How to Be Wrong 56 Times in a Row

David Ding

June 11, 2026

In a previous post, we showed that the integral of \(\text{sinc}(x)\) over the entire real line equals 1. We also showed that the integral of \(\text{sinc}^2(x)\) equals 1. That already feels suspicious: multiplying by another copy of sinc apparently does not shrink the area at all. But then a Fourier transform calculation shows something that makes it worse:

\[ \int_{-\infty}^{\infty} \text{sinc}(x) \cdot \text{sinc}\left(\frac{x}{3}\right)\,dx = 1 \]

Still one. And if you keep adding factors with odd denominators, it keeps being one. Three factors, four factors, five, six, seven. Always one.

\[ \int_{-\infty}^{\infty} \text{sinc}(x) \cdot \text{sinc}\left(\frac{x}{3}\right) \cdot \text{sinc}\left(\frac{x}{5}\right) \cdot \text{sinc}\left(\frac{x}{7}\right)\,dx = 1 \]

At this point you start to suspect the universe is trolling you. So here is the question. Define:

\[ I_n = \int_{-\infty}^{\infty} \prod_{k=1}^{n} \text{sinc}\left(\frac{x}{2k-1}\right)\,dx \]

Is there any \(n\) for which \(I_n < 1\)? My first instinct was "probably not." Turns out the answer is yes, and the first failure happens in a place so absurd that it looks like a bug in whatever software you use to compute it. To see why, we are going to weaponize convolution.

Scaling in the Frequency Domain

We established last time that \(\text{sinc}(t) \leftrightarrow \text{rect}(f)\). But now we need the Fourier transforms of \(\text{sinc}(x/3)\), \(\text{sinc}(x/5)\), and so on. What does stretching a sinc in time do to its spectrum? We need the scaling property, and the derivation is quick. Suppose \(x(t) \leftrightarrow X(f)\). Then:

\begin{align} \mathcal{F}\{x(at)\} &= \int_{-\infty}^{\infty} x(at)\,e^{-j2\pi ft}\,dt \end{align}

Substitute \(\tau = at\), so \(dt = d\tau/a\):

\begin{align} &= \frac{1}{a}\int_{-\infty}^{\infty} x(\tau)\,e^{-j2\pi (f/a)\tau}\,d\tau = \frac{1}{a}\,X\!\left(\frac{f}{a}\right) \end{align}

(For \(a < 0\) you pick up an absolute value from the flipped limits.) The scaling property:

\[ \boxed{x(at) \longleftrightarrow \frac{1}{|a|}\,X\!\left(\frac{f}{a}\right)} \]

The intuition here is physical. If you compress a signal by a factor of \(a\) along the time axis, everything happens faster. The signal wiggles more rapidly, transitions become sharper, and the frequencies needed to represent those rapid changes are higher. A signal that used to fit within some bandwidth now requires \(a\) times as much bandwidth to describe. Compress in time, expand in frequency. Conversely, stretching a signal in time makes it lazier and slower, so it occupies less bandwidth. The area under the spectrum is preserved because the amplitude shrinks as the bandwidth grows.

Applied to sinc-rect, we get \(\text{sinc}(t/a) \leftrightarrow |a|\,\text{rect}(af)\): a rectangle of width \(1/a\) and height \(a\), always with area 1. So each of our sinc factors corresponds to a rectangle in the frequency domain. They get progressively narrower and taller:

\begin{align} \text{sinc}(x) &\longleftrightarrow \text{rect}(f) & &\text{width 1, height 1} \\ \text{sinc}(x/3) &\longleftrightarrow 3\,\text{rect}(3f) & &\text{width } 1/3\text{, height 3} \\ \text{sinc}(x/5) &\longleftrightarrow 5\,\text{rect}(5f) & &\text{width } 1/5\text{, height 5} \\ \text{sinc}(x/7) &\longleftrightarrow 7\,\text{rect}(7f) & &\text{width } 1/7\text{, height 7} \end{align}

The Plateau

Now here is where things click. Multiplication in time corresponds to convolution in frequency. And integrating a function over all of time is the same as evaluating its Fourier transform at \(f = 0\). Put those together:

\[ I_n = \Big[\text{rect} * 3\,\text{rect}(3f) * 5\,\text{rect}(5f) * \cdots * (2n{-}1)\,\text{rect}\big((2n{-}1)f\big)\Big]\bigg|_{f=0} \]

The entire integral reduces to one question: what is the value of this iterated convolution at zero? Convolve a bunch of rectangles, then read the height at the center. Let's trace through what happens.

