February 2026 – Approximaths

If we want to calculate $\zeta(-2)$ for example we need to have a proper value for $\Gamma(s/2)$ in for $Re(s) \leq 0$ . The problem is that the Bernoulli representation of the Gamma function presented above:

$\displaystyle \Gamma(s) = \int_{0}^{\infty} e^{-t} t^{s-1}\,dt$

is only valid for $Re(s) > 0$ . We have to provide another representation of the Gamma function for negative values of s:

$\Gamma(s) = \int_{0}^{\infty} e^{-t}\, t^{s-1}\, dt$

$= \int_{1}^{\infty} e^{-t}\, t^{s-1}\, dt + \int_{0}^{1} e^{-t}\, t^{s-1}\, dt$

$= \int_{1}^{\infty} e^{-t}\, t^{s-1}\, dt + \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} t^{n+s-1}\, dt$

$= \int_{1}^{\infty} e^{-t}\, t^{s-1}\, dt + \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int_{0}^{1} t^{n+s-1}\, dt$

$= \int_{1}^{\infty} e^{-t}\, t^{s-1}\, dt + \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \left[ \frac{t^{n+s}}{n+s} \right]_{0}^{1}$

$= \int_{1}^{\infty} e^{-t}\, t^{s-1}\, dt + \sum_{n=0}^{\infty} \frac{(-1)^n}{n!\,(n+s)}$

The sum $\sum_{n=0}^{\infty} \frac{(-1)^n}{n!(n+s)}$ exists for negative $s$ except for $s= -1,-2,-3,..$ . This implies that $\frac{1}{\Gamma(s/2)}$ is entire with simple zeros at $s= -2,-4,-6,...$ . The simple pole of $\xi(s)$ at zero is cancelled by the corresponding zero of $\frac{1}{\Gamma(s/2)}$ . As a consequence the only singularity of $\zeta(s)$ is a single pole at $s=1$ . Since