= Solution
Near zero the integrand is asymptotic to $t^{z-1}/2$, and at infinity it decays exponentially. The defining integral therefore converges and is analytic for \b[$\operatorname{Re}z>0$], by locally uniform domination, also allowing logarithmic factors for differentiation. Multiplication by the <entire function> $1/\Gamma(z)$ does not enlarge the domain of the unevaluated integral itself.
On $\operatorname{Re}z>1$, subtract the two convergent integrals and use
$$
\frac1{e^t+1}-\frac1{e^t-1}=-\frac2{e^{2t}-1}.
$$
The substitution $s=2t$ gives $I(z)-\zeta(z)=-2^{1-z}\zeta(z)$. Thus
$$
\boxed{I(z)=(1-2^{1-z})\zeta(z),}
$$
first there and then wherever the two analytic continuations agree. The <Riemann zeta function> is <meromorphic> on the plane with its only pole a simple pole at one. The prefactor has a simple zero there, $1-2^{1-z}=(z-1)\log2+O((z-1)^2)$. It cancels the pole, giving $I(1)=\log2$. Elsewhere both factors are <holomorphic>. \b[The <analytic continuation> of $I$ is entire]; it is the <Dirichlet eta function>.
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