The n-Category Café

You can show that

$\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, d t = \frac{\pi}{2} }$

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

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

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

and so on.

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

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

holds when

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

but fails for all

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

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 $\sin(c x)/(c x)$ is a step function supported on the interval $[-c,c]$.

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

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

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:

• Patterns that eventually fail.