# How $\zeta(0)=-\frac12$ is true?

I have proved the following four identities already but I can't prove $$\zeta(0)=-\frac12$$. The book has stated it as "an easy corollary". I couldn't find any complete proof in the internet because either there is none or uses formulas not from the following four (and not based on this): How $$\zeta(0)=-\frac12$$ is true?

• Try k=0 as a substitution into the equation photo. Sep 25 at 19:54
• @TymaGaidash, k are 1,2,3,... Sep 25 at 19:59
• Are you allowed to use that $\Gamma(s)$ and $\zeta(1-s)$ both have simple poles with residue $1$ at $s=0$? Because then $\zeta(0) = \pi^{-1/2}\cdot \Gamma(1/2)\cdot \lim_{s\to 0}\frac{\zeta(1-s)}{\Gamma(s/2)} = \pi^{-1/2}\cdot \pi^{1/2}\cdot-\frac12=-\frac12$ Sep 25 at 20:12
• @Mastrem, I know that Γ(s) and ζ(1−s) both have simple poles with residue 1 at s=0, but how $\lim_{s\to 0}\frac{\zeta(1-s)}{\Gamma(s/2)}=-1/2$ holds? Sep 25 at 20:23
• $-s\zeta(1-s) \rightarrow 1$ as $s$ goes to $0$, because $\zeta$ has a simple pole at $1$ with residue $1$. For the same reason, $\frac{s}{2}\Gamma(s/2) \rightarrow 1$. Now take the quotient. Sep 25 at 20:32

Take $$s \rightarrow 0$$. The LHS is equal to $$\frac{2}{s}(\zeta(0)+o(1))$$ while the RHS is equivalent to $$-\pi^{-1/2}\Gamma(1/2)\frac{1}{s}$$. But $$\Gamma(1/2)=\pi^{1/2}$$, thus $$2\zeta(0)+o(1) = -1$$, hence the conclusion.
• $\zeta(1+s) = \frac{1+o(1)}{s}$ as $s \rightarrow 0$, so $\zeta(1-s) = -\frac{1+o(1)}{s}$. For LHS, $\zeta(s) = \zeta(0)+o(1)$, $\pi^{-s/2} = 1+o(1)$, $\Gamma(s/2) = \frac{2}{s}\gamma(1+s/2) = \frac{2}{s}(1+o(1))$. Sep 25 at 20:01