I've been trying to play around with the product: $$\prod_{p \text{ prime}}\frac{1}{1-(-p)^{-1.5}}$$ Where the product runs over all the prime numbers. The product is similar in appearance to the famous Euler product for the zeta function, evaluated at $1.5$: $$ \prod_{p \text{ prime}}\frac{1}{1-p^{-1.5}}=\zeta(1.5)$$

However, the sign of $p$ is switched, so instead of staying in the reals, as the zeta function does, this new product enters the complex numbers. I have tried approaching it in an interesting way, however my results seem to differ greatly from numerical data, so I would like to know where the issue is and if an answer is known. My work is very largely based off the derivation of Euler's product formula:

We begin by defining: $$\zeta_{\mathbb{Z}}(s)=\ldots+(-3)^{-s}+(-2)^{-s}+(-1)^{-s}+1^{-s}+2^{-s}+3^{-s}+\ldots$$ Now, if we multiply by $(-2)^{-s}$, we get: $$ (-2)^{-s}\zeta_{\mathbb{Z}}(s)=\ldots+6^{-s}+4^{-s}+2^{-s}+(-2)^{-s}+(-4)^{-s}+(-6)^{-s}+\ldots$$ $$= \ldots+(-6)^{-s}+(-4)^{-s}+(-2)^{-s}+2^{-s}+4^{-s}+6^{-s}+\ldots$$ And by subtracting, we get: $$ \left( 1-(-2)^{-s} \right)\zeta_{\mathbb{Z}}(s)=\ldots+(-5)^{-s}+(-3)^{-s}+(-1)^{-s}+1^{-s}+3^{-s}+5^{-s}+\ldots$$ Similarly, by multiplying by $\left(1-(-p)^{-s} \right)$, we remove all the multiples of $p$, so by multiplying by $\left(1-(-p)^{-s} \right)$ for all prime numbers, we get rid of everything except for $(-1)^{-s}+1^{-s}$. Thus we have: $$ \prod_{p \text{ prime}}\left( 1-(-p)^{-s} \right) \cdot \zeta_{\mathbb{Z}}(s)=(-1)^{-s}+1^{-s}=(-1)^{-s}+1$$ Thus, whenever $(-1)^{-s}+1$ is not equal to zero, we can say: $$ \prod_{p \text{ prime}}\left(\frac{1}{ 1-(-p)^{-s} }\right)=\zeta_{\mathbb{Z}}(s)\cdot \frac{1}{(-1)^{-s}+1}$$

Now, on the other hand, by definition, we have: $$ \zeta_{\mathbb{Z}}(s) = \sum_{n=1}^{\infty}{n^{-s}}+\sum_{n=1}^{\infty}{(-n)^{-s}} =\sum_{n=1}^{\infty}{n^{-s}}+(-1)^{-s}\sum_{n=1}^{\infty}{n^{-s}}$$ $$ =\left( 1+(-1)^{-s} \right)\sum_{n=1}^{\infty}{n^{-s}}=\left( 1+(-1)^{-s} \right)\zeta(s)$$

Therefore we have: $$ \prod_{p \text{ prime}}\left(\frac{1}{ 1-(-p)^{-s} }\right)=\zeta_{\mathbb{Z}}(s)\cdot \frac{1}{(-1)^{-s}+1}= \left( 1+(-1)^{-s} \right)\zeta(s)\cdot \frac{1}{(-1)^{-s}+1}=\zeta(s)$$

By plugging in $s=1.5$, we get: $$\prod_{p \text{ prime}}\left(\frac{1}{ 1-(-p)^{-1.5} }\right)=\zeta(1.5)$$

The reason I don't believe this result is because it is an entirely real number, and also WolframAlpha tells me that the product over the first $50,000$ primes is approximately $0.62+0.68i$.

My question also extends to values of $s$ other than $1.5$. When $s$ is an even integer, the product formula is identical to the Euler product formula, so it is clear that the product is equal to the zeta function. When $s$ is an odd integer, my technique of analysis does not apply because $(-1)^{-s}+1$ does equal zero in this case, and in fact the product certainly is not equal to the zeta function. For example (according to WolframAlpha): $$ \prod_{p \text{ prime}}\left(\frac{1}{ 1-(-p)^{-3} }\right) = \frac{\pi^6}{945 \zeta(3)}\neq \zeta(3) $$

My question is, what about all the non-integer values (for example $1.5$)? I cannot find an issue with my work, however much of it is informal and I have never actually taken a course in complex analysis (which I assume would help). None of this is any sort of homework; it is just a piece of curiousity. Any help is appreciated.

  • 2
    $\begingroup$ What does it mean to raise a negative number to the $-\frac{3}{2}$ power?! $\endgroup$ Jun 24, 2015 at 19:26

1 Answer 1


Well, the first thing is that the quantity $(-p)^{-s}$ does not immediately make sense. However, one could interpret this as $e^{-\pi is}p^{-s}$ for $s>1$, and then it does make sense. The problem with your work then arises right from the beginning, when you multiply the series for $\zeta_\mathbb{Z}$ by $(-2)^{-s}$. This does not flip the series because, for instance, $(-2)^{-s}\cdot(-2)^{-s}=e^{-2\pi i s}4^{-s}\ne 4^{-s}$, unless $s$ is an integer.

Beyond this, I supppose if $s$ is rational, then one can get product formulas similar to the one you have above which says $\zeta(3)\cdot\prod(1-(-p)^{-3})^{-1}=\pi^6/945=\zeta(6)$. For instance, $$\zeta(3/2)\prod(1-e^{-(3/2)\pi i}(p)^{-3/2})^{-1}\prod(1-e^{-3\pi i}(p)^{-3/2})^{-1}\prod(1-e^{-(9/2)\pi i}(p)^{-3/2})^{-1}=\zeta(6).$$ These follow from the formula $$\prod_{i=0}^{n-1}(x-\zeta_n^i y)=x^n-y^n$$ where $\zeta_n$ is a primitive $n$th root of unity. Hope this helps.


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