Which entire function $f\left(x\right)$ goes asymptotically to $\dfrac{e^{-x}}{x}$ as $x$ goes to infinity with $x$ positive? That is, $\left(e^{-x}/x \right)/f \left(x \right) \rightarrow 1$.
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2Hint: What's wrong with $e^{-x}/x$ by itself? Can you think of a way to subtract out that problem without affecting the behaviour at infinity? – Erick Wong Feb 17 '13 at 15:50
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I think that $e^ \left(-x \right)/x$ is not entire – spirtalio34 Feb 17 '13 at 16:00
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Can you be more specific about why it isn't entire? What $x$ is it not differentiable at? – Erick Wong Feb 17 '13 at 16:07
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Is it differentiable at $x = 0$? – spirtalio34 Feb 17 '13 at 16:13
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1What is $x$? Is it complex? Real? When you say "as $x$ goes to infinity" do you mean "as $|x|$ goes to infinity" or are you asking about a particular direction (e.g. as $x \to \infty$ with x < 0)? – Antonio Vargas Feb 17 '13 at 17:01
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I would like to know what happens when $x$ is complex. – spirtalio34 Feb 18 '13 at 08:49
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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.
Jakub Konieczny
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1As $x \to \infty$ with $x > 0$ your function $f(x)$ is asymptotic to $-1/x$, not $e^{-x}/x$. – Antonio Vargas Feb 17 '13 at 16:59
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2Fair point. I assumed the author of the question wanted $e^{-x}/x - f(x) \to 0$ as $x \to \infty$, and that's what this function does. Also, it is unique with this property. I guess it also makes sense to ask for $(e^{-x}/x)/f(x) \to 1$. – Jakub Konieczny Feb 17 '13 at 17:17
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1Ah, I see how you were interpreting it now. That makes sense too. – Antonio Vargas Feb 17 '13 at 17:21
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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.
Antonio Vargas
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