Proof of Euler's Representation of the Sinc Function
Definition
Unnormalized Sinc Function
The following function $\sinc : \mathbb{R} \to \mathbb{R}$ is called the sinc function.
$$ \sinc x := \begin{cases} \displaystyle {{\sin x} \over {x}} & , \text{if } x \ne 0 \\ 1 & , \text{if } x = 0 \end{cases} $$
Normalized Sinc Function
$$ \sinc x := \begin{cases} \displaystyle {{\sin \pi x} \over {\pi x}} & , \text{if } x \ne 0 \\ 1 & , \text{if } x = 0 \end{cases} $$
Theorem
Euler’s Representation
$$ \sinc x = {{\sin \pi x} \over {\pi x}} = \prod_{n=1}^{\infty} \left( 1 - {{x^2} \over { n^2}} \right) $$
Explanation
A sinc function is a function that divides $\sin x$ by $x$, and it is quite useful since it has its own name. Although not always mentioned by name in textbooks, it often appears in the sections on limits and continuity.
Essentially, the unnormalized and normalized sinc functions are the same, so they are not strictly distinguished. The definition used usually depends on the context.
For reference, the ideal integral of the sinc function is calculated as $\displaystyle \int_{- \infty}^{\infty} {{\sin x} \over {x} } dx = \pi$.
Proof 1
Strategy: The proof to be introduced is not intuitive and contains many technical parts, making it quite difficult to understand. However, it is relatively more straightforward and has the advantage of not using complex analysis.
Part 1. Even Function $r(x)$ Periodicity
Let the ratio of the sinc function $\sinc x$ and Euler’s representation $\displaystyle f(x) := \prod_{n=1}^{\infty} \left( 1 - {\frac{ x^{2} }{ n^{2} }} \right)$ be a function of $x$ as follows. $$ r(x) := {\frac{ \sinc x }{ f(x) }} $$ The relationship between $\sinc x$ and $\sinc (x+1)$ is as follows. $$ \begin{align*} \sinc (x+1) =& {\frac{ \sin \pi (x+1) }{ \pi \cdot (x+1) }} \\ =& {\frac{ - \sin \pi x }{ \pi x }} {\frac{ x }{ x+1 }} \\ =& \left( - {\frac{ x }{ x+1 }} \right) \cdot \sinc x \end{align*} $$ Meanwhile, $f(x)$ and $f(x+1)$ also have the following relationship. $$ \begin{align*} & \left( - {\frac{ x }{ x+1 }} \right) f(x) \\ =& \left( - {\frac{ x }{ x+1 }} \right) \prod_{n=1}^{\infty} \left( 1 - {{x^2} \over { n^2}} \right) \\ =& \left( - {\frac{ x }{ \cancel{x+1} }} \right) \left( 1 - x \right) \left( 2 - x \right) \cdots \prod_{n=2}^{\infty} {\frac{ 1 }{ n^{2} }} \cancel{\left( 1 + x \right)} \left( 2 + x \right) \cdots \\ =& - \left( 0 - x \right) \left( 1 - x \right) \left( 2 - x \right) \cdots \prod_{n=2}^{\infty} {\frac{ 1 }{ n^{2} }} \left( 2 + x \right) \cdots \\ =& - f(x+1) \end{align*} $$ Thus, $r(x)$ is a $1$-periodic function, and considering only $-1/2 < x \le 1/2$ is sufficient to show that $\sinc$ and $f$ are equal. In fact, both $\sinc x$ and $f(x)$ are even functions and so is $r(x)$ defined by their ratio, making $0 < x \le 1/2$ sufficient.
