Let $D= \{\,a_1x^{r_1} + \cdots + a_n x^{r_n} \, \vert \, a_i \in \mathbb{C} \text{ for } i= 1,2,\dots,n \text{ and each $r_i$ is a real number such that } 0 \le r_1 < r_2 < \cdots <r_n \,\}$
It is very clear that $D$ is an integral domain. It is also easy to see that the units in $D$ are a set of non-zero complex numbers. I am trying to see whether each element of $D$ is irreducible. It is possible to show that if each $r_i$ is rational then the element $a_1x^{r_1} + \cdots + a_n x^{r_n}$ with $r_n \ne 0$ is reducible. To see this let $f(x) = a_1x^{r_1} + \cdots + a_n x^{r_n}$ and observe that there exist an integer $s$ such that $f(x^s) \in \mathbb{C}[x]$ so we can apply fundamental theorem of algebra. There are complex numbers $b_1, b_2, \dots b_{r_n}$ such that $f(x^s) = a_n (x-b_1)\cdots (x- b_{r_n})$. So we have $f(x) = a_n (x^{\frac1s} -b_1)\cdots (x^{\frac1s} - b_{r_n})$. Since $(x^{\frac1s} -b_i)$ is reducible and not a product of irreducibles in $D$ so is $f(x)$.
If the indices are rationals then the above method works. What if the indices are any real numbers? For example the element $x^r + 1$ is reducible for every non zero $r$ since $x^r + 1 = (x^{\frac r2} - 1 )(x^{\frac r2} + 1 )$ and in this case also the above method works. Also if the indices are rational multiple of a fixed irrational number then the above method works. In general for $r_i \in \mathbb{R}$ if I can find a real $s$ such that $sr_i \in \mathbb{Q}$ for each $i$ then $f(x)$ is reducible.
If I take $g(x) = x^{\pi} + x^e + 1$. In this case, I do not know whether there exists a $s \in \mathbb{R}$ such that $s\pi$ and $se$ are rationals (If such $s$ exists then $\pi / e$ will be rational which is not known as of now). Does there exist any other method to factorize $g(x)$ in $D$? In general, do there exist methods to check an element is irreducible or not?
Update: I have asked an almost similar question on Mathoverflow and it has an answer also.