calculus - How does one prove $int_0^infty prod_{k=1}^infty operatorname{rm sinc}left( frac{t}{2^{k+1}} right) mathrm{d} t = 2 pi$

Looking into the distribution of a Fabius random variable:
$$
X := \sum_{k=1}^\infty 2^{-k} u_k
$$
where $u_k$ are i.i.d. uniform variables on a unit interval, I encountered the following expression for its probability density:
$$
f_X(x) = \frac{1}{\pi} \int_0^\infty \left( \prod_{k=1}^\infty \operatorname{\rm sinc}\left( \frac{t}{2^{k+1}} \right) \right) \cos \left( t \left( x- \frac{1}{2} \right) \right) \mathrm{d} t
$$
It seems, numerically, that $f\left(\frac{1}{2} \right) = 2$, but my several attempts to prove this were not successful.

Any ideas how to approach this are much appreciated.

Answer

From Theorem 1 (equation (19) on page 5) of Surprising Sinc Sums and Integrals, we have
$$\frac{1}{\pi} \int_0^\infty \left( \prod_{k=1}^N \operatorname{\rm sinc}\left( \frac{t}{2^{k+1}} \right) \right) \mathrm{d} t=2$$
for all $N<\infty$. I suppose you can justify letting
$N\to \infty$ to get your result.

---

One of the surprises in that paper concerns a similar integral
$$ \int_0^\infty \left( \prod_{k=0}^N \operatorname{\rm sinc}\left( \frac{t}{2{k+1}} \right) \right) \mathrm{d} t.$$ This turns out to be equal to $\pi/2$ when $0\leq N\leq 6$, but is slightly less than $\pi/2$ when $N=7$.