Part 2. Recurrence Relation $(n^2 - c^2) I_{n} (c) = ( n^{2} - n) I_{n-2} (c)$
$$ I_{n} (c) := \int_{0}^{ { {\pi} \over {2} } } \cos ^{n} t \cos ct dt $$ Define it, and then $$ I_{0} (0) = \int_{0}^{ { {\pi} \over {2} } } \cos 0 dt = { {\pi} \over {2} } \\ I_{0} (2x) = \int_{0}^{ { {\pi} \over {2} } } \cos 2xt dt = \left[ {{\sin 2xt} \over {2x}} \right]_{0}^{{ {\pi} \over {2} }} = {{\sin \pi x} \over {2 x}} $$ Therefore, $$ {{I_{0} (2x)} \over {I_{0} (0)}}= {{\sin \pi x} \over {\pi x}} = \sinc x $$ Consequently, we need to show $$ {{I_{0} (2x)} \over {I_{0} (0)}}= \prod_{n=1}^{\infty} \left( 1 - {{x^2} \over { n^2}} \right) $$ First, let’s express $I_{n} (c)$ as a recurrence relation. $$ \begin{align*} I_{n} (c) =& \int_{0}^{ { {\pi} \over {2} } } \cos ^{n} t \cos ct dt \\ =& \left[ {1 \over c} \cos^{n} t \sin c t \right]_{0}^{{ {\pi} \over {2} }} - \int_{0}^{ { {\pi} \over {2} } } {1 \over c} n \cos ^{n-1} t (-\sin t) \sin ct dt \\ =& {n \over c} \int_{0}^{ { {\pi} \over {2} } } \cos ^{n-1} t \sin t \sin ct dt \\ =& {n \over c} \left[ {1 \over c} \cos^{n-1} t \sin t (-\cos c t ) \right]_{0}^{{ {\pi} \over {2} }} \\ & - {n \over c} \int_{0}^{ { {\pi} \over {2} } } {1 \over c} \left\{ (n-1) \cos ^{n-2} t (-\sin^2 t) + \cos ^{n} t \right\} (-\cos ct) dt \\ =& {n \over {c^2} } \int_{0}^{ { {\pi} \over {2} } } \left\{ (n-1) \cos ^{n-2} t (\cos^2 t - 1) + \cos ^{n} t \right\} \cos ct dt \\ =& {n \over {c^2} } \int_{0}^{ { {\pi} \over {2} } } \left\{ n \cos ^{n} t - (n-1) \cos^{n-2} t \right\} \cos ct dt \\ =& {n \over {c^2} } \left\{ n I_{n}(c) - (n-1) I_{n-2}(c) \right\} \end{align*} $$ Rewriting it neatly, we get the following. $$ (n^2 - c^2) I_{n} (c) = ( n^{2} - n) I_{n-2} (c) $$
Dividing each side by the equation obtained by substituting $c=0$ into the above recurrence relation gives a new recurrence relation $$ { {(n^2 - c^2)} \over {n^2} } {{I_{n} (c)} \over {I_{n} (0)}} = { {I_{n-2} (c)} \over {I_{n-2} (0)} } $$ Repeating in the new recurrence relation until the right side becomes $\displaystyle { {I_{0} (c)} \over {I_{0} (0)} }$, we obtain $$ \prod_{k=1}^{m} { {(2k)^2 - c^2} \over {(2k)^2} } {{I_{2m} (c)} \over {I_{2m} (0)}} = { {I_{0} (c)} \over {I_{0} (0)} } $$ Substituting $c=2x$ here, $$ \begin{align*} & \prod_{k=1}^{m} { {(2k)^2 - (2x)^2} \over {(2k)^2} } {{I_{2m} (2x)} \over {I_{2m} (0)}} \\ = & {{I_{2m} (2x)} \over {I_{2m} (0)}} \prod_{k=1}^{m} { {k^2 - x^2} \over {k^2} } \\ = & {{I_{0} (2x)} \over {I_{0} (0)}} \end{align*} $$
Part 3. $\displaystyle \lim_{m \to \infty} {{I_{m} (2x)} \over {I_{m} (0)}}=1$
Our goal is to show $\displaystyle {{I_{0} (2x)} \over {I_{0} (0)}}= \prod_{n=1}^{\infty} \left( 1 - {{x^2} \over { n^2}} \right)$, so proving $\displaystyle \lim_{m \to \infty} {{I_{m} (2x)} \over {I_{m} (0)}}=1$ will complete the proof. Now, consider $$ I_{m} (2x) = \int_{0}^{ { {\pi} \over {2} } } \cos ^{m} t \cos 2xt dt $$ Since in Part 1 we have shown we can assume only $x \in (0, 1/2]$, for $t \in [0, 1/2]$ we obtain the following two inequalities. $$ \begin{align*} \cos 0 > \cos 2 x t \implies & I_{m} (0) > I_{m} (2x) \\ \cos 2 x t > \cos^{2} t \implies & I_{m} (2x) > I_{m+2} (0) \end{align*} $$ Simplified, $$ I_{m} (0) > I_{m} (2x) > I_{m+2} (0) $$ Dividing each side by $I_{m} (0)$, $$ 1 > {{I_{m} (2x)} \over {I_{m} (0)} } > {{I_{m+2} (0)} \over {I_{m} (0)}} $$ According to the recurrence relation obtained earlier, $$ (m+2)^2 I_{m+2}(0) = (m^2 + 3m + 2) I_{m}(0) $$ Thus, $$ \lim_{m \to \infty} {{I_{m+2} (0)} \over {I_{m} (0)}} = \lim_{m \to \infty} { {m+1} \over {m+2} } = 1 $$ According to the Sandwich Theorem, we obtain the following: $$ {{\sin \pi x} \over {\pi x}} = \prod_{n=1}^{\infty} \left( 1 - {{x^2} \over { n^2}} \right) $$
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Although the result is highly useful, the proof itself is not something you can memorize and use elsewhere. It is recommended to just acknowledge such a proof exists rather than thoroughly understanding and mastering it step by step.