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If that's not possible, consider moving to the Standards-compliant and open-source Mozilla browser. << Guillotine Partitions and the Hipparchus Operad | Main | Topos Institute Positions >> January 4, 2023 A Curious Integral Posted by John Baez MathML-enabled post (click for more details). On Mathstodon, Robin Houston pointed out a video where Oded Margalit claimed that it's an open problem why this integral: [?] 0 [?]cos(2x)[?] n=1 [?]cos(xn)dx \displaystyle{ \int_0^\infty\cos(2x)\ prod_{n=1}^\infty\cos\left(\frac{x}{n} \right) d x } is so absurdly close to p8\frac{\pi}{8}, but not quite equal. MathML-enabled post (click for more details). They agree to 41 decimal places, but they're not the same! [?] 0 [?]cos(2x)[?] n=1 [?]cos(xn)dx = 0.3926990816987241548078304229099378605246454... p8 = 0.3926990816987241548078304229099378605246461... \begin{array}{rcl} \ displaystyle{ \int_0^\infty\cos(2x)\prod_{n=1}^\infty\cos\left(\frac {x}{n}\right) d x } &=& 0.3926990816987241548078304229099378605246454... \\ \\ \displaystyle {\frac\pi8} &=& 0.3926990816987241548078304229099378605246461... \\ \ end{array} So, a bunch of us tried to figure out what was going on. Jaded nonmathematicians told us it's just a coincidence, so what is there to explain? But of course an agreement this close is unlikely to be "just a coincidence". It might be, but you'll never get anywhere in math with that attitude. We were reminded of the famous cosine Borwein integral [?] 0 [?]2cos(x)[?] n=0 Nsin(x/(2n+1))x/(2n+1)dx \displaystyle{ \int_0^\ infty 2 \cos(x) \prod_{n = 0}^{N} \frac{\sin (x/(2n+1))}{x/(2n+1)} \, d x} which equals p2\frac{\pi}{2} for NN up to and including 55, but not for any larger NN: [?] 0 [?]2cos(x)[?] n=0 56sin(x/(2n+1))x/(2n+1)dx[?]p2-2.3324[?]10 -138 \ 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} } But it was Sean O who really cracked the case, by showing that the integral we were struggling with could actually be reduced to an N=[?]N = \infty version of the cosine Borwein integral, namely [?] 0 [?]2cos(x)[?] n=0 [?]sin(x/(2n+1))x/(2n+1)dx \displaystyle{ \int_0^\ infty 2 \cos(x) \prod_{n = 0}^{\infty} \frac{\sin (x/(2n+1))}{x/ (2n+1)} \, d x} The point is this. A little calculation using the Weierstrass factorizations sinxx=[?] n=1 [?](1-x 2p 2n 2) \frac{\sin x}{x} = \prod_{n = 1}^\infty \ left( 1 - \frac{x^2}{\pi^2 n^2} \right) cosx=[?] n=0 [?](1-4x 2p 2(2n+1) 2) \cos x = \prod_{n = 0}^\infty \left( 1 - \frac{4x^2}{\pi^2 (2n+1)^2} \right) lets you show [?] n=1 [?]cos(xn)=[?] n=0 [?]sin(2x/(2n+1))x/(2n+1) \prod_{n = 1}^\infty \ cos\left(\frac{x}{n}\right) = \prod_{n = 0}^\infty \frac{\sin (2x/ (2n+1))}{x/(2n+1)} and thus [?] 0 [?]cos(2x)[?] n=1 [?]cos(xn)dx=[?] 0 [?]cos(2x)[?] n=0 [?]sin(2x/(2n+1))x/ (2n+1)dx \displaystyle{ \int_0^\infty \cos(2x) \prod_{n=1}^\infty \ cos\left(\frac{x}{n} \right) \; d x } = \displaystyle{ \int_0^\infty\ cos(2x) \prod_{n = 0}^\infty \frac{\sin (2x/(2n+1))}{x/(2n+1)} d x } Then, a change of variables on the right-hand side gives [?] 0 [?]cos(2x)[?] n=1 [?]cos(xn)dx=14[?] 0 [?]2cos(x)[?] n=0 [?]sin(x/(2n+1))x/ (2n+1)dx \displaystyle{ \int_0^\infty \cos(2x) \prod_{n=1}^\infty \ cos\left(\frac{x}{n} \right) \; d x } = \frac{1}{4} \displaystyle{ \ int_0^\infty 2\cos(x) \prod_{n = 0}^\infty \frac{\sin (x/(2n+1))}{x/ (2n+1)} d x } So, showing that [?] 