Sometimes you check just a few examples and decide something is always true. But sometimes even 1.5 × 10 43 1.5 \times 10^{43} examples is not enough.

You can show that

∫ 0 ∞ sin t t d t = π 2 \displaystyle{ \int_0^\infty \frac{\sin t}{t} \, d t = \frac{\pi}{2} }

∫ 0 ∞ sin t t sin ( t 101 ) t 101 d t = π 2 \displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, d t = \frac{\pi}{2} }

∫ 0 ∞ sin t t sin ( t 101 ) t 101 sin ( t 201 ) t 201 d t = π 2 \displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, \frac{\sin \left(\frac{t}{201}\right)}{\frac{t}{201}} \, d t = \frac{\pi}{2} }

∫ 0 ∞ sin t t sin ( t 101 ) t 101 sin ( t 201 ) t 201 sin ( t 301 ) t 301 d t = π 2 \displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, \frac{\sin \left(\frac{t}{201}\right)}{\frac{t}{201}} \, \frac{\sin \left(\frac{t}{301}\right)}{\frac{t}{301}} \, d t = \frac{\pi}{2} }

and so on.

It’s a nice pattern. But it doesn’t go on forever! In fact, Greg Egan showed the identity

∫ 0 ∞ sin t t sin ( t 101 ) t 101 sin ( t 201 ) t 201 ⋯ sin ( t 100 n + 1 ) t 100 n + 1 d t = π 2 \displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, \frac{\sin \left(\frac{t}{201}\right)}{\frac{t}{201}} \cdots \, \frac{\sin \left(\frac{t}{100 n +1}\right)}{\frac{t}{100 n + 1}} \, d t = \frac{\pi}{2} }

holds when

n < 15,341,178,777,673,149,429,167,740,440,969,249,338,310,889 n < 15,341,178,777,673,149,429,167,740,440,969,249,338,310,889

but fails for all

n ≥ 15,341,178,777,673,149,429,167,740,440,969,249,338,310,889 . n \ge 15,341,178,777,673,149,429,167,740,440,969,249,338,310,889 .

It’s not as hard to understand as it might seem; it’s a special case of the infamous ‘Borwein integrals’. The key underlying facts are:

The Fourier transform turns multiplication into convolution.

The Fourier transform of sin ( c x ) / ( c x ) \sin(c x)/(c x) is a step function supported on the interval [ − c , c ] [-c,c] .

The sum ∑ k = 1 n 1 100 k + 1 \displaystyle{\sum_{k = 1}^n \frac{1}{100k + 1}} first exceeds 1 1 when

n = 15,341,178,777,673,149,429,167,740,440,969,249,338,310,889 . n = 15,341,178,777,673,149,429,167,740,440,969,249,338,310,889.

For Greg’s more detailed explanation, based on that of Hanspeter Schmid, and for another famous example of a pattern that eventually fails, go here: