The Borwein integrals

Posted on by John

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.

2 thoughts on “The Borwein integrals”

  1. Joe Kelleher

    If you don’t append sinc(x/15), are there instead other sinc(x/(2k+1)) terms you could append to the integral, for other values of k, that still keep the result at exactly pi/2 ?

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