Seven times half of π, and then not

The function $\operatorname{sinc} x = \sin x / x$ is a sine wave squeezed by a factor that shrinks like $1/x$: one tall central hump and ripples that die away. Its zeros are the sine’s except for the one at the origin, where the division fills the hole with the value 1. Its area, counting the ripples below the axis as negative and integrating all the way out, is a classical result:

$$\int_0^\infty \frac{\sin x}{x}\, dx = \frac{\pi}{2}.$$

In 2001 David and Jonathan Borwein published a sequence of integrals built from it. Multiply sinc x by sinc(x/3), the same wave stretched three times wider. The area of the product is still $\pi/2$. Multiply in sinc(x/5): still $\pi/2$. Then sinc(x/7), sinc(x/9), sinc(x/11), sinc(x/13) — seven factors in all — and every one of the seven integrals is exactly $\pi/2$. The eighth, with sinc(x/15) multiplied in, is

$$\frac{467807924713440738696537864469}{467807924720320453655260875000}\cdot\frac{\pi}{2},$$

short of $\pi/2$ by about $2.3 \times 10^{-11}$.

A pattern that holds seven times and fails on the eighth, by a margin in the eleventh decimal place, is the kind of thing that looks like an error in the computation. It is not, and the reason it holds — and the reason it fails exactly when it does — takes two lines once the right picture is drawn.

An area that exists only on the way out #

The first integral is already delicate. The function $\sin x / x$ is positive on its first hump and then alternates, each ripple a little smaller than the last, and the total area of the ripples without their signs is infinite: the ripples shrink like $1/x$, and $\int dx/x$ diverges, as the harmonic series does. So the area $\pi/2$ exists only as the limit of the signed area out to a growing bound — a conditionally convergent integral, of exactly the kind that Lebesgue’s theory of integration refuses to call integrable, though Riemann’s improper integral accepts it.

The products are better behaved. With two factors the ripples shrink like $1/x^2$, and from there on the integrals converge absolutely, as an endless region can have a finite area when its height falls fast enough. That is one more reason the Borwein integrals look as if they should be a smooth, stable family: from the second on they are absolutely convergent integrals of nicely decaying functions, each a small perturbation of the one before. Their values are computed without difficulty by any careful numerical method — which is how the failure at the eighth was first noticed, as a discrepancy in a numerical check rather than as a theorem.

Eight curves that look alike #

The curves change gradually. Each additional factor is close to 1 over the region where the product is large, because a wave stretched by a factor of fifteen has barely begun to fall where sinc x has finished its first hump. It flattens the ripples further out, where the product is already small. Looking at the curves gives no reason to expect anything special at the eighth factor, and no reason to expect seven exact equalities in the first place: each product is a different function, and each area is computed over an infinite range where the function oscillates for ever.

The equalities are not a coincidence of the curves. They are a coincidence of something the curves are hiding, and the way to see it is to take the Fourier transform.

The other side of the transform #

Fourier’s idea, that a function can be rebuilt from waves, has a sharp form for sinc. The function $\operatorname{sinc}(x/a)$ is, up to a constant, the transform of a box: a function equal to a constant on an interval of half-width $1/a$ and nought outside. And the transform turns multiplication into convolution — the operation that, for probability densities, adds independent random quantities.

Put the two facts together. The product $\operatorname{sinc}(x)\operatorname{sinc}(x/3)\cdots\operatorname{sinc}(x/(2n+1))$ is the transform of the convolution of $n + 1$ boxes, of half-widths $1, 1/3, 1/5, \ldots, 1/(2n+1)$. The area under a function is its transform’s value at nought. Working it through, the Borwein integral is

$$\int_0^\infty \prod_{k=0}^{n} \operatorname{sinc}\frac{x}{2k+1}\, dx = \frac{\pi}{2}\cdot P\left(\,\left|\,\frac{U_1}{3} + \frac{U_2}{5} + \cdots + \frac{U_n}{2n+1}\right| \le 1\right),$$

where the $U_k$ are independent random numbers spread evenly between $-1$ and $1$. The first factor, sinc x, became the window $[-1, 1]$; each later factor became a random step.

The box and its transform are a pair already met in another essay. A box repeated periodically is a square wave, and its Fourier coefficients fall like 1/n, with the signs and the four over π that a square wave’s series carries; those coefficients are samples of a sinc function. Here the box is not repeated, so the transform is not a list of coefficients but the whole sinc curve, and the dictionary between the two sides — a product on one side is a convolution on the other — is what turns an integral of a product of waves into a statement about adding boxes.

The steps reach past the window #

So the question is no longer about oscillating integrals. It is about a sum of independent steps of shrinking sizes and whether that sum stays within distance 1 of nought.

