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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.

For $\Re s>1$, absolute convergence permits separation into odd and even terms:Using the defining integral of the Gamma function and the substitution $u=nt$ givesTermwise integration is justified by absolute convergence when $\Re s>1$, and the geometric sum isConsequently the Dirichlet eta function satisfies

Near zero the integrand is $O(t^{\Re s-1})$, while at infinity it decays exponentially. The integral therefore defines a holomorphic function for $\Re s>0$. Thusprovides the desired continuation wherever the displayed denominator is nonzero. At a nonreal zero of $1-2^{1-s}$, use insteadwith an integer $k$ for which $1-k^{1-s}\ne 0$; the bounded partial sums of $a_n$ give convergence for $\Re s>0$. This shows that those apparent singularities are removable and yields the Analytic continuation of the Riemann zeta function to the right half-plane.

At $s=1$,whereasHenceso $s=1$ is a simple pole and its residue is