# Are polylogarithms the simplest functions that decay exponentially in one limit, and grow polynomially in another limit?

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.

## 1 Answer

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 $x\to-\infty$.

• Thanks, but perhaps I failed to be clear enough. The functions should "asymptotically behave like" an exponential in one limit and a polynomial in the other. They should be asymptotically equivalent. $c_{\infty}=$ and $c_{-\infty}=$ should be neither infinity nor zero. – Wouter Apr 7 '16 at 15:39
• @Wouter: Edited with a new suggestion, then. – Henning Makholm Apr 7 '16 at 15:42