Wallis Formula

i) Show that for any $n\in \Bbb N$, $\int_0^{\pi/2}\sin^n x\;dx=\dfrac{n-1}{n}\int_0^{\pi/2}\sin^{n-2} x\;dx$.
ii) Show that for any $k\in \Bbb N$, $$\int_0^{\pi/2}\sin^{2k} x\;dx=\dfrac{(2k-1)(2k-3)\cdots 3\cdot 1}{(2k)(2k-2)\cdots 4\cdot 2}\dfrac{\pi}{2}=\dfrac{(2k)!}{(2^k k!)^2}\dfrac{\pi}{2}$$ and $$\int_0^{\pi/2}\sin^{2k+1} x\;dx=\dfrac{2k(2k-2)\cdots 4\cdot 2}{(2k+1)(2k-1)\cdots 3\cdot 1}=\dfrac{(2^k k!)^2}{(2k+1)!}.$$ iii) For $k\in \Bbb N$, let $$\mu_k=\dfrac{\int_0^{\pi/2}\sin^{2k} x\;dx}{\int_0^{\pi/2}\sin^{2k+1} x\;dx}.$$ Show that $1\leq \mu_k\leq \dfrac{2k+1}{2k}$ for each $k\in \Bbb N$ and consequently that $\mu_k\to 1$ as $k\to \infty$. Deduce that $$\sqrt{\pi}=\lim_{k\to \infty} \dfrac{(k!)^2 2^{2k}}{(2k)!\sqrt{k}}.$$ Thus $\pi\sim \dfrac{(k!)^4 2^{4k}}{((2k)!)^2 k}$. (Hint: $\sin^{2k+1}x\leq \sin^{2k}x\leq \sin^{2k-1}x$ for all $x\in[0,\pi/2]$.)

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i) $$\begin{align}\int_0^{\pi/2}\sin^n x\;dx&=-\int_0^{\pi/2}\sin^{n-1} x\;d(\cos x)\\
&=-\sin^{n-1}\cos x\bigg|_0^{\pi/2}+(n-1)\int_0^{\pi/2}\cos^2 x\sin^{n-2} x\;dx\\
&=0+(n-1)\int_0^{\pi/2}\sin^{n-2} x\;dx-(n-1)\int_0^{\pi/2}\sin^n x\;dx\\
\Rightarrow \int_0^{\pi/2}\sin^n x\;dx&=\dfrac{n-1}{n}\int_0^{\pi/2}\sin^{n-2} x\;dx
\end{align}$$ ii) $$\begin{align} \int_0^{\pi/2}\sin^{2k} x\;dx&=\dfrac{2k-1}{2k}\int_0^{\pi/2}\sin^{2k-2} x\;dx\\
&=\dfrac{2k-1}{2k}\dfrac{2k-3}{2k}\int_0^{\pi/2}\sin^{2k-4} x\;dx\\
&=\dfrac{2k-1}{2k}\dfrac{2k-3}{2k}\cdots \int_0^{\pi/2}\sin^4 x\;dx\\
&=\dfrac{2k-1}{2k}\dfrac{2k-3}{2k}\cdots \dfrac{3}{4} \int_0^{\pi/2}\sin^2 x\;dx\\
&=\dfrac{2k-1}{2k}\dfrac{2k-3}{2k}\cdots \dfrac{3}{4} \dfrac{1}{2}\int_0^{\pi/2}\sin^0 x\;dx\\\\
&=\dfrac{(2k-1)(2k-3)\cdots 3\cdot 1}{(2k)(2k-2)\cdots 4\cdot 2}\dfrac{\pi}{2}\\
&=\dfrac{(2k)(2k-1)(2k-2)(2k-3)\cdots 3\cdot 2\cdot 1}{[(2k)(2k-2)\cdots 4\cdot 2]^2}\dfrac{\pi}{2}\\
&=\dfrac{(2k)!}{(2^k\cdot [k(k-1)\cdots 2\cdot 1])^2}\dfrac{\pi}{2}\\
&=\dfrac{(2k)!}{(2^k k!)^2}\dfrac{\pi}{2}
\end{align}$$ The other case is similar.
Question: How come the integral of 'odd sine function' is the 'reciprocal' of that of 'even sine function'?

iii) It is simple to prove the claim with the hint (which is true because $\sin x\leq 1$ when $x\in [0,\pi/2]$). For all $x\in[0,\pi/2]$, $$\sin^{2k+1}x\leq \sin^{2k}x\leq \sin^{2k-1}x\\  \int_0^{\pi/2}\sin^{2k+1}x\;dx\leq \int_0^{\pi/2}\sin^{2k}x\;dx\leq \int_0^{\pi/2}\sin^{2k-1}x\;dx\\ \dfrac{\int_0^{\pi/2}\sin^{2k+1}x\;dx}{\int_0^{\pi/2}\sin^{2k+1}x\;dx}\leq \dfrac{\int_0^{\pi/2}\sin^{2k}x\;dx}{\int_0^{\pi/2}\sin^{2k+1}x\;dx}\leq \dfrac{\int_0^{\pi/2}\sin^{2k-1}x\;dx}{\int_0^{\pi/2}\sin^{2k+1}x\;dx}\\ 1\leq \mu_k\leq \dfrac{\int_0^{\pi/2}\sin^{2k-1}x\;dx}{\int_0^{\pi/2}\sin^{2k+1}x\;dx}$$ We have $\dfrac{\int_0^{\pi/2}\sin^{2k-1}x\;dx}{\int_0^{\pi/2}\sin^{2k+1}x\;dx}=\dfrac{[2^{k-1}(k-1)!]^2}{(2k-1)!}\dfrac{(2k+1)!}{(2^k k!)^2}=\dfrac{(2k+1)2k}{4k^2}=\dfrac{2k+1}{2k}$, hence the claim.

By sandwich theorem, since $\lim_{k\to \infty} \dfrac{2k+1}{2k}=1$, we have $\mu_k\to 1$ as $k\to \infty$. But $\mu_k\to 1$ implies $\sqrt{\mu_k}\to 1$.
Question: Why does $\mu_k$ tend to $1$ intuitively?
We can look at the graphs of powers of $\sin x$. When the powers are close to each other, their graphs are very similar. That's why the ratio of the two integrals tends to $1$.

Now, $$\mu_k=\dfrac{\pi}{2}\dfrac{(2k)!(2k+1)!}{(2^k k!)^4}\\ \sqrt{\mu_k}=\dfrac{\sqrt{\pi}}{\sqrt{2}}\dfrac{\sqrt{(2k)!(2k+1)!}}{(2^k k!)^2}=\sqrt{\pi}\dfrac{\sqrt{2k+1}}{\sqrt{2}}\dfrac{(2k)!}{(2^k k!)^2} \to 1.$$ It follows that $$\sqrt{\pi}=\lim_{k\to \infty} \dfrac{(2^k k!)^2}{(2k)!\sqrt{k}}$$ For the approximation of $\pi$, just take the square of the limit.

Wallis Formula is one of the many formulas that approximates $\pi$. There are other proofs of this formula. See the notes by Steven R. Dunbar.

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