The study of mathematics has long been steadfastly driven towards patterns. They seem to appear out of nowhere, and defy all sense and reason; we depend on the minds of curious thinkers to give them any logical backing at all. After all, human brains are designed to look for order and structure in our environments to predict what will come next. It gives us an edge over those that can’t. Stocking grain for winter becomes incredibly useful when you know that you cannot harvest wheat in December, and following herds is a lot easier when you know that they move up north for summer every year.
Mathematicians function in much the same way—we can’t deny that we do have biases towards pretty and unexpected results! Little mind is given to the patterns that don’t fit into our box of mathematical tricks, and we overlook what beauty they can hide in their spontaneity. Sometimes the patterns that break hold new secrets entirely, and show us new ways of viewing problems that may not have been apparent in the first place. But the last place we expect to see something like this is a normal looking integral.
That sinc-ing feeling
We can start with the sinc function across the real line, which is commonly used in signal analysis for signal reconstruction and frequency filtering.
$$
\operatorname{sinc}x=\frac{\sin(x)}{x}.
$$
We can use a nice trick to integrate sinc over the whole real line:
\[ \int_{-\infty}^\infty \operatorname{sinc}x \mathrm{d} x = \int_{-\infty}^\infty \frac{\sin(x)}{x} \mathrm{d}{x}.\]
Since both sine and $1/x$ are odd functions, multiplying them gives an even function that is symmetric across the $y$-axis. This means we can split our integral into two parts, which are both equal:
\begin{align*}
\int_{-\infty}^{\infty} \frac{\sin (x)}{x} \mathrm{d}{x} &=\int_{-\infty}^{0} \frac{\sin (x)}{x} \mathrm{d}{x}+\int_{0}^{\infty} \frac{\sin (x)}{x} \mathrm{d}{x}\\ &= 2\int_{0}^{\infty} \frac{\sin (x)}{x} \mathrm{d}{x}.
\end{align*}
This is a pretty well known integral, which we can evaluate to be $\pi/2$ using contour integration; so our whole integral comes out as $\pi$.
The next term in our pattern comes from multiplying $\operatorname{sinc}(x)$ by $\operatorname{sinc}(x/3)$: \[ \int_{-\infty}^\infty \frac{\sin(x)}{x} \frac{\sin(x/3)}{x/3}\mathrm{d}{x}.\] We can use the same trick of multiplying odd functions to end up with an even function, and once again we end up with $\pi$!
Now consider the whole integral class of this form where we have products of $\sin(x/a)/(x/a)$. Our pattern-prone brains immediately latch on to something that seems like it could be the start of a beautiful result in the making. We can even write out our integrals in a lovely triangular pattern:
\begin{align*} \int_{-\infty}^{\infty} \frac{\sin (x)}{x}\mathrm{d}{x}&=\pi, \\[2mm] \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{\sin (x / 3)}{x / 3}\mathrm{d}{x}&=\pi, \\[2mm] \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{\sin (x / 3)}{x / 3} \frac{\sin (x / 5)}{x / 5} \,\mathrm{d}{x}&=\pi, \\[2mm] \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{\sin (x / 3)}{x / 3} \frac{\sin (x / 5)}{x / 5} \frac{\sin (x / 7)}{x / 7}\mathrm{d}{x}&=\pi, \\[2mm] \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{\sin (x / 3)}{x / 3} \frac{\sin (x / 5)}{x / 5} \frac{\sin (x / 7)}{x / 7} \frac{\sin (x / 9)}{x / 9}\mathrm{d}{x}&=\pi, \\[2mm] \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{\sin (x / 3)}{x / 3} \frac{\sin (x / 5)}{x / 5} \frac{\sin (x / 7)}{x / 7} \frac{\sin (x / 9)}{x / 9} \frac{\sin (x / 11)}{x / 11}\mathrm{d}{x}&=\pi, \\[2mm] \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{\sin (x / 3)}{x / 3} \frac{\sin (x / 5)}{x / 5} \frac{\sin (x / 7)}{x / 7} \frac{\sin (x / 9)}{x / 9} \frac{\sin (x / 11)}{x / 11} \frac{\sin (x / 13)}{x / 13}\mathrm{d}{x}&=\pi. \end{align*}
They all equal $\pi$! And now the next term…
\begin{align*} &\int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{\sin (x / 3)}{x / 3} \frac{\sin (x / 5)}{x / 5} \frac{\sin (x / 7)}{x / 7} \frac{\sin (x / 9)}{x / 9} \frac{\sin (x / 11)}{x / 11} \frac{\sin (x / 13)}{x / 13} \frac{\sin (x / 15)}{x / 15}\mathrm{d}{x}\\[2mm] &=\frac{467807924713440738696537864469}{467807924720320453655260875000} \pi \approx \pi-4.62 \times 10^{-11}. \end{align*}
Our perfect pattern reaches an abrupt and rather anticlimactic end. It’s just off, but why?
