 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 $\mathrm{Re}z>0$, 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 $\mathrm{Re}z>1$, subtract the two convergent integrals and useThe substitution $s=2t$ gives $I(z)-\zeta (z)=-2^{1-z}\zeta (z)$. Thusfirst 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, $1-2^{1-z}=(z-1)\log 2+O((z-1{)}^2)$. It cancels the pole, giving $I(1)=\log 2$. Elsewhere both factors are holomorphic. The analytic continuation of $I$ is entire; it is the Dirichlet eta function.