Can $\sin (x)$ be considered zero at the point at $\infty$ in some context?

It is understood that $$\lim_{x \to\infty} \sin (x)$$ is not defined, but I have tricked myself into thinking it might be identifiable as zero at the point at infinity after thinking about continuity, differentiability, and symmetry and then reading J.G.'s answer to why there is a factor of $$2\pi$$ in $$\delta(k)=\frac{1}{2\pi}\int_{-\infty}^\infty \exp\left(ikx\right)dx$$.

Let's start with J.G.'s answer to the $$\delta$$-function question.

$$$$\tag{1}\label{eq1} \int_{-\frac{1}{\epsilon}}^\frac{1}{\epsilon}e^{ikx}dx=\frac{1}{ik}\left(e^{ik/\epsilon}-e^{-ik/\epsilon}\right)=\frac{2}{k}\sin\left(\frac{k}{\epsilon}\right)$$$$

Assuming k is nonzero, the integral in \ref{eq1} is supposed to be zero in the $$\lim_{\epsilon\to 0^+}$$, otherwise the definition of Dirac delta would need to be changed (please correct me if I am wrong).

But would this not then imply that:

$$$$\tag{2}\label{eq2} \lim_{\epsilon \to 0^+,k\neq0} \frac{2}{k}\sin\left(\frac{k}{\epsilon}\right)\stackrel{?}{=}0$$$$

Now, I have also convinced myself that $$\sin(x)$$ might be zero at the point at infinity using ideas of continuity, differentiability, and symmetry.

$$\sin (x)$$ is odd in $$x$$ so $$\sin (x) = \sin(-x)\Rightarrow \sin(x)=0\Rightarrow x=n\pi$$ and so $$x$$ and $$-x$$ are separated by $$2n\pi$$ which is a whole number of wavelengths which means we can "wrap the function around and connect the two ends together" and the shape would be continuous and differentiable. The idea of a function being continuous and differentiable at the point at infinity makes so much sense.

So I am wondering if there is any context where $$\sin(x)$$ can be identified as zero at the point at infinity.

Now, though I did not mention $$\cos(x)$$ in the question, I would appreciate commentary on that as well.

Since $$\cos(x)=\sin\left(x-\frac{\pi}{2}\right)$$, one would think that if all the above ideas about $$\sin(x)$$ were true, then $$\cos(x)$$ would be $$\pm 1$$ at the point at $$\infty$$ ( you could also use continuity, differentiability, and symmetry for $$\cos$$ on its own ). Either way (plus or minus) the area under the curve is zero and the idea that there should/should not be be a whole number of wavelengths between zero and $$\infty$$ does not seem relevant enough to decide between plus/minus 1. So if anyone finds a trick which supports $$\left|\lim_{x\to\infty}\cos(x)\right|=1$$ and/or suggests the sign should be one way or another, that would be much appreciated.

• I don't think so. It's already bad enough that the limit doesn't exist when working over the real numbers, but working over the complex numbers makes things even worse; $\sin z$ as a holomorphic function has an essential singularity at infinity, which is reflected in its much worse behavior if you approach infinity from complex directions. Commented Jun 1 at 21:54
• Focusing only on $~x \in \Bbb{R},~$ suppose that I specify $~\epsilon = (1/2).~$ Can you identify any specific positive number $~X_0,~$ such that for all $~x \geq X_0,~$ you have that $~| ~\sin(x) - 0 ~| < \epsilon ~?$ Commented Jun 1 at 23:20
• "The idea of a function being continuous and differentiable at the point at infinity makes so much sense." Perhaps in some contexts, but I don't understand your justification for this part. Commented Jun 2 at 1:19
• A nascent delta needs to be of the form $\tfrac{1}{\epsilon}f(\tfrac{k}{\epsilon})$ with $\int_{-\infty}^\infty f(t)dt=1$. In this case, $f(t):=\tfrac{\sin t}{\pi t}$. It's $f(t)$, not $tf(t)$, which $\to0$ as $t\to\infty$, as happens when $\epsilon\to0^+$ for fixed $k>0$. Note, however, that the sinc function only $\to0$ as $t\to\infty$ in the real case, not the complex one. which is irrelevant to nascent deltas.
– J.G.
Commented Jun 2 at 12:51

I've posted about the following papers several times, originally (I think) in this 14 January 2009 sci.math post and later in two or three Stack Exchange comments, but probably this wasn't visible enough to gain any traction with those who ask questions such as yours, which I've seen occasionally here. Although these papers don't really answer your question, since they deal with mathematicians' attempts to make sense of certain manipulations before current convergence meanings had widespread acceptance in mathematics (at least in the case of [1]; probably not for [2] and [3]) and before various theories of divergent series were put on a more mathematically rigorous basis, I'm using my answer as a way of making these papers more visible for those who might be interested in them for historical or other reasons.