Start with \(\text{rect}(f)\), a rectangle of width 1 and height 1. Now convolve it with \(3\,\text{rect}(3f)\), a narrow rectangle of width 1/3 and height 3. Remember what convolution does: flip one function, slide it across the other, and measure the overlap. When the narrow rectangle is sitting entirely inside the wide one, the overlap area is \(3 \times 1/3 = 1\). It does not matter exactly where the narrow rectangle is, as long as it fits. So the convolution has a flat plateau at height 1, extending as far as the narrow rectangle can slide while staying inside. That plateau has half-width \(1/2 - 1/6 = 1/3\).

The narrow rectangle (width 1/3, height 3) slides across the wide one. Whenever it fits completely inside, the overlap area is exactly 1.

Now convolve the result with the next rectangle: width 1/5, height 5. Same logic. As long as this even narrower rectangle fits within the existing flat plateau, the value at zero does not change. The plateau shrinks again (new half-width: \(1/3 - 1/10 = 7/30\)) but the height stays locked at 1. And the pattern continues: each new convolution nibbles at the edges of the plateau, making it narrower, but does not touch the center. The value at \(f = 0\) is protected by the flatness.

Each new factor smooths the edges and narrows the flat top. But the center stays stubbornly at 1.

The plateau half-width after \(n\) factors is:

\[ h_n = \frac{1}{2} - \frac{1}{2}\left(\frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cdots + \frac{1}{2n-1}\right) \]

The integral equals 1 as long as \(h_n > 0\). In other words, the magic holds as long as:

\[ \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cdots + \frac{1}{2n-1} < 1 \]

That is the whole game. The convolution keeps getting fancier, the shape keeps getting smoother, but the center stays flat. As long as the sum of half-widths of the nibbling rectangles has not eaten through the original rectangle's half-width, nothing changes at zero. The geometry is doing all the work.

So when does this finally fail? Let's run the numbers:

The first failure is at \(n = 8\). The first seven values are exactly 1. Not approximately. Not "close to." Exactly 1, with every decimal place locked in place forever. Then the eighth one quietly decides to become \(1 - 2.31 \times 10^{-11}\). The product of sincs through \(\text{sinc}(x/13)\) integrates to exactly 1. But the moment we include \(\text{sinc}(x/15)\):

\[ \int_{-\infty}^{\infty} \text{sinc}(x) \cdot \text{sinc}\left(\frac{x}{3}\right) \cdot \text{sinc}\left(\frac{x}{5}\right) \cdots \text{sinc}\left(\frac{x}{15}\right)\,dx < 1 \]

About twenty billionths. The drop is so small that most numerical computations would never notice.

Geometrically there is no mystery. For seven steps the nibbling stays within budget. On the eighth, the plateau gets chewed through, and the value at zero finally starts to drop. But the budget was so close to being met (the sum at \(n = 7\) is 0.9551, leaving just 4.5% of headroom) that the failure is almost invisibly small. You could stare at this integral for years and never notice.

A Fatter Rectangle

Now here is a natural follow-up question. The plateau broke because the first rectangle (width 1, half-width 1/2) ran out of room to absorb all the nibbling. What if we started with a wider rectangle? Can we make the pattern survive longer?

We can. And the tool is the modulation property. If \(x(t) \leftrightarrow X(f)\), then multiplying by a complex exponential just shifts the spectrum: \(x(t)\,e^{j2\pi f_c t} \leftrightarrow X(f - f_c)\). The derivation is one line (the exponentials just combine in the Fourier integral). Now use Euler's formula:

\[ x(t) \cdot 2\cos(2\pi f_c t) \longleftrightarrow X(f - f_c) + X(f + f_c) \]

Multiplying by \(2\cos\) takes the spectrum and places two copies at \(\pm f_c\). This is how AM radio works: your audio gets shifted up to a carrier frequency. But we are going to use it for something else entirely.

