t^t\,dt$; the others differ from this one by a constant. Like other functions, this antiderivative cannot be expressed in terms of “elementary” functions. There's nothing strange in the fact you can compute the derivative in terms of elementary functions, because this is an elementary function: $x^x=e^{x\log x}$. It's a similar situation to $f(x)=1/(1+x^2)$ where you have to go to a transcendental function and the antiderivative cannot be expressed in terms of polynomial functions. $\endgroup$ – egreg Feb 28 '17 at 18:44
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    $\begingroup$ Related: Sophomore's dream $\endgroup$ – Clement C. Feb 28 '17 at 18:47
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    $\begingroup$ $x^x$ is better written as $e^{x\log x}$ which shows that for $x>0$ it is well-defined and continuous (since it is composition of continuous ones.) So, its antiderivative does exist by FTC. However, its calculation is some other story... or a novel $\endgroup$ – Behnam Esmayli Feb 28 '17 at 18:48
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    $\begingroup$ The answer to the question in the title is yes by the FTC: $x^x$ has an antiderivative on $[0,\infty)$ because $x^x$ is continuous on this interval. $\endgroup$ – zhw. Feb 28 '17 at 19:16
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There is no exact. But you can integrate the Taylor series of the function. $$x^x=\sum_{\text{n}=0}^\infty\frac{x^n\ln^n\left(x\right)}{n!}$$

And you will get an infinite series as the integral as well.

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    $\begingroup$ @Jan Eerland: I wanted to ask why did you make those edits, but i could find it out: i is not lucky, because we get complex number for negative x values, and appropirate to use the same notation as the questioner (ln instead of log). Thanks. $\endgroup$ – user145014 Feb 28 '17 at 18:58

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