TITLE: Infinite Product $\prod\limits_{k=1}^\infty\left({1-\frac{x^2}{k^2\pi^2}}\right)$
QUESTION [25 upvotes]: I've been looking at proofs of Euler's sine expansion, that is 
$$\frac{\sin x}{x}=\prod_{k=1}^\infty\left({1-\frac{x^2}{k^2\pi^2}}\right)$$
All the proofs seem to rely on complex analysis and Fourier series. 
Is there any more elementary proof?

REPLY [16 votes]: Here's a proof using elementary (albeit not very well known)
trigonometric integrals.
We may assume $0 < x \leq 1/2$ because the ratio of $\sin(\pi x) / (\pi x)$
to $\prod_{k=1}^\infty \big(1 - (x/k)^2\bigl)$ is readily seen to be
an even function of period $1$.
For $n \geq 0$ and $c \in {\bf R}$ define
$$
I_n(c) := \int_0^{\pi/2} \cos^n t \, \cos ct \, dt.
$$
Then $I_0(0) = \pi/2$ and $I_0(2x) = \sin(\pi x) / (2x)$.
Hence
$$
\frac{I_0(2x)}{I_0(0)} = \frac{\sin \pi x}{\pi x}.
$$
I claim that also
$$
\frac{I_0(2x)}{I_0(0)} = \prod_{k=1}^\infty \left( 1 - \frac{x^2}{k^2} \right),
$$
which will prove the desired product formula.
The key is a recursion for the ratios $I_n(2x) / I_n(0)$
obtained from the following integration by parts:
Lemma. For $n \geq 2$,
$$
(n^2 - c^2) I_n(c) = (n^2-n) I_{n-2}(c).
$$
Proof: Integrate by parts twice to find
$$
c^2 I_n(c) = c \int_0^{\pi/2} \cos^n t \, d(\sin ct)
= n c \int_0^{\pi/2} \cos^{n-1} t \, \sin t \, \sin ct \, dt
$$ $$
= -n \int_0^{\pi/2} \cos^{n-1} t \, \sin t \, d(\cos ct)
= n \int_0^{\pi/2} (\cos^n t - (n-1) \cos^{n-2} t \sin^2 t) \, \cos ct \, dt.
$$
But $\sin^2 t = 1 - \cos^2 t$, so
$$
c^2 I_n(c) = n \int_0^{\pi/2} (n \cos^n t - (n-1) \cos^{n-2} t) \, \cos ct \, dt
= n^2 I_n(c) - (n^2-n) I_{n-2}(c),
$$
and the recursion follows. $\Box$
Therefore
$$
\frac{I_{n-2}(c)}{I_{n-2}(0)} = \frac{n^2-c^2}{n^2} \frac{I_n(c)}{I_n(0)},
$$
whence by induction on $m=1,2,3,\ldots$ 
$$
\frac{I_0(c)}{I_0(0)} =
\prod_{k=1}^m \frac{(2k)^2-c^2}{(2k)^2} \cdot \frac{I_{2m}(c)}{I_{2m}(0)}.
$$
Taking $c=2x$ we find
$$
\frac{\sin \pi x}{\pi x} =
\prod_{k=1}^m \frac{k^2-x^2}{k^2} \cdot \frac{I_{2m}(2x)}{I_{2m}(0)}
= \prod_{k=1}^m \left( 1 - \frac{x^2}{k^2} \right) \cdot \frac{I_{2m}(2x)}{I_{2m}(0)},
$$
and it remains to prove that
$I_n(2x) / I_n(0) \rightarrow 1$ as $n \rightarrow \infty$.
But this follows from the inequalities
$$
I_n(0) > I_n(2x) > I_{n+2}(0)
$$
(the second inequality is where we use $x \leq 1/2$: we need
$\cos 2xt \leq \cos^2 x$), since by our Lemma
$I_{n+2}(0) / I_n(0) = (n+1)/(n+2)$, which approaches $1$ as
$n \rightarrow \infty$.  Taking $m \rightarrow \infty$ we deduce
$$
\frac{\sin \pi x}{\pi x} = \prod_{k=1}^\infty \left( 1 - \frac{x^2}{k^2} \right),
$$
QED.
I noticed this argument some years back (see for instance #8 of
this
problem set from 2000).  Naturally the formula for integrals $I_n(c)$
is known (even for non-integer $n$, see integral 3.631#9 in
Gradshteyn and Ryzhik), but I don't know if and where it was
previously noticed that one can use these integrals to prove
the product formula for the sine.