The Riemann zeta function can be defined by the counter integral (25) in herė: https://mathworld.wolfram.com/RiemannZetaFunction.html
Now, by using the Maclaurin series of$\frac{z}{e^z-1}$ in (25), I can compute $\zeta(-n)$ for $n\in\Bbb{N}$ by Residue theorem: $$\zeta(-n)=\frac{\Gamma(n+1)}{2\pi i}∮_{C}\frac{z^{-n-1}}{e^{-z}-1}dz=\frac{n!}{2\pi i}∑_{k=0}^{∞}∮_{C}(-1)^{k-1}\frac{B_k}{k!}z^{k-n-2}dz$$ $$=n! (-1)^n \frac{B_{n+1}}{(n+1)!}=(-1)^n\frac{B_{n+1}}{n+1}.\tag{1}$$
I can also compute $\zeta(n)$ for $n\in\Bbb{Z}^{+}$ in some cases by using the functional equation (14) in Mathworld's page: $$\zeta(n)=\Gamma(\frac{n}{2})^{-1}\pi^{\frac{n}{2}}\Gamma(\frac{1-n}{2})^{-1}\pi^{-\frac{1-n}{2}}\zeta(1-n)=\pi^{\frac{2n-1}{2}}\frac{\Gamma(\frac{1-n}{2})}{\Gamma(\frac{n}{2})}(-1)^{n-1}\frac{B_{n}}{n}.\tag{2}$$
$(2)$ makes sense only for $n=2k$. For odd $n$, we have $∞.0$ uncertainity. So, let $n=2k$. Then $$\zeta(2k)=\pi^{\frac{4k-1}{2}}\frac{\Gamma(\frac{1}{2}-k)}{\Gamma(k)}(-1)^{2k-1}\frac{B_{2k}}{2k}=\pi^{\frac{4k-1}{2}}\frac{(-1)^{k}2^{2k} k! \sqrt{\pi}}{(2k)!(k-1)!}(-\frac{B_{2k}}{2k})$$ And hence, $$\zeta(2k)=\frac{\pi^{2k}2^{2k-1}(-1)^{k+1}B_{2k}}{(2k)!}. $$
My questions are:
(Main question) Can we compute the values of the Riemann zeta function at positive even integers directly form a contour integral. But, this contour integral is not allowed to carry the functional equation information implicitly in it.
(Hard question) What can be done for the $∞.0$ uncertainity in $(2)$ mentioned above? Is it possible to get information about the values of Riemann zeta function at positive odd integers from this uncertainity?
I found this identity: For $m∈ℤ^+$, $$ζ(2m+1)=(-1)^{m}2^{2m+1}π^{2m}\left(∫_1^{∞}\frac{z^{-2m-1}}{e^{z}-1}dz+∑_{k=0}^{∞}\frac{B_{k}}{k!}\frac{1}{k-2m-1}\right)$$ by counter integral of $ζ'(s)$ and using this identity: $$ζ′(-2m)=\frac{(-1)^{m}(2m)!}{2^{2m+1}π^{2m}}ζ(2m+1)$$ for all $m∈ℤ^+$.