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Consider $\Omega = \{0,1\}^\mathbb{N} $. Let $U(\omega) =\sum_{i=1}^\infty \frac{\omega_i}{2^i} $ where all $\omega_i \in \{0,1\}$. We can show, that $2U+1$ is uniformly distributed in $[-1,1]$.Therefore we know, that its characteristic function is $\frac{\sin(t)}{t}.$

We now want to establish, that sinus is an infinte product. Therefore we have to calculate $E[e^{it (2U-1)}]$ in a different way as a infinite product $E[e^{it (2U-1)}] = \prod_{i=1}^\infty...$ How can one do that? Does anyone know the ansatz for that?

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    @Raskolnikov: yes it does!2011-10-25

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Use the fact that the characteristic function of a sum with independent increments is the product of the characteristic functions. After that, you might need the characteristic function of the random variable $X_n$ defined by $X_n(\omega)=2\omega_n-1$ hence you may begin by computing this.

As mentioned by @Raskolnikov, the result involves Viète's infinite product.