= Solution
For $x>0$, $(e^x-1)^{-1}=\sum_{n\ge1}e^{-nx}$. If $\sigma=\operatorname{Re}s>1$, 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 $u=nx$ gives
$$
J(s)=\sum_{n\ge1}\int_0^\infty x^{s-1}e^{-nx}dx
=\boxed{\Gamma(s)\zeta(s).}
$$
On 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 $\operatorname{Re}s>1$ 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>, giving
$$
C(-k)=\frac{2\pi i(-1)^k}{k!}.
$$
At $s=-k$, 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, so
$$
I(-k)=2\pi i\operatorname{Res}_{z=0}\frac{z^{-k-1}}{e^z-1}
=2\pi i\,[z^{k+1}]\frac z{e^z-1}.
$$
The 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 $m\ge1$. For $k=2m-1$, the given generating function gives the coefficient $(-1)^{m-1}B_m/(2m)!$. Division by $C(-(2m-1))=-2\pi i/(2m-1)!$ yields
$$
\boxed{\zeta(1-2m)=\frac{(-1)^mB_m}{2m}\qquad(m\ge1).}
$$
The 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.
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