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Show that the identity

$$\int_0^\infty \prod_{k=0}^N \text{sinc}\left(\frac{x}{2k+1}\right)\,{\rm d}x – \sum_{n=1}^\infty \prod_{k=0}^N \text{sinc}\left(\frac{n}{2k+1}\right) = \frac{1}{2}$$ where $\text{sinc}(x) = \frac{\sin(x)}{x}$ holds for $N=0,1,2,\ldots,40000$ but fails for all larger $N$.

I remember seeing this strange identity a few years ago and it stuck to my mind, but unfortunately I can’t find the source of this right now and it’s reconstructed from memory (and checked with a computer for small $N$ although $40000$ might not be accurate). This is why I’m asking it here.

If I remember correctly it’s closely linked to Fourier transforms and for $N$ larger than $\sim 40000$ the difference between the left and right hand side should be smaller than $\sim 10^{-10000}$ so the agreement is extremely good.

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Do anyone know the source of this problem or otherwise how to solve it? What is the theory behind it (i.e. why does it break down at some finite value)?

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There is a theorem stated on page 2 of this paper, which states that for $N+1 > 1$ positive numbers $a_0,a_1,\ldots,a_N > 0$ the identity

$$\dfrac{1}{2}+\sum_{n = 1}^{\infty}\prod_{k = 0}^{N}\text{sinc}(a_kn) = \int_{0}^{\infty}\prod_{k = 0}^{N}\text{sinc}(a_kx)\,dx$$

holds provided that $\displaystyle\sum_{k = 0}^{N}a_k \le 2\pi$. (Also, for $N = 0$, the identity holds if $a_0 < 2\pi$).

Since $\displaystyle\sum_{k = 0}^{40248}\dfrac{1}{2k+1} \approx 6.283175 < 2\pi < 6.283188 \approx \displaystyle\sum_{k = 0}^{40249}\dfrac{1}{2k+1}$, the identity holds for $1 \le N \le 40248$ but fails for $N \ge 40249$.

The proof of this theorem has to do with Fourier transforms. Evaluating the Fourier transform of a function at $0$ gives you the integral of the function over $\mathbb{R}$. The Fourier transform of the product of sinc functions is a convolution of “rectangle” functions whose widths are proportional to $a_k$. If you convolve enough rectangle functions together, the value of the result at $0$ changes. This is an intuitive/non-rigorous explanation. I’m sure a more rigorous explanation can be found online.

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