Consider what happens when we multiply \(\text{sinc}(x)\) by \(2\cos(\pi x)\). The spectrum of \(\text{sinc}(x)\) is \(\text{rect}(f)\), which lives on \([-1/2,\, 1/2]\). Modulating by \(2\cos(\pi x) = 2\cos(2\pi \cdot \frac{1}{2} \cdot x)\) places two copies at \(\pm 1/2\):

\[ 2\cos(\pi x) \cdot \text{sinc}(x) \longleftrightarrow \text{rect}(f - \tfrac{1}{2}) + \text{rect}(f + \tfrac{1}{2}) \]

The first copy lives on \([0,\,1]\) and the second on \([-1,\,0]\). Together they tile perfectly into a single rectangle of width 2, height 1, centered at zero. In other words:

\[ \boxed{2\cos(\pi x) \cdot \text{sinc}(x) \longleftrightarrow \text{rect}(f/2)} \]

This is the key insight. Multiplying sinc by \(2\cos(\pi x)\) doubles the width of its spectrum from 1 to 2. We have built ourselves a fatter rectangle. And a fatter rectangle means a bigger budget for the plateau.

Same game as before, but the first rectangle is now twice as wide. The narrow rect (width 1/3, height 3) fits comfortably inside for much longer. The flat region spans \(|f| \le 5/6\).

Now consider the integral:

\[ J_n = \int_{-\infty}^{\infty} 2\cos(\pi x) \cdot \prod_{k=1}^{n} \text{sinc}\left(\frac{x}{2k-1}\right)\,dx \]

We can group \(2\cos(\pi x) \cdot \text{sinc}(x)\) as a single factor whose spectrum is \(\text{rect}(f/2)\): width 2, half-width 1. The remaining factors are \(\text{sinc}(x/3)\), \(\text{sinc}(x/5)\), etc., with the same narrow rectangles as before. By the exact same convolution argument, the plateau holds as long as:

\[ \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cdots + \frac{1}{2n-1} < 2 \]

The budget doubled. Instead of the sum needing to stay below 1 (which it barely exceeded at \(n = 8\)), it now needs to stay below 2. That harmonic-like sum grows slowly, so it takes far longer to reach 2 than it took to reach 1.

Starting from the fat rectangle (width 2), the plateau narrows with each factor but takes far more steps to collapse. Compare with the original above.

The pattern survives until \(n = 56\). Fifty-six factors! The integral of \(2\cos(\pi x)\) times the product of sincs from \(\text{sinc}(x)\) through \(\text{sinc}(x/111)\) equals exactly 1. Add one more factor, \(\text{sinc}(x/113)\), and it finally drops. The budget that ran out at 8 factors before now lasts for 56.

And the margin of failure? If you thought \(I_8\) missing by \(10^{-11}\) was small, the cosine version makes that look enormous. At \(n = 57\), the half-width overshoot past the budget is only \(\delta \approx 0.0016\). But this tiny overshoot gets processed through 56 convolution steps, each one compounding the smallness. The deficit involves the factor \(\delta^{56} \cdot (3 \cdot 5 \cdot 7 \cdots 113) / 56!\), and when you work it out, the integral misses 1 by approximately \(10^{-138}\). That is not a typo. The integral misses 1 by a number with 138 zeros after the decimal point. For context, the number of atoms in the observable universe is about \(10^{80}\). This deficit is closer to the reciprocal of that number squared than to anything you could ever measure. You could compute this integral to a hundred decimal places and still see nothing but ones and zeros.

This is the same phenomenon, the same geometry, the same running-sum argument. The only thing that changed is the width of the first rectangle. The \(2\cos(\pi x)\) factor doubled it from 1 to 2, and that single change carried the plateau from 8 factors to 56. The pattern now holds for so long, and fails by so little, that calling it "approximately true" does not begin to do justice to how spectacularly close it is to being exactly true.

Closing Thoughts

What I love about this result is how the entire mystery collapses into a picture you could draw on the back of a napkin. A stack of rectangles getting convolved together, a flat plateau slowly being eaten from the edges, and a running sum that quietly ticks upward until it crosses a threshold. There is no trick, no contour integration, no clever substitution. Just convolution.

The first version gives you seven successes before failing on the eighth. The cosine version gives you fifty-six successes before failing on the fifty-seventh. Both are the same mechanism. Both are explained by the same one-line inequality. And both are examples of a pattern that looks like a theorem until it suddenly isn't one. Which is maybe the most dangerous kind of pattern in mathematics: the kind that's true often enough to fool you.

What still amazes me is how little of the sinc function actually mattered. The oscillations, the alternating signs, the infinite tails that make these integrals look so terrifying on paper: none of that determined when the miracle broke. In the end, the answer came from adding up fractions:

\[ \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cdots \]

Somewhere underneath all those oscillations was a geometry problem pretending to be analysis.