 Solution

For $x>0$, $(e^x-1{)}^{-1}={\sum }_{n\ge 1}{e}^{-nx}$. If $\sigma =\mathrm{Re}s>1$, the integral of the sum of absolute values is $\Gamma (\sigma )\sum n^{-\sigma }<\infty$. Therefore interchange of sum and integral is justified, and the substitution $u=nx$ givesOn the two unbounded rays of the specified Hankel contour, $x\ge \pi$ and $1/(e^x-1)$ decays exponentially, dominating every fixed power $x^{\sigma -1}$. The branch multiplier on the lower ray is a fixed factor $e^{2\pi i(s-1)}$. On the circular part, the chosen branch is continuous along the parametrized arc, and $e^z-1$ never vanishes because its nonzero zeros have modulus at least $2\pi$. Thus $I(s)$ exists for every $s$. The estimates are uniform when $s$ ranges over a compact set, including after differentiation in $s$, so $I$ is an entire function.

Let $C(s)=(e^{2\pi is}-1)\Gamma (s)$. The given identity shows that $I(s)/C(s)$ agrees with $\zeta (s)$ for $\mathrm{Re}s>1$ wherever the quotient is directly defined. It supplies the meromorphic continuation elsewhere wherever $C$ is finite and nonzero, with removable points filled by limits. At a nonpositive integer $-k$, the simple zero of the exponential factor cancels the supplied gamma pole, givingAt $s=-k$, the integrals on the two rays cancel because the integrand is single-valued. The remaining positively oriented circle has only the pole at zero inside it, soThe coefficient at $k=0$ is $-1/2$, hence $\boxed{\zeta (0)=-1/2}$. For positive even $k$, the relevant odd coefficient has degree at least three and is zero, giving $\boxed{\zeta (-2m)=0}$ for $m\ge 1$. For $k=2m-1$, the given generating function gives the coefficient $(-1{)}^{m-1}{B}_m/(2m)!$. Division by $C(-(2m-1))=-2\pi i/(2m-1)!$ yieldsThe symbols $B_m$ here follow the printed generating function: they are the positive magnitudes of the usual even-index Bernoulli numbers, so $B_1=1/6$, $B_2=1/30$. This convention accounts for the alternating sign in the final expression.