The integrals were originally observed by late father–son pair of mathematicians Jonathan and David Borwein, who first evaluated it with an early model of some integral calculator software in 2001. Naturally, the small discrepancy in the calculation value looked like a numerical error on the side of the integral calculator, which, at the time, was in early stages of development and highly prone to errors. It was brushed off as a mistake and sent to the integral calculator company as a bug, although some stories suggest that Borwein knew the pattern, and sent it in as a practical joke (perhaps mathematicians have a sense of humour after all).
The whole pattern goes off in a completely different direction, getting further and further from $\pi$ each time.
Everything but the kitchen sinc
To understand how this works, we employ the Fourier transform. Essentially, this deconstructs our function into a sequence of sine and cosine waves, which is really helpful when we want to analyse a function further, or make certain parts of it easier to manipulate.
The sinc function.
To find the Fourier transformation of a function, $f$, we calculate the integral $$ \mathcal{F}(\xi)=\int_{-\infty}^{\infty} f(t) \mathrm{e}^{\mathrm{i} 2 \pi \xi}\mathrm{d}{\hspace{0.5pt}t}. $$ To get the integral of our original function back, we set $\xi=0$, which makes our exponent term 1. $$ \mathcal{F}(0)=\int_{-\infty}^{\infty} f(t) \cdot 1\mathrm{d}{\hspace{0.5pt}t} $$ The convolution theorem tells us that if two functions are multiplied in the integral inside a Fourier transform, we can deal with them as a convolution under the right conditions:
$$ \mathcal{F}(f_1\cdot f_2)=\mathcal{F}(f_1)*\mathcal{F}(f_2).$$
A really interesting geometrical implication of convolving two functions is that we get what we can essentially describe as a ‘moving average’ of the two functions, which is the result of a weighted blend of the two functions. A popular way to visualise it for the sinc function is as a window whose width is determined by $f_1$ passing over the graph of $f_2$, and the average of all the values inside the window at position $x$ becomes the height at $x$ on the graph of the convolution.
So, to grasp what’s actually happening to our sinc functions as we multiply them together inside the transform, we need to understand what happens once we take their transforms and convolve them with other sinc functions.
When we decompose our sinc function using a Fourier transform, we get the rectangle function: it’s a constant between $-1$ and $1$, and zero everywhere else.
When we transform $\operatorname{sinc}(x / 3)$, we get another rectangle that is three times as narrow and three times as tall, preserving the area under it. \[\mathcal{F}\left(\frac{\sin(ax/k)}{ax/k}\right)= {k(\operatorname{rect}(akx))}.\]
The Fourier transform of
$\operatorname{sinc}(x)$.
Let’s apply the convolution to both our $\operatorname{sinc}$ functions, and see where we can go from there. When we blend our two functions together, the width of our window corresponds to $\operatorname{sinc}(x / 3)$ and it slides across the rectangle function, averaging out the rectangle and smoothing it into a trapezium.
The Fourier transform of
$\operatorname{sinc}(x/3)\operatorname{sinc}(x/5)$.
For the next integral, we convolve this trapezium with $\operatorname{sinc}(x / 5)$, which translates to taking its moving average with another window. This time its width corresponds to $\operatorname{sinc}(x / 5)$, and is a little bit narrower, which takes the graph from the trapezium above to a rounded trapezium with a wider base that still fits inside the $(-1,1)$ interval from $\mathcal{F}( \operatorname{sinc}(x / 3))$, because \[\frac{1}{3}+\frac{1}{5}=\frac{8}{15}<1.\]
The Fourier transform of $\operatorname{sinc}(x / 3) \cdots \operatorname{sinc}(x / 15)$.
This works when we bring in $\operatorname{sinc}(x / 7)$, because \[\frac{1}{3}+\frac{1}{5}+\frac{1}{7}=\frac{71}{105}<1;\] but once we get up to $\operatorname{sinc}(x / 15)$, we have \[ \frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{15}=\frac{46027}{45045} > 1.\] Now our graph is suddenly smoothed out so much that it overflows beyond $(-1,1)$, and spills out of the limits that support the integral being equal to $\pi$.
