Proving Euler's spiral is an isometric embedding with bounded image

Question

$\newcommand{\al}{\alpha}$

I am trying to prove Euler's spiral is an isometric embedding of $\mathbb{R}$ into $\mathbb{R}^2$ with bounded image. Here is the definition of the spiral:

$(*) \, \,\al(t)= (\int_0^{t} \cos(s^2)ds,\int_0^{t} \sin(s^2)ds)$

$\al$ is clearly a smooth isometric immersion.

Questions:

(1) Why is $\al$ injective? (There is a picture, but I want a proof...)

Edit: Christian Blatter (see answer below) gave a proof based on Kneser's Nesting Theorem. The theorem also implies the boundedness of $\operatorname{Image}(\al)$. I still wonder whether there is a "direct" proof based only on the formula $(*)$.

Perhaps one can prove that the distance of $\al(t)$ from the (suitable) limit point decreases with $|t|$? (A naive attempt to do so didn't work).

(2) Why is $\al$ an embedding? (i.e homeomrphism onto its image)

Intuitively, there are "no troubles" from a topological point of view, since the path approaches its accumulation points only "from one direction" (Compare for the "figure-eight" in contrast).

(3) Why is the image bounded? That is, how to show the integrals are bounded?

Edit: This follows from Dirichlet's test for improper integrals. (See levap's answer)

Any reference (or self-contained proof of course) would be welcome. I have tried googling for some time but in vain.

Answer 1

Regarding your third question, set $x(t) = \int_0^t \cos(s^2) \, ds$. The function $x$ is continuous so to show that $x$ is bounded it is enough to show that $\lim_{x \to \pm \infty} x(t) = \pm \int_0^{\infty} \cos(s^2) \, ds$ exists. We have

$$ \int_0^{\infty} \cos(s^2) \, ds = \int_0^1 \cos(s^2) \, ds + \int_1^{\infty} \cos(s^2) \, ds = \int_0^1 \cos(s^2) \, ds + \int_1^{\infty} \frac{\cos(u)}{2\sqrt{u}} \, du$$

and the latter integral converges using Dirichlet's test for improper integrals. Similarly for $y(t)$.

Now, I'm not sure how to easily answer your first and second question. However, if your purpose is to construct more or less explicitly an isometric embedding of $\mathbb{R}$ into $\mathbb{R}^2$ with a bounded image, I can suggest the following way which I find more direct:

Consider the map $\pi \colon \mathbb{R} \times \mathbb{R}_{+} \rightarrow \mathbb{R}^2$ given by $\pi(\theta, r) = (r \cos \theta, r \sin \theta)$. This map is a local diffeomorphism and in particular an immersion. Now, choose any bounded monotone increasing function $r \colon (-\infty, \infty) \rightarrow \mathbb{R}_{+}$ and consider the curve $\gamma(t) = \pi(t, r(t))$. The curve $\gamma$ is injective (since $r$ is injective), an immersion (since $\pi$ is a local diffeomorphism) and an embedding (since $$ \gamma((t_0 - \delta, t_0 + \delta)) = \gamma(\mathbb{R}) \cap \pi((t_0 - \delta, t_0 + \delta) \times (r(t_0 - \delta), r(t_0 + \delta))) $$ and the right hand side is open as $\pi$ is open). If $|r(t)| \leq M$ then the image of $\gamma$ lies inside the ball of radius $M$ inside $\mathbb{R}^2$. Now take a unit-length reparametrization of $\gamma$ and you obtain an isometric embedding.

For example, if $r(t) = \arctan(t) + \frac{\pi}{2}$ then the curve $\gamma$ looks like a two-sided spiral that spirals toward $(0,0)$ as $t \to -\infty$ and toward a limit circle of radius $\pi$ centered around $(0,0)$ as $t \to \infty$ (drew using wolfram alpha):

Answer 2

From $$\alpha(t)=\int_0^t e^{is^2}\>ds$$ it follows that $\dot \alpha(t)=e^{it^2}$ has absolute value $1$, hence $\alpha$ is a locally isometric immersion. From ${\rm arg}\bigl(\dot\alpha(t)\bigr)=t^2$ it then follows that the curvature $$\kappa(t)={d\over dt}{\rm arg}\bigl(\dot\alpha(t)\bigr)=2t$$ is strictly monotonically increasing for $-\infty<t<\infty$. Kneser's Nesting Theorem then implies (i) that all points $\alpha(t)$ with $t>1$ are contained in the oscillating circle at $\alpha(1)$, and (ii) that the map $\alpha(\cdot)$ is globally injective.