0 [?]cos(2x)[?] n=1 [?]cos(xn)dx \displaystyle{ \int_0^\infty\cos(2x)\ prod_{n=1}^\infty\cos\left(\frac{x}{n} \right) d x } is microscopically less than p8\frac{\pi}{8} is equivalent to showing that [?] 0 [?]2cos(x)[?] n=0 [?]sin(x/(2n+1))x/(2n+1)dx \displaystyle{ \int_0^\ infty 2\cos(x) \prod_{n = 0}^\infty \frac{\sin (x/(2n+1))}{x/(2n+1)} d x } is microscopically less than p2\frac{\pi}{2}. This sets up a clear strategy for solving the mystery! People understand why the cosine Borwein integral [?] 0 [?]2cos(x)[?] n=0 Nsin(x/(2n+1))x/(2n+1)dx \displaystyle{ \int_0^\ infty 2 \cos(x) \prod_{n = 0}^{N} \frac{\sin (x/(2n+1))}{x/(2n+1)} \, d x} equals p2\frac{\pi}{2} for NN up to 5555, and then drops ever so slightly below p2\frac{\pi}{2}. The mechanism is clear once you watch the right sort of movie. It's very visual. Greg Egan explains it here with an animation, based on ideas by Hanspeter Schmid: * John Baez, Patterns that eventually fail, Azimuth, September 20, 2018. Or you can watch this video, which covers a simpler but related example: * 3Blue1Brown, Researchers thought this was a bug (Borwein integrals). So, we just need to show that as N-+[?]N \to +\infty, the value of the cosine Borwein integral doesn't drop much more! It drops by just a tiny amount: about 7x10 -437 \times 10^{-43}. Alas, this doesn't seem easy to show. At least I don't know how to do it yet. But what had seemed an utter mystery has now become a chore in analysis: estimating how much [?] 0 [?]2cos(x)[?] n=0 Nsin(x/(2n+1))x/(2n+1)dx \displaystyle{ \int_0^\ infty 2 \cos(x) \prod_{n = 0}^{N} \frac{\sin (x/(2n+1))}{x/(2n+1)} \, d x} drops each time you increase NN a bit. At this point if you're sufficiently erudite you are probably screaming: "BUT THIS IS ALL WELL-KNOWN!" And you're right! We had a lot of fun discovering this stuff, but it was not new. When I was posting about it on MathOverflow, I ran into an article that mentions a discussion of this stuff: * Eric W. Weisstein, Infinite cosine product integral, from MathWorld--A Wolfram Web Resource. and it turns out Borwein and his friends had already studied it. There's a little bit here: * J. M. Borwein, D. H. Bailey, V. Kapoor and E. W. Weisstein, Ten problems in experimental mathematics, Amer. Math. Monthly 113 (2006), 481-509. and a lot more in this book: * J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics: Computational Paths to Discovery, Wellesley, Massachusetts, A K Peters, 2004. In fact the integral [?] 0 [?]2cos(x)[?] n=0 [?]sin(x/(2n+1))x/(2n+1)dx \displaystyle{ \int_0^\ infty 2 \cos(x) \prod_{n = 0}^{\infty} \frac{\sin (x/(2n+1))}{x/ (2n+1)} \, d x} was discovered by Bernard Mares at the age of 17. Apparently he posed the challenge of proving that it was less than p4\frac{\pi}{4}. Borwein and others dived into this and figured out how. But there is still work left to do! As far as I can tell, the known proofs that p8-[?] 0 [?]cos(2x)[?] n=1 [?]cos(xn)dx[?]7.4073[?]10 -43 \displaystyle{ \frac{\ pi}{8} - \int_0^\infty\cos(2x)\prod_{n=1}^\infty\cos\left(\frac{x}{n} \right) d x } \; \approx \; 7.4073 \cdot 10^{-43} all involve a lot of brute-force calculation. Is there a more conceptual way to understand this difference, at least approximately? There is a clear conceptual proof that p8-[?] 0 [?]cos(2x)[?] n=1 [?]cos(xn)dx>0 \displaystyle{ \frac{\pi}{8} - \ int_0^\infty\cos(2x)\prod_{n=1}^\infty\cos\left(\frac{x}{n} \right) d x } \;\; > \;\; 0 That's what Greg Egan explained in my blog article. But can we get a clear proof that p8-[?] 0 [?]cos(2x)[?] n=1 [?]cos(xn)dx