16
$\begingroup$

According to the paper "Ten Problems in Experimental Mathematics",

$$\int_0^\infty \cos(2x)\prod_{n=1}^\infty \cos\left(\frac{x}{n}\right)dx \quad = \quad \frac{\pi}{8}\color{blue}{-7.407 \times 10^{-43}}$$

The article goes into some detail on how to compute the integral numerically in order to verify that the LHS is not strictly equal to $\pi/8$, but no theoretical explanation is given for why they are so close.

The extremely high accuracy to which this relation holds leaves a strong feeling that it is more than a mere "mathematical coincidence", in the same sense that it is not a coincidence that $e^{\pi\sqrt{163}}$ is almost an integer.

I am looking for an insight that can support that feeling.

$\endgroup$
1
  • $\begingroup$ Unless someone computes an infinite series for the expression, I don't think it can really be proved. it may help if you know any other similar example. $\endgroup$
    – ghosts_in_the_code
    Oct 31, 2014 at 15:48

2 Answers 2

Reset to default
7
$\begingroup$

That is known as a "Borwein Integral", named after one of the authors of the paper you linked.

https://en.wikipedia.org/wiki/Borwein_integral

Wikipedia has some references which explain what is going on. Here is one of them.

http://schmid-werren.ch/hanspeter/publications/2014elemath.pdf

I would have left this as a comment but this site requires 50 reputation for that.

$\endgroup$
1
  • $\begingroup$ Nice! That paper explains also how to handle the extra factor $\cos 2x$. $\endgroup$
    – Hans Lundmark
    Nov 1, 2014 at 10:35
5
$\begingroup$

This is not a complete answer, but I followed their advice "see [16, chap. 2] for additional details" (that's the book Experimentation in Mathematics). In section 2.5.2 they show that the cosine product (without the factor $\cos 2x$) equals $\prod_{k=0}^{\infty} \operatorname{sinc}\left(\frac{2x}{2k+1}\right)$, whose integral can be computed using the Fourier transform, giving an answer $\pi_1$ which is a bit less than $\pi/4$.

(Ultimately, this is because the Fourier transform of the sinc function is the characteristic function of a symmetric interval around the origin, and if you take the convolution product of several such functions, it will "gnaw" its way into the origin if you have sufficiently many factors with sufficiently big intervals, making the value there less than $1$. See page 22 in the paper, and this answer on MO; don't miss the comments!).

For the integral $\pi_2$ (with the factor $\cos 2x$), they don't give any details (it's just an exercise on page 124 to prove that $\pi_2 < \pi/8$), but I presume it's pretty similar. To see just why the deviation from $\pi/8$ becomes so very tiny, one would have to work out exactly what the corresponding sinc product is in that case. Perhaps if you wait a while, someone will do it and give a fuller answer here.

$\endgroup$
1

You must log in to answer this question.