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I have this function:

$$f(x) = \prod_{p\text{ is prime}} \left(1 - \frac{x^2}{p^2}\right)$$

Now, this function can be said to be an infinite degree polynomial with zeros on each of the primes and their negatives and nowhere else.

Properties of $f(x)$ that I'm aware of:

  • $f(x)$ is an even function.
  • The zeroes of $f(x)$ are the set of prime numbers union the set of their negatives.
  • $f(x)$ is continuous.
  • $-\infty < f(x) < \infty \quad {\small \text{This stems from the previous two properties.}}$

Now, what I have so far is the following:

$$ \hat{f}(\xi) = \int_{-\infty}^{\infty} e^{-2\pi i x\xi}\left(\prod_{p\text{ is prime}} \left(1 - \frac{x^2}{p^2}\right)\right)\ dx $$

I can't manage to get any further.

Is there a way to get a closed form for the Fourier transform of this function?

Additional information:

$f(x)$ is known to have at least one non-zero point.

$$f(1) = \zeta(2)^{-1} = \frac{6}{\pi^2} \approx 0.6079$$

Axoren
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  • That's not a function define for any $x$. What is $f(0)$? – Thomas Andrews Feb 01 '15 at 02:44
  • Ugh, my mistake. It's technically only defined on the primes and it's undefined on everything else. I'm going to be correcting the question, there's a missing factor. – Axoren Feb 01 '15 at 02:47
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    The general form would be something like $\prod_p \left(1-\frac{x^2}{p^2}\right)$. – Thomas Andrews Feb 01 '15 at 02:49
  • That function works and still has the same properties that I outlined, I believe. I'll use that instead in the question rather than the mess I actually have (which is probably also still wrong). – Axoren Feb 01 '15 at 02:51
  • Hmm. Do you know if this function is in $L^1$ or $L^2$? – Cameron Williams Feb 01 '15 at 02:53
  • I'm not sure how you ensure that $f(x)\to 0$ as $x\to\infty$. – Thomas Andrews Feb 01 '15 at 03:11
  • I don't ensure that it's limit is $0$ as it approaches $\infty$. What I do ensure is that it will never be $\infty$. The reason for this is that for any zero-point, there is another zero-point on either side of it. If it were to be $\infty$ or undefined at any point in between, it would cease to be continuous. – Axoren Feb 01 '15 at 03:14
  • I guess I forgot to quote you and now it's too late to edit the post, @ThomasAndrews . There was some discussion in chat that led to the discovery of a non-zero point of the function. I'll be adding that to the question's text sometime today. – Axoren Feb 02 '15 at 13:21
  • Perhaps this can help: lmfdb.org – Klangen Dec 17 '18 at 11:21

0 Answers0