The largest the total can possibly be is the sum of the steps’ half-widths: $1/3$ for one step, $1/3 + 1/5 = 0.533$ for two, and $0.955$ for six. As long as that maximum is at most 1, the total lands in the window with certainty, the probability is exactly 1, and the integral is exactly $\pi/2$. Nothing approximate is involved; the integral equals $\pi/2$ because a certain event has probability one.

The seventh step changes that. $1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/15 = 1.0218$, just past 1. Now the total can leave the window — when all seven steps fall near their upper ends — and the probability is a little less than one. How little is the volume of a corner of a seven-dimensional box: the region where the seven uniform steps all fall so near their maxima that their sum exceeds 1, which is a tiny simplex with edges of order $0.0218$ divided by each step’s scale. Its volume is about $1.47 \times 10^{-11}$ of the whole, and that is the shortfall.

The densities in the figure also show something the integrals hide. With each step the density of the total becomes rounder, approaching a bell: the sum of independent steps tends to the normal distribution even when the steps have different sizes, as long as no one step dominates, and the rate at which the bell arrives is set by how unequal the steps are. Seven steps of sizes $1/3$ to $1/15$ already give a curve close to a bell, which is why its tail beyond 1 is so thin. The tail of a bell would never be exactly zero; the tail of a finite sum of bounded steps is exactly zero up to the maximum reach and only then begins — which is the whole difference between “exactly $\pi/2$” and “very nearly $\pi/2$”. Uniform steps that must stay below a threshold have appeared before, in the number of uniform draws needed to pass one, where the same corner simplices, of volume $t^n/n!$, gave the constant $e$.

The exact values #

The probabilities can be computed exactly, as fractions, because the distribution of a sum of uniform steps is a piecewise polynomial: the chance that it is at most $t$ is an alternating sum, over every pattern of signs, of $(t \pm a_1 \pm a_2 \pm \cdots)_+^n$, the positive parts raised to the power $n$. With the half-widths $1/3, 1/5, \ldots$ every term is a fraction, and the result is exact.

The eighth value, as a fraction of $\pi/2$, is the ratio of two thirty-digit integers given above, and it agrees with the value the Borweins published. After the threshold the shortfall grows quickly: the corner that pokes out of the window grows with each step, and its volume grows like a power of how far past 1 the reach has gone. The ninth integral is short by a part in a hundred million, the tenth by a part in a million.

The exact computation also explains why the first failure is so small. When the reach first exceeds 1 by a little, only one corner of the box — the one where every step is at its maximum — crosses the window, and its volume is $\delta^n / (n!\, \prod 2a_k)$ for an overshoot $\delta$: a high power of a small number, divided by a factorial. At seven steps $\delta = 0.0218$, and $0.0218^7$ is already about $10^{-12}$.

Doubling the window #

The Borweins found a second family that hides its failure much longer. Multiply every product by $2\cos x$ as well. The cosine’s transform is a pair of spikes at $\pm 1$, which shifts the window, and the integral becomes $\pi/2$ times the chance that the steps’ total stays within distance 2 of nought: the window is twice as wide.

The reach is a tail of the harmonic series, so it grows without bound, but slowly — about half the natural logarithm of the number of steps. It takes fifty-six steps to pass 2. So the integrals of $2\cos x$ times the first $k + 1$ sinc factors equal $\pi/2$ exactly for every $k$ from nought to fifty-five, and fail at fifty-six. The failure at fifty-six is the volume of one corner of a fifty-six-dimensional box with an overshoot of $0.0033$: about $10^{-138}$ of $\pi/2$. A numerical check to a hundred decimal places would find fifty-seven equalities and no failure. The cosine’s role is purely to move the window. Its transform is a pair of spikes at $\pm 1$, and convolving the boxes with those spikes slides the sum’s density sideways by one in each direction; averaging the two slid copies over the original window $[-1, 1]$ is the same as asking the unslid density to stay inside $[-2, 2]$. The trick depends on the slide being exactly the window’s half-width: the two slid windows, $[-2, 0]$ and $[0, 2]$, fit together into the doubled window with no gap and no overlap. Other multipliers produce other windows and other run lengths — evidence that the length of a run of equalities says nothing about the integrals themselves, only about where a hidden sum is placed relative to a boundary.

The rule, applied to other waves #

The explanation predicts which families of products have this property and for how long, without computing anything oscillatory.

With the factors $\operatorname{sinc}(x/2), \operatorname{sinc}(x/3), \ldots$ the steps are $1/2 + 1/3 = 0.833$ and then $1/2 + 1/3 + 1/4 = 1.083$, so the pattern fails at the fourth factor and was never convincing. With squares, $1/4 + 1/9 + 1/16 + \cdots$ never exceeds $\pi^2/6 - 1 \approx 0.645$, so every product in that family has area exactly $\pi/2$ — infinitely many equalities, proved by the same two lines. With powers of two, the steps $1/2 + 1/4 + \cdots$ approach 1 without ever passing it, and the area is $\pi/2$ for every finite product. The odd numbers are not special; they are merely the family with the longest run before the steps’ sum crosses the window — long enough to look like a law.

