= Solution
The <Dirichlet series> for the <Riemann zeta function> converges for $\Re z>1$. In the Hankel <integral representation> the branch is a branch of $t^{z-1}$, so the argument restriction concerns $t$, not the parameter $z$. Use the principal branch cut along the negative real axis and a positive Hankel loop with a fixed small indentation around zero. Its integral $I(z)$ is an entire function of $z$, since the tails have exponential decay and the fixed contour avoids zero. Multiplication by $\Gamma(1-z)/(2\pi i)$ gives the <meromorphic continuation> of $\zeta$: it is directly defined away from positive integers, has a pole at $1$, and the apparent singularities at $2,3,\ldots$ are interpreted by limits.
For a check in the initial half-plane, collapsing the indentation for $\Re z>1$ gives
$$
I(z)=2i\sin(\pi z)\int_0^\infty\frac{r^{z-1}}{e^r-1}\,dr
=2i\sin(\pi z)\Gamma(z)\zeta(z).
$$
Expansion of $(e^r-1)^{-1}$ into exponentials justifies the last equality there, and the <Gamma reflection formula> recovers the printed <integral representation>.
Take the upper semicircle contour with the straight segment oriented from left to right, making the total orientation positive. Its enclosed poles are $t=2\pi ij$, $j=1,\ldots,N$; each has <residue> $-(2\pi ij)^{z-1}$. Thus
$$
\boxed{\int_\gamma\frac{t^{z-1}}{e^{-t}-1}\,dt
=-2\pi i(2\pi)^{z-1}e^{i\pi(z-1)/2}\sum_{j=1}^N j^{z-1}}.
$$
The reverse orientation changes the sign.
Initially take $\Re z<0$, so the series on the right converges as $N\to\infty$ and the large arcs can be discarded under the given assumption. Denote by $L_+$ and $L_-$ the integrals along horizontal lines just above and below the real axis, both from left to right. The <residue> computation and its lower-half-plane clockwise counterpart give
$$
L_+=-2\pi i(2\pi i)^{z-1}\zeta(1-z),\qquad
L_-=2\pi i(-2\pi i)^{z-1}\zeta(1-z).
$$
Contour deformation about the negative-axis branch cut gives $I=L_--L_+$. Hence
$$
I(z)=4\pi i(2\pi)^{z-1}\sin(\pi z/2)\zeta(1-z).
$$
Multiplying by $\Gamma(1-z)/(2\pi i)$ yields
$$
\boxed{\zeta(z)=2^z\pi^{z-1}\sin(\pi z/2)\Gamma(1-z)\zeta(1-z)}.
$$
It extends by <meromorphic continuation> to all $z\ne1$, with removable products evaluated by limits where individual factors diverge. The derivation does not assert ordinary convergence of the original positive-term series outside $\Re z>1$.
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