\[\int_{-\infty}^\infty \frac{\sin(x)}{x}dx = \pi\] \[\int_{-\infty}^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}dx = \pi\] \[\int_{-\infty}^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\frac{\sin(x/5)}{x/5}dx = \pi\] \[\int_{-\infty}^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\frac{\sin(x/5)}{x/5}\frac{\sin(x/7)}{x/7}dx = \pi\] \[\int_{-\infty}^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\frac{\sin(x/5)}{x/5}\frac{\sin(x/7)}{x/7}\frac{\sin(x/9)}{x/9}dx = \pi\] \[\int_{-\infty}^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\frac{\sin(x/5)}{x/5}\frac{\sin(x/7)}{x/7}\frac{\sin(x/9)}{x/9}\frac{\sin(x/11)}{x/11}dx = \pi\] \[\int_{-\infty}^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\frac{\sin(x/5)}{x/5}\frac{\sin(x/7)}{x/7}\frac{\sin(x/9)}{x/9}\frac{\sin(x/11)}{x/11}\frac{\sin(x/13)}{x/13}dx = \pi\] \[\int_{-\infty}^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\cdots \frac{\sin(x/13)}{x/13}\frac{\sin(x/15)}{x/15}dx \approx \pi -4.62 \cdot 10^{-11}\]
But why?
Let's define \(X_n \sim \text{Unif}(-a_n, a_n),\) \(S = \sum_{n=1}^N X_n,\) then the characteristic function is given by \[\mathbb{E}\left[\exp(i \omega S)\right] = \prod_{n=1}^N \frac{\sin(a_n \omega)}{a_n \omega}.\]
We obtain the pdf by inverse Fourier transform \[p_S(x) = \frac{1}{2\pi}\int_{-\infty}^\infty \prod_{n=1}^N \frac{\sin(a_n \omega)}{a_n \omega} \exp(-i \omega x) d\omega.\]
We get the integral of interest at 0, \[p_S(0) = \frac{1}{2\pi}\int_{-\infty}^\infty \prod_{n=1}^N \frac{\sin(a_n \omega)}{a_n \omega}d\omega.\]
Lemma. Let \(U \sim \text{Unif}(a,b)\) and \(\text{supp}(X) = [c,d] \subseteq [a,b],\) then \[p_{U+X}(z) = \frac{1}{b-a}, \quad a+d \le z \le b+c.\]
For \(a+d \le z \le b+c,\) \(c \le x \le d\) it holds that \(a \le z-x \le b\).
\begin{align} p_{U+X}(z) &= \int_{c}^{d} p_U(z-x) p_X(x)dx\\ &= \frac{1}{b-a} \int_{c}^{d} p_X(x)dx\\ &= \frac{1}{b-a} \end{align}
\(\Box\)
For \(X_n \sim \text{Unif}(-\frac{1}{n}, \frac{1}{n})\), since \(\frac{1}{3} + \frac{1}{5}+\cdots + \frac{1}{13} < 1\) \[\frac{1}{2} = p_{X_1 + (X_3 + X_5 +\cdots + X_{13})} = \frac{1}{2\pi} \int_{-\infty}^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\cdots \frac{\sin(x/13)}{x/13}dx.\]
But \(\frac{1}{3} + \frac{1}{5}+\cdots + \frac{1}{15} > 1\) and the equality does not hold anymore.