# 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.

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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 $N=7$.

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Since $u_k$ and $1-u_k$ are equal in distribution, it follows that $$X \stackrel{d}{=} \sum_{k=1}^\infty 2^{-k} \left(1-u_k\right) = \sum_{k=1}^\infty 2^{-k} - X = 1 - X$$ Since $u_k \geq 0$, it follows that $\mathbb{P}\left(0 \leqslant X \leqslant 1\right) = 1$.

Therefore $f_X(x) = 0$ for $x < 0$ or $x>1$. It also shows that $f_X(x) = f_X(1-x)$ for $0<x<1$.

Notice that $$\begin{eqnarray} f_X^\prime(x) &=& \frac{1}{\pi} \int_0^\infty \left(- t \sin\left( t \left(x-\frac{1}{2} \right) \right) \right) \operatorname{\rm sinc}\left( \frac{t}{4} \right) \prod_{k=2}^\infty \operatorname{\rm sinc}\left( \frac{t}{2^{k+1}} \right) \mathrm{d} t \end{eqnarray}$$ Now using $t \cdot \operatorname{\rm sinc}\left( \frac{t}{4} \right) = 4 \sin\left( \frac{t}{4} \right)$, and $$\sin\left( t \left(x-\frac{1}{2} \right) \right) \sin\left(\frac{t}{4} \right) = \frac{1}{2} \left[ \cos\left( \frac{t}{2} \left( (2 x-1) -\frac{1}{2} \right) \right) - \cos\left( \frac{t}{2} \left( 2 x -\frac{1}{2} \right) \right) \right]$$ which gives $$\begin{eqnarray} f_X^\prime(x) &=& 4 f_X(2 x) - 4 f_X(2x-1) \end{eqnarray}$$

Now let $0 < z \leqslant \frac{1}{2}$. Then $f_X^\prime(z) = 4 f_X(2z)$ and thus $$f_X(z) - f_X(0) = \int_0^{z} 4 f_X(2x) \mathrm{d} x = 2 \left(F_X(2z) - F_X(0)\right)$$

Clearly $F(0) = 0$, $F(1) = 1$ and $F_X\left(\frac{1}{2} \right) = \frac{1}{2}$, since the distribution is symmetric about $x=\frac{1}{2}$, thus $$f_X\left(\frac{1}{2}\right) = f_X\left(0\right) + 2 \qquad f_X\left(\frac{1}{4}\right) = f_X\left(0\right) + 1$$

If I now assume that $f_X(0) = 0$, the result follows.

Added A missing proof of $f_X(0)=0$, as well as alternative proof of $f_X\left(\frac{1}{2}\right) = 2$ results from writing: $$X \stackrel{d}{=} \frac{u_1}{2} + \frac{1}{2} \sum_{k=2}^\infty 2^{-k+1} u_k \stackrel{d}{=} \frac{u_1+X}{2}$$ This equality in distribution implies the following integral equation for $f_X$: $$f_X(x) = 2 \int_0^1 f_X(2x -u) \mathrm{d} u$$ Substituting $x=0$ we get $f_X(0) = 2 \int_0^1 f_X(-u) \mathrm{d} u =0$, since $f_X(x)=0$ for $x<0$. Incidentally, using $x=\frac{1}{2}$ gives the desired identity as well: $$f_X\left(\frac{1}{2}\right) = 2 \int_0^1 f_X(1-u) \mathrm{d} u = 2 \int_0^1 f_X(u) \mathrm{d} u = 2$$

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The probabilist in me likes this kind of argument! –  Byron Schmuland Jan 6 '12 at 21:01