It is known that $\int_0^\infty \frac{\sin x}{x} \mathrm dx=\dfrac{\pi}{2}$ (proof) and $\int_0^\infty\left(\frac{\sin x}{x}\right)^2\mathrm dx=\dfrac{\pi}{2}$ (proof).
Numerical investigation suggests that we also have:
$$\prod\limits_{k=1}^\infty\left(1+\int_{k}^{k+1}\left(\frac{\sin (\pi x)}{x}\right)^2\mathrm dx\right)=\dfrac{\pi}{2}$$
But I do not know how to prove this. $\int\left(\frac{\sin (\pi x)}{x}\right)^2\mathrm dx$ cannot be expressed in terms of elementary functions.
Is this infinite product really equal to $\dfrac{\pi}{2}$?
Best Answer
The answer is no, the product does not reach $\pi/2$ in the limit. We prove this with comparison techniques, given the numerical result for $k=100000$.
For every $k$ we may render by comparison:
$\int_k^{k+1}\left(\dfrac{\sin(\pi x)}{x}\right)^2 dx<\int_k^{k+1}\left(\dfrac{\sin(\pi x)}{k}\right)^2 dx=\dfrac{1}{2k^2}$
So
$\ln\left(1+\int_k^{k+1}\left(\dfrac{\sin(\pi x)}{x}\right)^2 dx\right)<\dfrac{1}{2k^2}.$
Summing this from $k=k_0$ to $k=\infty$ must give less than $1/(k_0-1)$, thus for $k_0=100001$ this sum is less than $1.000005$. With the numerical result for $k<100000$ we are forced to accept
$\prod_{k=1}^\infty\left(1+\int_k^{k+1}\left(\dfrac{\sin(\pi x)}{x}\right)^2 dx\right)<\dfrac{\pi}2(1-0.00008)\exp(0.000005)$
and the reciprocal of $1-0.00008$ must be greater than $\exp(0.00008)$.
The last part of this comparison is rather loose (essentially comparing two numbers differing by a factor of $16$); a tighter bound would indicate that in fact the limiting ratio can't be far from $0.99992$. Which is what you found from the further numerical result.