Yeah I wanted to keep the question within the bounds of VCE terminology so I tried just doing the R^3 case, but now I see that was a mistake lol.
Just gonna prove the convergence because ceebs looking up how to solve that integral right now.
\[ x = \sqrt{u} \implies dx = \frac1{2\sqrt{u}}\,du.\\ u = \frac{1}{s} \implies ds = -\frac{1}{s^2}\,ds.\\ \begin{align*} \int_0^\infty \cos (x^2)\,dx &= \int_0^\infty\frac{1}{2}u^{-1/2}\cos (u)\,du\\ &= \int_0^1 \frac12 u^{-1/2} \cos (u^2)\,du + \int_1^\infty \frac12 u^{-1/2}\cos u\,du \\ &= \int_1^\infty \frac12 s^{1/2} \cos \left( \frac{1}{s^2} \right) s^{-2}\,ds +\int_1^\infty \frac12 u^{-1/2}\cos u\,du \end{align*} \]
\[ \text{For the second integral}\\ \begin{align*} \int_1^\infty u^{-1/2} \cos u\,du &= u^{-1/2} \sin u \big|_1^\infty + \frac12 \int_1^\infty u^{-3/2} \sin u\,du. \end{align*}\\ \text{The former is finite and equals }\sin 1\\ \text{and the latter converges via comparison with the }p\text{-integral }\int_1^\infty u^{-3/2}\,du. \]
Squeeze theorem limit used
\[ 0 = \lim_{u\to \infty} -u^{-1/2} \leq \lim_{u\to \infty} u^{-1/2}\sin u \leq \lim_{u\to \infty} u^{-1/2} = 0\\ \implies \lim_{u\to \infty} u^{-1/2}\sin u = 0 \]
The first integral converges using the same comparison. So puzzling everything together the whole thing converges.
Was thinking I might get baited because the answer seemed a bit too clean, but here is what I got anyway. 
It's harder for the VCE students because they don't have that nice formula that we do. In the HSC your approach would be the intended approach
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Continuing,
\begin{align*}e^t &= \frac{\tan \frac{y}{2}}{\tan \frac{\alpha}{2}} \\ y &= 2 \arctan \left( e^t \tan \frac{\alpha}{2} \right) + 2m\pi \end{align*}
where \(m \in \mathbb{Z}\).
\[ \text{As }\alpha\text{ is not an integer multiple of }\pi,\\ \tan \frac{\alpha}{2} \text{ is well defined and not equal to 0.}\\ \text{If }\tan \alpha > 0,\, 2\arctan \left( e^t \tan \frac{\alpha}{2} \right) \to 2\times \frac\pi2 = \pi\text{ and we're done.}\\ \text{Otherwise it just approaches }-\frac\pi2\text{ instead, but then sub }m=n+1\text{ and we're also done.} \]
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Next question:
High school:
\[ \text{1. Prove that }\sum_{k=1}^n k^2 = \frac16 n (n+1)(2n+1)\\ \text{2. Now derive it from scratch.} \]
First year/Second year university:
\[ \text{Show that in general, }\lim_{x\to a} \lim_{y\to b} f(x,y) \neq \lim_{y\to b} \lim_{x\to a} f(x,y)\\ \text{where }a, b \in \mathbb{R} \cup \{\infty\} \]
Try to construct a routine counterexample, and also a creative counterexample.
Beyond:
Label the vertices of \(K_5\) 1, 2, 3, 4 and 5, where \(K_n\) is the complete graph on \(n\) vertices. Suppose that two Euler circuits are the same if their sequences of edges are up to the same rotational symmetry. (For example 1-2-3-4-5-1-3-5-2-4-1 would be the same as 2-3-4-5-1-3-5-2-4-1-2).
1. Explain why \(K_3\) has only two distinct Euler circuits (easy warm-up)
2. Carefully prove that \(K_5\) has
264 distinct Euler circuits.
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