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Keep the circular portion of a Hankel contour at fixed nonzero radius. Its numerator $I(z)={\int }_H{t}^{z-1}{e}^{t}dt$ is an entire function of $z$, because the contour avoids zero and its tails decay exponentially, uniformly on compact parameter sets. At a positive integer $n$, the integrand is single-valued and entire in $t$, so the bank integrals cancel and the circular integral is zero. Thus $I(n)=0$, canceling the simple zero of $2i\sin (\pi z)$ in the Hankel representation of the gamma function. At a nonpositive integer $-n$, the circular integral instead gives $2\pi i/n!$, yielding residue $(-1{)}^n/n!$.

For $x>0$, $(e^x-1{)}^{-1}={\sum }_{n\ge 1}{e}^{-nx}$. If $\sigma =\mathrm{Re}s>1$, 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 $u=nx$ givesOn 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 $\mathrm{Re}s>1$ 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, givingAt $s=-k$, 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, soThe 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 $m\ge 1$. For $k=2m-1$, the given generating function gives the coefficient $(-1{)}^{m-1}{B}_m/(2m)!$. Division by $C(-(2m-1))=-2\pi i/(2m-1)!$ yieldsThe 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.

Fix a Hankel contour with a positive-radius circle around zero and a branch $-\pi <\arg t<\pi$, traversing the lower bank inwards and the upper bank outwards. Write its numerator as $I(z)={\int }_H{t}^{z-1}{e}^{t}dt$. For fixed circle radius, $I$ is an entire function of $z$: the finite contour pieces avoid zero, and the tails converge uniformly on compact $z$ sets by exponential decay, also after differentiation with respect to $z$.

At $z=-n$, the two straight-bank contributions cancel because $t^{-n-1}{e}^t$ is single-valued. The remaining circle is positively oriented, so the residue theorem and the exponential series giveSince $\sin (\pi z)=(-1{)}^n\pi (z+n)+O((z+n{)}^3)$, the quotient has the simple poleFor cancellation of positive-integer gamma singularities on a Hankel contour, at a positive integer $n$, the integrand $t^{n-1}{e}^t$ is entire and its bank contributions again cancel; its circular integral is zero. Hence $I(n)=0$. The numerator is analytic near $n$, while the denominator has a simple zero with nonzero derivative. Factoring $I(z)=(z-n)J(z)$ therefore cancels that zero, proving there is no pole at a positive integer using only this representation. Cancellation of the numerator, not merely the zero of the sine, decides the issue. One can also shrink the contour for $\Re z>0$, obtaining $I(z)=2i\sin (\pi z){\int }_0^{\infty }{u}^{z-1}{e}^{-u}du$ and hence the finite value $(n-1)!$.

Take a Hankel contour $C_{\eta }$ around the positive real axis, going from infinity on its lower side toward a circle of radius $0<\eta <1$, clockwise around that circle, then out along the upper side. Use $0\le \arg z\le 2\pi$ and defineThis is an entire function of $t$: the small circle stays away from zero and the pole at $-1$, while the exponentially decaying rays converge uniformly for $t$ in compact subsets. Contour deformation shows independence of $\eta$. For $\Re t>-1$, let $\eta \to 0$. The upper ray contributes $F(t)$, the lower ray contributes $-e^{2\pi it}F(t)$, and the small-circle contribution vanishes. Thus the analytic continuation isThe only apparent singularities are at integers. At the nonnegative integers they are removable because the original integral is holomorphic throughout $\Re t>-1$. Hence singularities can occur only at $-1,-2,\dots$.

In fact all of those points are simple poles. At $t=-n-1$ the two rays cancel and the clockwise circle gives $H(-n-1)=-2\pi i{c}_n$, whereSince the denominator has derivative $-2\pi i$ at each integer, the residue at $-n-1$ is $c_n$. The parameter has been called $t$ consistently; the final use of $z$ for this parameter in the printed question is harmless notation.

For $\mathrm{Re}s>1$, expand $(e^t-1{)}^{-1}={\sum }_{n\ge 1}{e}^{-nt}$. Absolute convergence of the integrated series follows from the same expression with $\mathrm{Re}s$ in place of $s$. Integrating termwise and substituting $u=nt$ givesThis proves the Mellin transform representation of the Riemann zeta function.

Take the Hankel contour along the lower side of the negative axis towards zero, around zero counterclockwise, and back along the upper side. For $\mathrm{Re}s>1$ the small-circle contribution tends to zero. On the lower and upper rays, respectively, $\arg t=-\pi ,+\pi$, so the two integrals combine toThe gamma reflection formula now proves agreement of the two representations. At an integer $s=m\ge 2$, the contour integral has a zero while $\Gamma (1-s)$ has a pole: their product is interpreted by its finite limit, equal to $\zeta (m)$, not by multiplying literal zero and infinity.

