# A Curious Integral | The n-Class Café

*by*Phil Tadros

On Mathstodon, Robin Houston identified a video the place Oded Margalit claimed that it’s an open downside why this integral:

$displaystyle{ int_0^inftycos(2x)prod_{n=1}^inftycosleft(frac{x}{n} proper) d x }$

is so absurdly near $frac{pi}{8}$, however not fairly equal.

They comply with 41 decimal locations, however they’re not the identical!

$start{array}{rcl} displaystyle{ int_0^inftycos(2x)prod_{n=1}^inftycosleft(frac{x}{n}proper) d x } &=& 0.3926990816987241548078304229099378605246454… displaystyle{fracpi8} &=& 0.3926990816987241548078304229099378605246461… finish{array}$

So, a bunch of us tried to determine what was occurring.

Jaded nonmathematicians advised us it’s only a coincidence, so what’s there to clarify? However in fact an settlement this shut is unlikely to be “only a coincidence”. It is likely to be, however you’ll by no means get anyplace in math with that angle.

We had been reminded of the well-known cosine Borwein integral

$displaystyle{ int_0^infty 2 cos(x) prod_{n = 0}^{N} frac{sin (x/(2n+1))}{x/(2n+1)} , d x}$

which equals $frac{pi}{2}$ for $N$ as much as and together with 55, however not for any bigger $N$:

$displaystyle{ int_0^infty 2 cos(x) prod_{n = 0}^{56} frac{sin (x/(2n+1))}{x/(2n+1)} , d x ; ; approx ;; frac{pi}{2} – 2.3324 cdot 10^{-138} }$

However it was Sean O who actually cracked the case, by exhibiting that the integral we had been fighting might truly be lowered to an $N = infty$ model of the cosine Borwein integral, particularly

$displaystyle{ int_0^infty 2 cos(x) prod_{n = 0}^{infty} frac{sin (x/(2n+1))}{x/(2n+1)} , d x}$

The purpose is that this. A little bit calculation utilizing the Weierstrass factorizations

$frac{sin x}{x} = prod_{n = 1}^infty left( 1 – frac{x^2}{pi^2 n^2} proper)$

$cos x = prod_{n = 0}^infty left( 1 – frac{4x^2}{pi^2 (2n+1)^2} proper)$

enables you to present

$prod_{n = 1}^infty cosleft(frac{x}{n}proper) = prod_{n = 0}^infty frac{sin (2x/(2n+1))}{x/(2n+1)}$

and thus

$displaystyle{ int_0^infty cos(2x) prod_{n=1}^infty cosleft(frac{x}{n} proper) ; d x } = displaystyle{ int_0^inftycos(2x) prod_{n = 0}^infty frac{sin (2x/(2n+1))}{x/(2n+1)} d x }$

Then, a change of variables on the right-hand aspect offers

$displaystyle{ int_0^infty cos(2x) prod_{n=1}^infty cosleft(frac{x}{n} proper) ; d x } = frac{1}{4} displaystyle{ int_0^infty 2cos(x) prod_{n = 0}^infty frac{sin (x/(2n+1))}{x/(2n+1)} d x }$

So, exhibiting that

$displaystyle{ int_0^inftycos(2x)prod_{n=1}^inftycosleft(frac{x}{n} proper) d x }$

is microscopically lower than $frac{pi}{8}$ is equal to exhibiting that

$displaystyle{ int_0^infty 2cos(x) prod_{n = 0}^infty frac{sin (x/(2n+1))}{x/(2n+1)} d x }$

is microscopically lower than $frac{pi}{2}$.

This units up a transparent technique for fixing the thriller! Folks perceive why the cosine Borwein integral

$displaystyle{ int_0^infty 2 cos(x) prod_{n = 0}^{N} frac{sin (x/(2n+1))}{x/(2n+1)} , d x}$

equals $frac{pi}{2}$ for $N$ as much as $55$, after which drops ever so barely beneath $frac{pi}{2}$. The mechanism is obvious when you watch the appropriate form of film. It’s very visible. Greg Egan explains it right here with an animation, based mostly on concepts by Hanspeter Schmid:

Or you possibly can watch this video, which covers an easier however associated instance:

So, we simply want to point out that as $N to +infty$, the worth of the cosine Borwein integral doesn’t drop way more! It drops by only a tiny quantity: about $7 instances 10^{-43}$.

Alas, this doesn’t appear simple to point out. A minimum of I don’t know easy methods to do it but. However what had appeared an utter thriller has now turn out to be a chore in evaluation: estimating how a lot

$displaystyle{ int_0^infty 2 cos(x) prod_{n = 0}^{N} frac{sin (x/(2n+1))}{x/(2n+1)} , d x}$

drops every time you improve $N$ a bit.

At this level for those who’re sufficiently erudite you’re most likely screaming: *“BUT THIS IS ALL WELL-KNOWN!”*

And also you’re proper! We had quite a lot of enjoyable discovering these items, nevertheless it was not new. After I was posting about it on MathOverflow, I bumped into an article that mentions a dialogue of these things:

and it seems Borwein and his buddies had already studied it. There’s a little bit bit right here:

and much more on this ebook:

- J. M. Borwein, D. H. Bailey and R. Girgensohn,
*Experimentation in Arithmetic: Computational Paths to Discovery*, Wellesley, Massachusetts, A Ok Peters, 2004.

The truth is the integral

$displaystyle{ int_0^infty 2 cos(x) prod_{n = 0}^{infty} frac{sin (x/(2n+1))}{x/(2n+1)} , d x}$

was found by Bernard Mares on the age of 17. Apparently he posed the problem of proving that it was lower than $frac{pi}{4}$. Borwein and others dived into this and discovered how.

However there may be nonetheless work left to do!

So far as I can inform, the recognized proofs that

$displaystyle{ frac{pi}{8} – int_0^inftycos(2x)prod_{n=1}^inftycosleft(frac{x}{n} proper) d x } ; approx ; 7.4073 cdot 10^{-43}$

all contain quite a lot of brute-force calculation. Is there a extra conceptual strategy to perceive this distinction, no less than roughly? There *is* a transparent conceptual proof that

$displaystyle{ frac{pi}{8} – int_0^inftycos(2x)prod_{n=1}^inftycosleft(frac{x}{n} proper) d x } ;; > ;; 0$

That’s what Greg Egan defined in my weblog article. However can we get a transparent proof that

$displaystyle{ frac{pi}{8} – int_0^inftycos(2x)prod_{n=1}^inftycosleft(frac{x}{n} proper) d x } ; ; < ; ; C$

for some small fixed $C$, say $10^{-40}$ or so?

One can argue that till we do, Oded Margalit is true: there’s an open downside right here. Not an issue in proving that one thing is true. An issue in understanding *why* it’s true.