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Is there anything meaningful one can say about the limit

$$\lim_{x\to\infty}e^{i a x}$$

for $a\in\mathbb C$ ? Thanks for any suggestion!

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    You can rewrite that by Eulers formula so you get $cos(ax) +isin(ax)$ which doesn't converge unless the real part of $a$ is zero. – 2017-02-28
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    Depends. What if $a=i$? What if $a=-i$? What if $\Im a < 0$? $>0$? $=0$? – 2017-02-28
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    @lordoftheshadows you're assuming $a$ is real; the OP is not. – 2017-02-28
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    Thanks for the suggestions so far! I am mainly interested in the case of generic $a$, not assuming any special cases. Is there any meaning in the infinite traversal of a circle? Maybe in the sense that taking an average of all these points one gets the limit $0$? Does this make any sense? – 2017-02-28
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    If $a$ has real part $0$ then it's boring. If it has imaginary part $0$ it's also boring. If it has both then $a = m + ni$ so we get $e^{mix} * e^{-nx}$. Since $m$ is a real number we get that $e^{mix}$ is bounded so we only care about $e^{-nx}$ which either goes to $1$ in which case it doesn't converge. It goes to $0$ in which case the whole thing goes to $0$ or it diverges and the whole thing diverges. – 2017-02-28
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    You are right, only if $a\in\mathbb R$ we get a circle, but why do you say its boring? In other cases it definitely either diverges or vanishes, which suggests those other cases are boring (meaning, easily figured out). – 2017-02-28
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    @Kagaratsch somewhat but that wouldn't be a limit. If we look at the limit it becomes undefined as $x \to \infty$ for two reasons 1) both $\cos$ and $\sin$ lie in the interval [0,1] so it would be impossible to pinpoint one specific number. 2) it oscillates period just like $\lim_{x\to 0}\sin (1/x)$ is undefined at 0. The same can be said about your situation. – 2017-02-28

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If we interpret the limit classically, then the limit exists if and only if $\text{Im}(a)<0$.

However, note that for any function $\phi(a)$ that is $L^1$ and with compact support, the Riemann-Lebesgue Lemma guarantees that

$$\lim_{x\to \infty}\int_{-\infty}^\infty \phi(a)e^{iax}\,da=0$$

Hence, for $a\in \mathbb{R}$, $\lim_{x\to \infty}e^{iax}\sim 0$ as a distribution.

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    Thank you! This is exactly what I was looking for. – 2017-03-01
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If we let $a = \alpha + \beta i$, then \begin{align*} e^{iax} &= e^{i(\alpha + \beta i)x} \\ &= e^{\alpha xi - \beta x} \\ &= e^{-\beta x} e^{\alpha x i} \end{align*} where the first factor $e^{-\beta x}$ is the modulus and scales the second factor $e^{\alpha x i}$ (the argument) which denotes a complex number on the unit circle that revolves around the origin as $x$ is increased with the rotational speed and direction determined by the value of $\alpha$ (positive means it revolves counter-clockwise, negative means it revolves clockwise, zero means it stays fixed at the real number $1$).

Hence if $\beta > 0$, we have the modulus decaying to zero as $x$ increases, and so $e^{iax}$ spirals toward zero, and we have limit $0$.

If $\beta < 0$, we have a positive coefficient for $x$ in $e^{-\beta x}$ and so the modulus grows without bound and we get $e^{iax}$ spiraling out to infinity.

If $\beta = 0$, we have the modulus fixed at $1$ and so $e^{iax}$ forever orbits the origin at unit distance and hence the limit is undefined (unless $\alpha = 0$ also, in which case $e^{iax}$ is trivially equal to $1$ for all $x$).

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    +1. I would clarify the $\beta < 0$ case though. If $\alpha=0$, for instance, the limit is $+\infty$. However, if $\alpha \neq 0$, there is no limit, due to the spiraling behavior you mentioned. – 2017-02-28
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    I was thinking of the extended complex plane that includes the point at infinity, in which case I think the $\beta < 0$ case has that point as a well-defined limit. But if we don't include the point at infinity, then you're right: $e^{iax}$ has no limit except if $\alpha = 0$. But I can change it if this interpretation is unhelpful. – 2017-02-28
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    even under that definition it does not converge for nonzero $\alpha$. – 2017-02-28
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    I did not mean that it converged. I only meant it in the same way it is sometimes treated in real analysis when one might write $\lim_{x \to \infty} e^x = +\infty$. Likewise, I could write $\lim_{x \to \infty} e^{iax} = \infty$ where $\infty$ denotes the complex point at infinity, and this would be true regardless of the value of $\alpha$. – 2017-02-28
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    The notion of "the" complex point at infinity just isn't valid. There isn't just one, after all. $e^x$ approaches $+\infty$, and $-e^x$ approaches $-\infty$---two different "points". Similarly, for fixed $y$, $e^{x+iy}$ tends towards infinity in a particular direction. But we cannot say something similar for the $\alpha\neq 0$ case above, because of the spiral nature of the trajectory. So there really is a material difference between the $\alpha=0$ case and the $\alpha\neq 0$ case. – 2017-02-28
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    I wouldn't say the notion isn't _valid_, it is one way to treat infinite limits (see [Riemann sphere](https://en.wikipedia.org/wiki/Riemann_sphere) and [Projectively extended real line](https://en.wikipedia.org/wiki/Projectively_extended_real_line)). "The" point at infinity is just meant to describe what happens as numbers grow large in absolute value. However, it is true this treatment has its drawbacks like you mention. You do lose directional information. But the reason I used this notion of infinity in this context is because direction seemed unimportant because of the spiraling behavior. – 2017-02-28
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    Well, i'll be darned, in some contexts people do refer to a single complex "point at infinity". I do maintain the distinction I've made, however. After all, in those contexts, the real positive and negative infinities are collapsed as well: https://en.m.wikipedia.org/wiki/Point_at_infinity – 2017-02-28