For nonpositive integers the integrand is single-valued, and the contour closes around zero. Sincethe residue of $t^{-1}/(e^{-t}-1)$ is $-1/2$, giving $\zeta (0)=-1/2$. Alsois odd. Its Laurent series therefore has only odd powers. Multiplication by $t^{-2n-1}$ cannot produce a $t^{-1}$ term, and the subtracted constant cannot do so when $n\ge 1$. Hence the contour integral vanishes and $\zeta (-2n)=0$ for every positive integer $n$.

For $\mathrm{Re}s>1$, expand $(e^t-1{)}^{-1}={\sum }_{n\ge 1}{e}^{-nt}$. Absolute integrability, bounded near zero by a constant times $t^{\mathrm{Re}s-2}$, permits termwise integration. The Gamma function integral then givesSpecify the Hankel contour to run from the negative real axis below the cut towards zero, circle zero counterclockwise with fixed radius $0<r<2\pi$, and return above the cut. Take $-\pi <\arg t<\pi$. The circle excludes every nonzero pole $2\pi i\mathbb{Z}$ of the denominator. For $\mathrm{Re}s>1$ its radius can shrink to zero, and the lower and upper rays give respectively $-e^{-i\pi s}\Gamma (s){\zeta }_R(s)$ and $e^{i\pi s}\Gamma (s){\zeta }_R(s)$. By the Gamma reflection formula,away from integer $s$ in that half-plane, with limiting equality at the removable points.

With the circle radius kept fixed, the contour integral is an entire function of $s$: the rays decay exponentially and converge locally uniformly, and the circular part stays away from zero. Multiplication by $\Gamma (1-s)$ is analytic for $\mathrm{Re}s<1$ and meromorphic globally. The apparent singularities at positive integers $s\ge 2$ are removable because the integral vanishes there and it already agrees with the original analytic zeta function nearby. Only $s=1$ remains a simple pole. Thus this constructs the meromorphic continuation to the whole plane and an analytic continuation on $\mathbb{C}\setminus \{1\}$. Keeping the circular segment is essential when the integral is not locally integrable at zero.

At $s=-1$ the two rays cancel, since $t^{-2}$ is single-valued. The Laurent expansionshows that the residue of $t^{-2}/(e^{-t}-1)$ is $-1/12$. Since $\Gamma (2)=1$, the residue theorem gives

Take the Hankel contour about the negative real axis: it starts at $-\infty$ just below the cut, circles the origin counterclockwise, and returns to $-\infty$ just above the cut. The notation $(0+)$ records this return side of the branch point. Choose the branch $-\pi <\arg t<\pi$ for $t^{z-1}$.

For $\mathrm{Re}z>1$, collapse the contour onto the two banks of the cut. Their phase difference is determined byand the resulting sine factor, together with the Gamma reflection formula $\Gamma (z)\Gamma (1-z)=\pi /\sin (\pi z)$, recovers the integral representation (‡). The Hankel integral converges exponentially at its unbounded end and, after the standard local subtraction of the singular part at zero, depends holomorphically on $z$. The apparent poles of $\Gamma (1-z)$ are cancelled by zeros of the contour integral except at $z=1$. It therefore gives the analytic continuation of the Riemann zeta function to $\mathbb{C}\setminus \{1\}$.

For $|z|<1$ on the negative-real portions of the Hankel contour, expandLocal uniform convergence permits termwise integration. The reciprocal Hankel formula givesso $I(z,s)={\sum }_{n\ge 1}{z}^n/n^s={\mathrm{Li}}_s(z)$ initially on the disc.

For $z\notin [1,\infty )$, choose the contour around the negative axis so that it passes between zero and every pole satisfying $e^{-t}=z$. On compact subsets of the slit $z$-plane it can be chosen uniformly, and differentiation under the integral proves holomorphy. Deforming without crossing a pole gives a single-valued analytic continuation. Near the cut, the pole $t=-\log z$ lies on opposite sides of the two admissible contours; this is why the contour must pass it consistently.

Use the Hankel contour about the negative real axis: it starts at $-\infty$ below the cut, circles the origin counterclockwise, and returns to $-\infty$ above the cut. Take $-\pi <\arg t<\pi$. For $\mathrm{Re}s>0$, the small circle vanishes and the two banks giveThe gamma reflection formula then gives

Near zero,Subtracting any desired number of these terms makes the local contour integral converge in successively larger left half-planes. The resulting apparent singularities are removable after multiplication by $\Gamma (1-s)$, so the Hankel formula analytically continues $\beta$ to all $s\in \mathbb{C}$.