Asymptotic behavior of entire functions Which entire function $f\left(x\right)$ goes asymptotically to $\dfrac{e^{-x}}{x}$ as $x$ goes to infinity with $x$ positive? That is, $\left(e^{-x}/x \right)/f \left(x \right) \rightarrow 1$.
 A: The reason $\frac{e^{-x}}{x}$ is not entire is that it has a pole at $0$. Note, however, that away from $0$, the function $x$ has no zeros, so you can divide $e^{-x}$ by $x$ and get a holomorphic function. Close to $0$, you can express $e^{-x} = 1 + xf(x)$ where $f(x) = \frac{ e^{-x} - 1}{x}$ is entire (just write out the series expansion of $e^{-x}$). You thus find $e^{-x} = \frac{1}{x} + f(x)$, where $\frac{1}{x}$ goes to $0$ as $x$ goes to infinity, and $f(x)$ is your sought entire function.
A: There are infinitely many entire functions which are asymptotic to $e^{-x}/x$ as $x \to \infty$ with $x > 0$.
Pick any finite $a > 0$ and any $\varphi \,\colon [0,a] \to \mathbb{C}\cup\{\infty\}$ such that


*

*$\varphi(0)$ is finite and nonzero,

*$\varphi$ is differentiable in a neighborhood of $0$, and

*$\int_0^a|\varphi(t)|\,dt < \infty$.


Then the function
$$
f(z) = e^{-z} \int_0^a \varphi(t) e^{-zt}\,dt
$$
is entire by Morera's theorem and
$$
f(z) \sim e^{-z}/z
$$
as $z \to \infty$ with $|\arg z| \leq \pi/2 - \delta$ for any fixed $\delta > 0$ by Watson's lemma.
I haven't the faintest idea for how to compile a list of all functions with the desired asymptotic character, though.
