Denote by $x_1,\dotsc, x_k$ the random real numbers that you chose. Define the random numbers

$$ y_j= \frac{1}{n-1}(x_j-1), $$

so that $y_j$ is uniformly distributed in $[0,1]$. We have

$$ x_j= (n-1) y_j+1. $$

Denote by $t_j$ the numbers $y_j$ rearranged in increasing order. Then

$$ \alpha_j =(n-1)t_j+1, $$

and for any $r>0$ we have

$$ r^{\alpha_j-j}= r^{(n-1)t_j+1-j}. $$

The random variables $t_j$ are distributed according to the Beta distribution

$$ f_j(t)dt =\frac{k!}{(k-1)!(k-j)!} t^{j-1}(1-t)^{k-j}dt. $$

We have $\newcommand{\bE}{\mathbb{E}}$

$$\bE(r^{\alpha_j-j})= \frac{r^{1-j}k!}{(j-1)! (k-j)!}\int_0^1 r^{(n-1)t} t^{j-1}(1-t)^{k-j}dt $$

$$= \frac{k!}{(j-1)! (k-j)!}\int_0^1 r^{(n-1)t} \left(\frac{t}{r}\right)^{j-1}(1-t)^{k-j}dt $$

$$=k\binom{k-1}{j-1}\int_0^1 r^{(n-1)t} \left(\frac{t}{r}\right)^{j-1}(1-t)^{k-j}dt. $$

$$\sum_{j=1}^k \bE(r^{\alpha_j-j})=k\int_0^1 r^{(n-1)t}\left(\sum_{j=1}^k \binom{k-1}{j-1}\left(\frac{t}{r}\right)^{j-1}(1-t)^{k-j}\right)dt $$

$$= k \int_0^1 r^{(n-1)t} \left(1 +\frac{t}{r}-t\right)^{k-1} dt. $$

Thus we have to compute an integral of the form

$$\int_0^1 e^{\lambda t}(1+at)^k-1 dt,\;\;\lambda, a\in\mathbb{R},\;\;a\neq 0. $$

If we make the change in variables $s= 1+at$, this integral becomes

$$ \frac{1}{a} \int_1^{a+1} e^{\lambda\frac{s-1}{a}} s^{k-1} ds=\frac{1}{ae^{\lambda/a} }\int_1^{a+1} e^{\frac{\lambda}{a} s} s^{k-1} ds. $$

The last integral can be computed integrating by parts.

**Remark.** When the numbers $x_1,\dotsc, x_k$ are integers one thing goes horrible wrong: the probability that two of the numbers are equal is not zero. However you can still find the distribution of $\alpha_j$. For $\ell=1,\dotsc, n$, $P(\alpha_j\leq \ell)$ is the probability that at least $j$ of the numbers $x_1,\dotsc, x_k$ are $\leq \ell$ which is

$$ P_j(\ell)=\frac{1}{n^k}\sum_{s\geq j} \binom{k}{s}\ell^s(n-\ell)^{k-s}. $$

This allows you to find a formula for the expectation of $r^{\alpha_j}$. Note that the function

$$ r\mapsto P_j(r):=\bE(r^{\alpha_j}) $$

is the probability generating function of the random variable $\alpha_j$. In this case it is a polynomial of degree $n$ in $r$. More precisely

$$ P_j(r)= \sum_{\ell=1}^n r^\ell\Bigl(P_j(\ell)-P_j(\ell-1)\Bigr). $$

Hence
$$ \bE(r^{\alpha_j-j})=\sum_{j=1}^k \sum_{\ell=1}^n r^{\ell-j}\Bigl( P_j(\ell)-P_j(\ell-1)\Bigr). $$