I have a function $f(x)$ which is defined as the solution to a certain differential equation. The boundary conditions are that in the $x\rightarrow \infty$ limit, it should be asymptotically equivalent to $\exp(-x)$, i.e. $$\lim_{x\rightarrow \infty} \frac{f(x)}{\exp(-x)}=c_{\infty}$$ where $c_{\infty}$ is some finite constant (not $\infty$, not 0).
In the $x\rightarrow -\infty$ limit, it should be asymptotically equivalent to a polynomial, say $x^n$: $$\lim_{x\rightarrow -\infty} \frac{f(x)}{x^n}=c_{-\infty}$$ where $c_{-\infty}$ is some finite constant (not $\infty$, not 0).
I hoped to approximate this function with elementary functions, or at least with fairly well-known special functions. So, I thought, which functions behave exponential-like in one limit but polynomial-like in the other? The sigmoid function $\frac{1}{1+\exp(-x)}$ is an example, it approaches $\exp(x)$ in one limit and $1$ in the other limit. By repeated integration of the sigmoid function, we can find a function that behaves like $\exp(x)$ in one limit, and like $x^n$ in the other.
This works fine, but, alas, repeatedly integrating $\frac{1}{1+\exp(-x)}$ gives rise to Polylogarithms, which are still somewhat more complicated than I would have hoped.
So: are there functions that can be expressed in terms of elementary functions, that behave like $\exp(-x)$ for very large positive $x$ and like $x^n$ for very large negative $x$?
Edit: Repeated integration of any sigmoid-like function (like tanh, arctan or the error function) would work, though I would be surprised if elementary sigmoid-like functions existed whose integrals are simpler than those of $\frac{1}{1+\exp(-x)}$. Multiplying a sigmoid with a polynomial does not work - doing that makes at least one of the limits diverge.
If you want $c_{\infty}, c_{-\infty}$ both in $(0,\infty)$, I suggest $$ f(x) = \frac{x^n}{a+x^ne^x} $$ with $a$ chosen large enough to avoid dividing by zero in case of odd $n$.
Then for $x\to\infty$ we have $$ \frac{f(x)}{e^{-x}} = \frac{x^n}{ae^{-x}+x^n} \to 1 $$ and for $x\to\infty$ $$ \frac{f(x)}{x^n} = \frac{1}{a+x^ne^x} \to \frac1a $$ since $x^ne^x\to 0$ for $x\to-\infty$.