[1] Samuel Earnshaw (1805-1888), On the values of the sine and cosine of an infinite angle, Transactions of the Cambridge Philosophical Society 8 (1849), pp. 255-268.

At Google Books and Internet Archive. In the same volume see also Young's paper on pp. 429-440 (especially 2nd footnote on p. 434) and De Morgan's paper on pp. 182-192.

[2] James Whitbread Lee Glaisher (1848-1928), On $$\sin \infty$$ and $$\cos \infty,$$ Messenger of Mathematics 5 (1871), 232-244.
[paper dated 31 March 1870]

[3] William Walton (1813-1901), Note on $$\sin \infty$$ and $$\cos \infty,$$ Quarterly Journal of Pure and Applied Mathematics 11 (1871), pp. 326-327.
[paper dated 25 July 1871]

To address your first example, the reason this does not work comes down to the fact that the Dirac delta is not a function which is $$\infty$$ at zero and zero elsewhere, and the precise nature of the convergence involved. The key property of the dirac delta is that for any continuous $$f$$,

$$\int_{\mathbb{R}} f(x)\delta(x) dx = f(0)$$

Technically this statement still doesn't make sense, because the dirac delta is actually a measure (or a distribution, depending on how you want to think about it). Really what we're doing is integrating against the measure $$\delta dx$$, which assigns a measure of $$1$$ to any set containing zero and a measure of zero to any set not containing zero. To make this more precise you'll need the machinery of Lebesgue integration.

The second key issue is the type of convergence involved. The functions $$g_\epsilon(k) = 2\sin(k/\epsilon)/k$$ do not converge pointwise to a dirac delta, they converge in the sense of measures (indeed, since the dirac delta is not a function, it doesn't make sense to think of pointwise convergence to it). This means that for every continuous $$f$$, we should have $$\lim_{\epsilon\to 0^+} \int_{\mathbb{R}}f(k)g_{\epsilon}(k)dk = f(0)$$

You can think of this like convergence in an averaged sense. The oscillations in $$g_\epsilon$$ cancel out the contributions away from $$k=0$$, even though the pointwise limit of the $$g_{\epsilon}$$ is not a dirac delta.

Notably, pointwise convergence to something which is $$\infty$$ at zero and zero elsewhere doesn't guarantee that the limiting object is a dirac delta, (e.g. $$g_n(x) = n$$ at $$x=0$$ and $$0$$ otherwise; integrating this against any continuous $$f$$ yields zero).

One related (and possibly cleaner) way you can make sense of this idea is through a related type of averaged convergence: the weak convergence of $$\sin(nx)$$ to zero in $$L^2([0,2\pi])$$ (note the similarity to the $$g_{\epsilon}$$ above).

To say that a sequence of functions $$g_n(x)$$ converges weakly in $$L^2([0,2\pi])$$ to another function $$g(x)$$ means that for all $$f\in L^2([0,2\pi])$$, we have

$$\lim_{n\to\infty} \int_0^{2\pi} f(x)g_n(x)dx = \int_0^{2\pi} f(x)g(x) dx$$

Which is essentially the same statement as the statement of convergence in the sense of measures, just with a different domain and set of $$f$$. Now, by the Riemann-Lebesgue lemma (which is a good exercise in measure theory), one can show that $$\sin(nx)$$ converges weakly to zero. The spirit is the same: even though we don't have pointwise convergence to zero, on average, the oscillations dominate and cancel out any contribution from $$f$$ upon integration. To gain some intuition, try proving the Riemann-Lebesgue lemma for $$f$$ which equals $$1$$ in some interval $$[a,b]\subset[0,2\pi]$$ and $$0$$ outside that interval.

• Thank you. I need more help using Riemann-Lebesgue Lemma (RLL). I follow the Wikipedia page for RLL (and use their definition of Fourier Transform (FT)). I get that sin(x) is the FT of $g=\frac{\delta(k+1)-\delta(k-1)}{4\pi i}$. Now $\int|g|dx<\infty$, but $\delta$ is not $L^1$ integrable [math.stackexchange.com/questions/522603/…. Also, the FT of the "unit box" function you described is identically zero and I'm wondering if this is a clever hinting that measurable functions can be built from many unit boxes which each have width $2\pi$. Commented Jun 3 at 23:48
• $g$ is not a function so it doesn’t make sense to write $\int |g|dx$! Also, the FT of the indicator function of an interval is not identically zero, but yes, this is hinting at the way to prove the RLL (by approximating measurable functions by simple functions). Commented Jun 4 at 0:12