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!
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.