The Borwein integrals introduced in [1] are a famous example of how proof-by-example can go wrong.

Define sinc(x) as sin(x)/x. Then the following equations hold.

begin{align*} int_0^infty text{sinc}(x) ,dx &= frac{pi}{2} \ int_0^infty text{sinc}(x) , text{sinc}left(frac{x}{3}right) ,dx &= frac{pi}{2} \ int_0^infty text{sinc}(x), text{sinc}left(frac{x}{3}right) ,text{sinc}left(frac{x}{5}right) ,dx &= frac{pi}{2} \ vdots &phantom{=} \ int_0^infty text{sinc}(x) , text{sinc}left(frac{x}{3}right) cdots text{sinc}left(frac{x}{13}right) ,dx &= frac{pi}{2} \ end{align*}

However

int_0^infty text{sinc}(x) , text{sinc}left(frac{x}{3}right) cdots text{sinc}left(frac{x}{15}right) ,dx = frac{pi}{2} - delta

where δ ≈ 2.3 × 10−11.

This is where many presentations end, concluding with the moral that a pattern can hold for a while and then stop. But I’d like to go just a little further.

Define

B(n) = int_0^infty prod_{k=0}^{n} text{sinc}left(frac{x}{2k+1}right) , dx.

Then B(n) = π/2 for n = 1, 2, 3, …, 6 but not for n = 7, though it almost holds for n = 7. What happens for larger values of n?

The Borwein brothers proved that B(n) is a monotone function of n, and the limit as n → ∞ exists. In fact the limit is approximately π/2 − 0.0000352.

So while it would be wrong to conclude that B(n) = π/2 based on calculations for n ≤ 6, this conjecture would be approximately correct, never off by more than 0.0000352.

[1] David Borwein and Jonathan Borwein. Some Remarkable Properties of Sinc and Related Integrals. The Ramanujan Journal, 3, 73–89, 2001.

The post The Borwein integrals first appeared on John D. Cook.

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