π/2, π/2, π/2 … and Then Not

The function $\mathrm{sinc}(x)=\sin (x)/x$ is a wave that starts at 1 and ripples away to nothing. The area under it, counted from 0 out to infinity, with the parts below the axis counted as negative, is exactly $\pi /2$. That is a classical result, and the curve in the laboratory's first view is the thing being measured.

Now multiply in a second wave, $\mathrm{sinc}(x/3)$, which is the same shape stretched out three times as wide. The product is a different curve, and the area under it is $\pi /2$ again. Multiply in $\mathrm{sinc}(x/5)$, then $\mathrm{sinc}(x/7)$, always the next odd denominator, and the area under the product is still exactly $\pi /2$. Each new factor reshapes the curve everywhere, and each time the area comes out the same.

It does that for a long run of factors, long enough that most people would bet it goes on for ever. It does not. At one particular factor the area is a hair less than $\pi /2$. The hair is about one and a half parts in a hundred billion of the answer, which is far smaller than a pixel, a printed digit or the precision of most calculators, so neither a picture nor an ordinary numerical integration can tell that anything happened.

These are the Borwein integrals. The laboratory computes every one of them exactly, as a fraction of $\pi$ in whole numbers, which is the only way to see the gap at all. It also shows what decides where the run ends, and the answer is not a coincidence of the numbers but a small, checkable piece of arithmetic.

Here is a way to read the integral that makes the answer almost visible. The product of waves corresponds, by a standard change of viewpoint, to adding up random numbers. Each factor $\mathrm{sinc}(x/d)$ stands for a random number chosen evenly between $-1/d$ and $1/d$. A factor with a large denominator is a very small wobble; the first factor, $\mathrm{sinc}(x)$, is the largest, a number between $-1$ and $1$.

The area under the whole product then turns out to be $\pi /2$ times a probability. Draw the later factors' random numbers, add them up, and ask how often the total is between $-1$ and $1$, the range the first factor allows. The area is $\pi /2$ multiplied by that fraction of the time.

Now see why it is exactly $\pi /2$ so often. If the later random numbers, each at its very largest, still add up to less than 1, then the total can never leave the range. It is between $-1$ and $1$ every single time, the fraction is exactly 1, and the area is exactly $\pi /2$. Adding another factor adds a smaller wobble, and so long as the biggest possible total stays under 1, nothing changes at all.

The biggest possible total is just $1/3+1/5+1/7+\dots$, which is the running total in the plate above. When it finally goes over 1, a very small fraction of draws, those where every random number lands near its extreme at once, push the total out of range. The fraction is no longer exactly 1, and the area falls short of $\pi /2$ by exactly that tiny chance.

An exact pattern can have an expiry date that is hard to see coming. Every integral up to the break is exactly $\pi /2$, with nothing approximate about it, and the pattern still ends.

The expiry is set by one slowly growing quantity. The total $1/3+1/5+1/7+\dots$ grows without limit, because the odd reciprocals add up like half the natural logarithm of the last denominator, and the logarithm is a very patient function. It must pass 1, so the pattern must end, and it does so only after a run long enough to look like a law.

The extra factor $2\cos (x)$ makes the length of the run easy to compare. It leaves the integral at $\pi /2$ for as long as the total stays under 2, not 1. Because the total grows like a logarithm, moving the line from 1 to 2 does not double the run; it multiplies the last denominator by roughly $e^2$, about seven and a half times. That is a long way to carry a pattern, and the laboratory shows it ending all the same.

A pattern that survives a hundred checks has told you a lot, and may still have told you nothing about the case that matters. Here the case that matters arrives at a factor where nothing in the picture, the area or the first ten digits looks any different.

Let $a_0=1$ and let $d_1,d_2,\dots ,d_n$ be positive numbers, with $n\ge 1$. Let $R=U_1+\cdots +U_n$ where the $U_j$ are independent and $U_j$ is uniform on the interval $(-1/d_j,1/d_j)$.

1. $${\int }_0^{\infty }\mathrm{sinc}(x)\prod _{j=1}^n\mathrm{sinc}(x/d_j)dx=\frac{\pi }{2}\mathrm{Pr}(|R|\le 1).$$ Proved, in layer 5.
2. The probability is exactly 1, so the integral is exactly $\pi /2$, if and only if $1/d_1+\cdots +1/d_n\le 1$. Proved, in layer 5, since $|R|$ cannot exceed that sum and can come as close to it as you like.
3. For the odd denominators $1,3,5,\dots$: the integral is exactly $\pi /2$ for the products that stop at 1, 3, 5, 7, 9, 11 and 13, and for the product that stops at 15 it equals $$\frac{467807924713440738696537864469}{935615849440640907310521750000}\pi ,$$ which is $\pi /2$ less about $2.31\times 10^{-11}$. Proved, from 1 and 2 and exact rational arithmetic on a finite list of fractions.
4. With the extra factor $2\cos (x)$ the same argument applies with the line at 2 in place of 1. The integral is $\pi /2$ for products stopping at any odd denominator up to 111, and falls short at 113. Exhaustively checked over the odd denominators 1 to 119, exactly, and cross-checked against numerical integration where it can be run.