The sum that equals the integral #

The same Fourier dictionary produces a stranger coincidence, which Robert Baillie and the two Borweins described in 2008. Replace the integral by a sum over the whole numbers: $\sum_{n \ge 1} \sin n / n$. A sum and an integral of the same function have no reason to agree, and they do not quite: the sum is $(\pi - 1)/2$ and the integral is $\pi/2$, so the sum is exactly the integral minus one half. The same holds, exactly, for the product $\operatorname{sinc} n \cdot \operatorname{sinc}(n/3)$: its sum over $n \ge 1$ is $1.0707963\ldots$, the integral minus a half, to every decimal a careful computation reaches.

The reason is the same box. Summing a function over the integers is integrating it against a comb of spikes, and on the transform side a comb is again a comb, with teeth $2\pi$ apart. As long as the product’s transform — the convolution of the boxes — fits between two teeth, only the central tooth touches it, and the sum equals the integral apart from the half-weight at nought. That holds while the boxes’ total half-width stays below about $2\pi$, a far wider window than the integrals’ 1, so the sums agree with the integrals for every product in the odd-number family until its reach passes $2\pi - 1$, which takes about forty thousand factors. A sum and an integral that agree for a structural reason, and then stop, is the Borwein phenomenon a second time.

Small cases that lie #

The Borwein integrals are a standard example of a pattern that holds for a while and then stops, and the example is better than most because the failure is explained completely and in advance. That distinguishes it from the cases where the explanation comes only after the counterexample. The prime race has one: primes leaving remainder 3 on division by 4 lead those leaving remainder 1 for every bound up to 26,860 and then lose the lead, which was found by computation before anyone could say why the lead must eventually change hands. The patterns on the Ulam spiral and the Mertens conjecture, refuted with no counterexample ever exhibited, are others.

Here the counterexample is a theorem rather than a search. Anyone who has seen the Fourier picture can predict the threshold — the first $n$ with $1/3 + \cdots + 1/(2n+1) > 1$ — and the size of the first failure, without evaluating a single integral. The pattern’s seven-fold success was never evidence of anything except that the steps had not yet added up to 1.

Exact fractions and drawn curves #

The values in the table are exact. Each is a finite sum of rational numbers, computed in whole-number arithmetic, and the eighth matches the published fraction digit for digit. The cosine family’s shortfall at fifty-six steps is computed exactly too, from the volume of the one corner that crosses the window — a formula valid because the overshoot, $0.0033$, is smaller than every step, so no second corner can cross.

The densities in the box figure are computed numerically, by repeated averaging on a fine grid, and are drawn to show the shape and the support; the thin sliver of the seven-step density beyond 1 is far too small to see, and nothing quantitative is read from that figure. The identity between the integral and the probability is a theorem of Fourier analysis, not something the figures demonstrate; the integrands in the first figure are drawn, not integrated.

Still open: products that are not boxes #

The Borwein phenomenon is completely understood for products of sinc functions, because every factor’s transform is a box and the whole question reduces to sums of uniform steps. For products of other functions whose transforms have compact support — any function band-limited in Fourier’s sense — the same reasoning applies, and the integral equals its leading value until the supports’ total width passes the first function’s. What is not systematically understood is the analogue for functions whose transforms do not have compact support, where there is no threshold and the “pattern” decays gradually rather than breaking; and the inverse question, of which sequences of exact equalities in mathematics are of this kind — true because some hidden quantity has not yet crossed a boundary — has no general method, only a growing catalogue.

There is also an arithmetic question that the exact values raise. The eighth integral’s ratio to $\pi/2$ is a fraction with a thirty-digit denominator, and the denominators for later integrals grow rapidly; their prime factorisations are products of the odd numbers involved raised to high powers, which is clear from the formula, but their numerators — which encode the corner volumes — have no known structure.

A sum that had not yet reached 1 #

Seven integrals of oscillating functions over an infinite range come out to exactly $\pi/2$, and the eighth comes out to $\pi/2$ minus $2.3 \times 10^{-11}$. Seen through Fourier’s transform, each integral is $\pi/2$ times the probability that a random walk with steps $1/3, 1/5, 1/7, \ldots$ stays within distance 1 of its start. That probability is exactly 1 until the steps can add up to more than 1, which they first can at the seventh step, and the failure is the volume of a single corner of a seven-dimensional box.

So the pattern was never about the integrals. It was about $1/3 + 1/5 + \cdots + 1/13 = 0.955$ being less than 1 and $1/3 + \cdots + 1/15 = 1.022$ being more — and with a cosine attached, about $1/3 + \cdots + 1/111$ staying under 2 while $1/3 + \cdots + 1/113$ does not.