= Solution
For $\Re s>1$, absolute convergence permits separation into odd and even terms:
$$
\begin{aligned}
\sum_{n=1}^\infty(-1)^{n-1}n^{-s}
&=\sum_{n=1}^\infty n^{-s}-2\sum_{n=1}^\infty(2n)^{-s}\\
&=(1-2^{1-s})\zeta(s).
\end{aligned}
$$
Using the defining integral of the <Gamma function> and the substitution $u=nt$ gives
$$
\Gamma(s)n^{-s}=\int_0^\infty t^{s-1}e^{-nt}\,dt.
$$
Termwise integration is justified by absolute convergence when $\Re s>1$, and the geometric sum is
$$
\sum_{n=1}^\infty(-1)^{n-1}e^{-nt}=\frac1{1+e^t}.
$$
Consequently the <Dirichlet eta function> satisfies
$$
(1-2^{1-s})\zeta(s)=\eta(s)
=\frac1{\Gamma(s)}\int_0^\infty\frac{t^{s-1}}{1+e^t}\,dt.
$$
Near zero the integrand is $O(t^{\Re s-1})$, while at infinity it decays exponentially. The integral therefore defines a holomorphic function for $\Re s>0$. Thus
$$
\boxed{\displaystyle
\zeta(s)=\frac1{(1-2^{1-s})\Gamma(s)}
\int_0^\infty\frac{t^{s-1}}{1+e^t}\,dt}
$$
provides the desired continuation wherever the displayed denominator is nonzero. At a nonreal zero of $1-2^{1-s}$, use instead
$$
(1-k^{1-s})\zeta(s)
=\sum_{n\geq1}a_n n^{-s},
\qquad
a_n=\begin{cases}1,&k\nmid n,\\1-k,&k\mid n,
\end{cases}
$$
with an integer $k$ for which $1-k^{1-s}\ne0$; 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 $s=1$,
$$
\eta(1)=\int_0^\infty\frac{dt}{1+e^t}=\log2,
$$
whereas
$$
1-2^{1-s}=1-e^{-(s-1)\log2}
=(s-1)\log2+O((s-1)^2).
$$
Hence
$$
\zeta(s)=\frac1{s-1}+O(1),
$$
so $s=1$ is a simple pole and its <residue> is
$$
\boxed{\operatorname{Res}_{s=1}\zeta(s)=1}.
$$
Back to article page
