A curious coincidence: $\prod\limits_{k=1}^\infty\left(1+\int_{k}^{k+1}\left(\frac{\sin (\pi x)}{x}\right)^2\mathrm dx\right)\overset{?}{=}\pi/2$

Question

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 $\dfrac{\pi}{2}$?

Answer 1

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.

Answer 2

(Self-answering)

Assuming that Desmos is reliable, I now think that the product in the OP is approximately $0.99992076\times\dfrac{\pi}{2}$, so the conjecture is false.

Why I think the conjecture is false

Letting $f(n)=\color{red}{\dfrac{2}{\pi}}\prod\limits_{k=1}^n\left(1+\int_{k}^{k+1}\left(\frac{\sin (\pi x)}{x}\right)^2\mathrm dx\right)$, I found that:

$\dfrac{f\left(10^N\right)}{f\left(10^{N-1}\right)}\approx 1+4.5\times10^{-N}$ for $N=3,4,5,6,7,8$

$f(10^8)\approx 0.999920757989$ (found on Desmos; it may take a while to load)

This suggests that $\lim\limits_{n\to\infty}f(n)\approx 0.99992076$, so the product in the OP is approximately $0.99992076\times\dfrac{\pi}{2}$.

Given that, as the OP notes, $\int_0^\infty \frac{\sin x}{x} \mathrm dx=\dfrac{\pi}{2}$ and $\int_0^\infty\left(\frac{\sin x}{x}\right)^2\mathrm dx=\dfrac{\pi}{2}$, it is indeed a curious coincidence that the product in the OP is so close to $\dfrac{\pi}{2}$.

I still wonder if the the product in the OP has a closed form expression.

Most small numbers are near to simple expressions with $\pi$ or $e$

Pick a small random real number, say between $1$ and $2$. Round it to say four decimal places, and type it into Wolfram. It is likely that Wolfram will suggest a simple closed form involving $\pi$ or $e$. For example, I tried $1.2345$, and Wolfram suggested $\log (2e-2)$. We find that $\dfrac{1.2345}{\log (2e-2)}\approx 1.00002$.

Other curious findings about $p(x)=\frac{\sin (\pi x)}{x}$

• $\prod\limits_{k=1}^{\color{red}{10000}}\left(1+2\int_{k}^{k+1}p(x)^2\mathrm dx\right)\approx0.99992\left(1+\frac{e}{2}\right)$ but $\prod\limits_{k=1}^{\color{red}{100000}}\left(1+2\int_{k}^{k+1}p(x)^2\mathrm dx\right)\approx1.000005\left(1+\frac{e}{2}\right)$
• $\prod\limits_{k=1}^{\color{red}{100000}}\left(1+\frac{\pi}{2}\int_k^{k+1}\frac{|p(x)|}{x}\mathrm dx\right)\approx0.9999995\left(\frac{\sqrt{e}}{\ln 2}\right)$, but $\prod\limits_{k=1}^{\color{red}{1000000}}\left(1+\frac{\pi}{2}\int_k^{k+1}\frac{|p(x)|}{x}\mathrm dx\right)\approx1.00001\left(\frac{\sqrt{e}}{\ln 2}\right)$
• We have $\int_0^\infty p(x)^2\mathrm dx=\frac{\pi^2}{2}$, and the product of the arc lengths between neighboring positive roots of $p(x)$ is $\prod\limits_{k=1}^\infty \int_k^{k+1}\sqrt{1+(p'(x))^2}\mathrm dx\approx 0.9985\left(\frac{\pi^2}{2}\right)$.

Mathematical imposters

The OP is not the first time I've found a "mathematical imposter". Here is another one.