Statement 3 is what the page is for; 2 is the reason; 4 shows the reason carrying over to a larger setting, and is labelled checked rather than proved only because the page does not write out the extra step in full.

The transform of sinc. For $a>0$, the Fourier transform of $\mathrm{sinc}(ax)$ is

a rectangle of height $\pi /a$ on $(-a,a)$. That is $2\pi$ times the density $u_a(\omega )=1_{|\omega |<a}/(2a)$ of a uniform random number on $(-a,a)$.

Products become convolutions. With two or more factors the product is absolutely integrable, since it falls off like $1/x^2$ or faster, and the transform of a product of integrable functions is $(2\pi {)}^{-n}$ times the convolution of their transforms. Each transform is $2\pi u_{a_j}$, so with $n+1$ factors

where $g=u_{a_0}\ast \cdots \ast u_{a_n}$ is the density of the sum of the independent uniform numbers. The product is an even function of $x$, so the integral over $(0,\infty )$ is half of that: $\pi g(0)$.

Isolating the first factor. Take $a_0=1$ and write $\rho$ for the density of $R$, the sum of all the other uniform numbers. Then $g(0)=\int u_1(-r)\rho (r)dr$, and $u_1=\frac{1}{2}{1}_{|r|<1}$, so $g(0)=\frac{1}{2}\mathrm{Pr}(|R|\le 1)$. The integral is $\frac{\pi }{2}\mathrm{Pr}(|R|\le 1)$, which is statement 1. ∎

The threshold. Each $U_j$ lies between $-1/d_j$ and $1/d_j$, so $|R|\le B=1/d_1+\cdots +1/d_n$, and if $B\le 1$ the probability is 1. If $B>1$, then the event that every $U_j$ is within a small distance of its upper end has positive probability, and on it $R>1$, so the probability is below 1. That is statement 2. ∎

The exact value. By symmetry the shortfall is $\mathrm{Pr}(|R|>1)=2\mathrm{Pr}(R>1)$, and for a sum of independent uniform numbers of widths $w_j=2/d_j$ the tail has the classical inclusion-exclusion form

where $(\cdot {)}_{+}$ is the part above zero. Only the subsets $E$ with ${\sum }_{j\in E}{w}_j<B-1$ contribute, and when $B$ has only just passed 1 that is only the empty set. Multiplying everything by the common denominator turns each term into a whole number, so the sum is exact. For the product ending at 1/15 the sum of the later reciprocals is $B=1.0218\dots$, only the empty set contributes, and the formula gives $\mathrm{Pr}(R>1)=(B-1{)}^7/(7!{\prod }_j{w}_j)$ in exact fractions. The integral is $\pi /2-\pi \mathrm{Pr}(R>1)$, which is the fraction in statement 3. ∎

The cosine. The transform of $2\cos (x)$ is two point masses of weight $2\pi$, one at each of $\omega =\pm 1$. That is $4\pi$ times the law of a fair coin flip between $-1$ and $1$, in place of $2\pi$ times a density. Following the proof through, the first factor's window is now centred on the coin's result, so the sum of the other numbers must land in $[0,2]$ rather than $[-1,1]$, and the integral is $\pi \mathrm{Pr}(0\le R\le 2)=\frac{\pi }{2}\mathrm{Pr}(|R|\le 2)$, with the threshold at 2. The laboratory checks this against numerical integration of products of sinc functions whose sums reach past 2; the check is a test, not the written proof, which is why statement 4 is marked checked. ∎

What the numbers on this page are, and are not. Every value in the table is an exact fraction of $\pi$ in whole-number arithmetic with no rounding at any stage. The curves in the first view are ordinary floating-point drawings, and the running area is a numerical integration that stops at a finite distance. Neither can show the gap, since the first shortfall is about $10^{-11}$, and a drawing that claimed to would be wrong. That is why the integral's exactness is stated from the table, never from the picture.

What the exact computation costs. The inclusion-exclusion sum has one term for every subset of the later factors whose widths add up to less than $B-1$. Just past the threshold that is a handful. Far past it, the number of subsets grows quickly, so the laboratory offers the plain product up to 18 factors and the version with $2\cos (x)$ up to 60, where the shortfall is still tiny, and refuses to start a sum that would take more than four hundred thousand steps. A second implementation that sums over every subset without that shortcut agrees with the fast one on every product up to ten factors.

What is not proved here. Two things. The cosine version is checked and not written out in full. And the table shows each shortfall larger than the one before, which is observed over the range the page computes and is not claimed for all products.

Where the mathematics goes next. The family does not stop at one extra cosine. There are whole sequences of related integrals and sums, and some of them break in surprising places; the cited papers set out how far the pattern can be extended and what governs each break. This page follows one thread from the start of that story to its first exact failure.

Read this next to 1, 2, 4, 8, 16, and Then Not 32. That page is the same lesson on a scale you can draw, with a count that parts company with doubling at a visible step. Here the step is invisible, and exactness is what lets you see it.