What is the series expansion of $ \sqrt[x] x $? What is the series expansion of $ \sqrt[x] x $ ? I want to find it because I want to find $ \int \sqrt[x] x dx$ but believe that the integral cannot be expressed in the form of elementary functions.
 A: Given $\,f(x) := x^{1/x},\,$ assuming $\,x>0.\,$
Let $\,t>0,\,$ let $\,u := -\log(t),\,$ and $\, z := (x-t)/t^2.\,$ Expand $\,f(x)\,$ into a power series at $\,x=t\,$ getting
$$ \frac{f(x)}{f(t)} = \sum_{n=0}^\infty a_n \frac{z^n}{n!} =
 1 + (1+u)z + (u^2 + 2(1-t)u +(1-3t))\frac{z^2}{2!} +\dots$$
where $\,a_n\,$ is a polynomial in $\,t\,$ and $\,u\,$ of degree
$\,n\,$ in $\,u.\,$ For example,
$$ a_3 = u^3 + 3(1-2t)u^2 +3(1-5t+2t^2)u +(1-9t+11t^2).$$
Note that the choice of $\,t\,$ is arbitrary and you can let it be $\,t=1,\,$ for example, but I have not yet determined the radius of convergence which is an important detail since it determines where the expansion is valid. It is known that all derivatives of $f(x)$ are $0$ at $x=0$ and so the Taylor series sums to $0$ and is not valid for $x>0.$ The plot of function $\,f(x)\,$ is:

Note also that the terms in the power series expansion of
the function $\,f(x) = \exp(\log(x)/x)\,$ all had a factor of $\,f(t)\,$ which is why I factored it out in my expansion. Verify this by finding
successive derivatives of $\,f(x)\,$ as follows.
Let $\,f(x) = \exp(g(x))\,$ where $\,g(x) = \log(x)/x.$
Let $\,d_n := \left(\frac{d}{dx}\right)^n f(x)\,$ be the $n$-th derivative of $\,f(x)\,$ so $\,d_0 = f(x)\,$ by definition. Next $\,d_1 = f(x)\,g'(x)\,$ where
$\,g'(x) = (1 - \log(x))/x^2.\,$ Next $\,d_2 = f(x)(g'(x)^2 + g''(x)),\,$ and so on. By induction $\,d_n\,$ is a polynomial in $\,x\,$ and $\,\log(x)\,$ all divided by $\,x^{2n}.\,$
A: Easier than I thought.
We can write $ x ^ \frac {1}{x} $ as:
$$ x ^ \frac {1}{x}= e ^ { \frac {ln x} {x}} $$
The expansion of $e^x$ is:
$$ e^x=\sum_{n=0}^\infty \frac {x^n}{n!}$$ 
Similarly the expansion of $e ^ { \frac {ln x} {x}}$ is:
$$ e ^ { \frac {ln x} {x}}=\sum_{n=0}^\infty \frac {(\ln x)^n}{x^n.n!} $$
A: $$x^{1/x}$$ doesn't have a series expansion around $x=0$. All derivatives are $0$ (assuming the domain $\mathbb R^+$).
