The zeta function satisfies the functional equation:

$$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s),$$

where $Γ(s)$ is the Gamma function. This is an equality of meromorphic functions valid on the whole complex plane. My question is about how to actually do the arithmetic of the operation where a pole cancels with a zero. For instance, we know $\zeta(-4)=0$ but $\zeta(1+4)\neq0$.

$$\zeta(-4) = 2^{-4}\pi^{-5}\ \sin\left(\frac{-4\pi }{2}\right)\ \Gamma(5)\ \zeta(5)=0$$ $$\zeta(-4) = 2^{-4}\pi^{-5}\ \times\left(0\right)\ \times(5!)\times \zeta(5)=0$$ $$\zeta(-4) = \left(0\right) \times \zeta(5)=0$$

This works out exactly as expected. However, if we write

$$\zeta(5) = 2^5\pi^{4}\ \sin\left(\frac{5\pi }{2}\right)\ \Gamma(-4)\ \zeta(-4),$$ $$\zeta(5) = 2^5\pi^{4}\ \sin\left(\frac{5\pi }{2}\right)\ \Gamma(-4)\ \times (0),$$ $$\zeta(5) = \Gamma(-4)\ \times (0),$$

then I am not sure how to complete the arithmetic. What is the formula for cancelling the pole with the zero? Thanks!

  • 1
    $\begingroup$ $\zeta(s)$ has a simple zero at $s=-4$, and $\Gamma(s)$ has a simple pole there; so $\Gamma(s)\zeta(s)$ has a removable singularity.... $\endgroup$ Commented Jun 30, 2019 at 16:48

1 Answer 1


The correct way is to take the limit: $$\zeta(5)=2^5\pi^4\sin(5\pi/2)\lim_{s\to-4}\Gamma(s)\zeta(s).$$ Now we know that $s=-4$ is a pole of $\Gamma(s)$, and $$\lim_{s\to-4}(s+4)\Gamma(s)=\lim_{s\to-4}\frac{\Gamma(s+5)}{s(s+1)(s+2)(s+3)}=\frac{1}{24},$$ thus all you get is $\zeta(5)=\dfrac{4}{3}\pi^4\lim\limits_{s\to-4}\dfrac{\zeta(s)}{s+4}=(4/3)\pi^4\color{blue}{\zeta'(-4)}$. (Nothing more.)

  • $\begingroup$ Thank yo so much! This is perfect! $\endgroup$ Commented Jun 30, 2019 at 16:52
  • $\begingroup$ Double thanks on this one actually friendo! $\endgroup$ Commented Jun 30, 2019 at 17:59

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