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$$\dfrac{\sin(x)}{x}=\displaystyle\prod_{i=1}^\infty\left(1-\dfrac{x^2}{i^2\pi^2}\right)= \left(1+\dfrac{x}{\pi}\right) \left(1-\dfrac{x}{\pi}\right) \left(1+\dfrac{x}{2\pi}\right) \left(1-\dfrac{x}{2\pi}\right) \left(1+\dfrac{x}{3\pi}\right) \ldots$$

They use this form for proving $\zeta(2),\zeta (4)$ etc., but how? Yes it has true root and $x=0$, but this function might be another.For example, let $f$ be, $f(x)=0$ for $x=nt,\quad n\in\mathbb Z,t\in\mathbb R$.$f(x)=n^x\dfrac{\sin x}{x},\quad n\in\mathbb R$

Of course we can say that ;

$$f(x)=\displaystyle\prod_{i=1}^\infty\left(1-\dfrac{x^2}{i^2\pi^2}\right)= \left(1+\dfrac{x}{\pi}\right) \left(1-\dfrac{x}{\pi}\right) \left(1+\dfrac{x}{2\pi}\right) \left(1-\dfrac{x}{2\pi}\right) \left(1+\dfrac{x}{3\pi}\right) \ldots$$

but $f(x)$ doesn't equal to $\dfrac{\sin x}{x}$

How can we show exactly what $\dfrac{\sin x}{x}$ equals to ? If you can explain clearly please give also advanced proofs.Thank you in advance.

Micheal Brain Hurts
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  • http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-limits-k-1-infty-frac1k2

    For example, in there.

     – Micheal Brain Hurts Apr 09 '17 at 09:48
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    It follows from $$\pi\cot(\pi z) = \frac{1}{z} - 2z\sum_{n=1}^{\infty}\frac{1}{n^2 - z^2}$$ which can be proved by complex integration. – Zaid Alyafeai Apr 09 '17 at 09:53
  • Thank you, but where this comes from, as well? – Micheal Brain Hurts Apr 09 '17 at 10:00
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    http://math.stackexchange.com/questions/208317/show-sum-n-0-infty-frac1a2n2-frac1a-pi-coth-a-pi2a2 – Zaid Alyafeai Apr 09 '17 at 10:04
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    Weierstrass's Factorization Theorem says that if $f(z)$ is entire and suppose that it has simple zeros at $a_1$, $a_2$, $\cdots$ where $0 < |a_n| < |a_{n+1}|$ and $\lim |a_n| =\infty$. Then $f(z)$ can be expressed as an infinite product of the form $$f(z)=f(0)e^{...}\prod(1-\frac{z}{a_n})e^{\frac{z}{a_n}}$$ – Nosrati Apr 09 '17 at 10:29
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    This Theorem applies for $\dfrac{1}{\Gamma(z+1)}$ which is entire with zeros $n<0$ and $n$ is integer. – Nosrati Apr 09 '17 at 10:32
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    From $\Gamma(z)\Gamma(1-z)=\dfrac{\pi}{\sin\pi z}$ relation obtains. – Nosrati Apr 09 '17 at 10:34

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