Just one sinc-ular sensation
Now we can see that the odd coefficients that we introduced at the beginning were somewhat convincing red herrings, and this integral isn’t alone in its unconventional character, with sudden pattern breaks rather commonplace in other integrals involving similarly structured products of the $\operatorname{sinc}$ function, where all the coefficients add to less than one.
In fact, there are many alterations of the integral which allow our pattern to continue for much, much longer before tapering off.
For example, we can simply prepend a factor of $2 \cos (x)$ to our integrand to create a pattern that continues up to 113:
\begin{align*} \int_{-\infty}^\infty 2 \cos(x) \frac{\sin(x)}{x}\mathrm{d}{x} &= \pi, \\[2mm] \int_{-\infty}^\infty 2 \cos(x) \frac{\sin(x)}{x} \frac{\sin(x/3)}{x/3} \,\mathrm{d}{x} &= \pi, \\[0mm] &\hspace{0.75mm}\vdots \\[0mm] \int_{-\infty}^{\infty} 2 \cos (x) \frac{\sin (x)}{x} \frac{\sin (x / 3)}{x / 3} \ldots \frac{\sin (x / 110)}{x / 110} \frac{\sin (x / 111)}{x / 111}\mathrm{d}{x}&=\pi, \\[2mm] \int_{-\infty}^{\infty} 2 \cos (x) \frac{\sin (x)}{x} \frac{\sin (x / 3)}{x / 3} \ldots \frac{\sin (x / 111)}{x / 111} \frac{\sin (x / 113)}{x / 113}\mathrm{d}{x}&=\pi-4.6648\times10^{-138}. \end{align*}
This works because the cosine term increases the overtip value to 2. Our sequence \[1, \; 1+ \frac13, \; 1+ \frac13 + \frac15, \;\dots\] does (eventually) diverge to infinity, but it does so incredibly slowly, allowing the integrals to equal $\pi$ for longer.
The hills are alive with the sound of sinc-ing
The physicists Satya N Majumdar and Emmanuel Trizac proposed an alternative way to understand these integrals: by employing ‘random walkers’. These lost souls make random jumps in any direction and over any distance, as long as they follow some very specific rules.
A random George Walker Bush (Image: 030114-O-0000D-001
President George W. Bush. Photo by Eric Draper, White House.)
These models are incredibly helpful when it comes to modelling the randomness of Brownian motion and other particularly volatile behaviour patterns, such as share prices, gene reconstructions, social behaviour, and even Google’s PageRank algorithm for sorting websites.
We start our walkers with some basic rules:
- They all start at the origin of a one-dimensional space.
- They can move in either direction (left or right).
- At each step, the maximum distance they can travel is one of our odd coefficients: on the first step, they move a random distance between 0 and 1; for the second, they move a random distance between 0 and $1/3$, and so on. Within the interval, the distance chosen is uniformly distributed.
Our Walker taking their first step.
So, on the first step, all the walkers start at 0, and can move anywhere in the interval $[-1,1]$, and on the second step they can move by a distance up to $1 / 3$, or negative, towards $-1 / 3$, and so on.
Now we ask, after the $n$th step, how many walkers remain close to the origin?
Rather strangely, it seems to behave like the value of our $n$th Borwein integral.
For the first seven steps, the probability density (the probability of finding a walker at 0) is $1 / 2$.
This is because some of the walkers hopped right out to the edges, and are close to $1$ or $-1$ with their first step, and it is not possible for them to get back to 0 in only seven steps.
As soon as our walkers have taken more than seven steps, this is no longer the case: some will be able to return from $1$ and $-1$. Other walkers may not return, but instead go further, allowing the group to become more spread out, and the number of walkers near the origin just slightly drops.
As the number of walkers we start with reaches infinity, the drop in density approaches the value of the deviation of the Borwein integral from $\pi$ at the eighth step.
Just sinc about it
Borwein integrals reveal how patterns can emerge and break in surprising ways, showing the depth and complexity hidden within simple mathematical expressions. We can conjure them from the inner workings of wave transmissions, optical signals, and the whimsical dance of random walkers.
The integrals really are a beautiful reminder of the depth and unpredictability of mathematics. It isn’t often we find beauty in a chaotic system, but when we do, it’s always eye-opening (as well as just being nice to look at!). Picking apart the maths behind the breaks in patterns like this lift the curtains to let us see behind the scenes of our clockwork universe and pick out the cogs. Perhaps all we can do is hope that we will one day find the manual.
Infinitely many walkers.
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