- Use the previous exercise to give another proof that \(\zeta ( s )\) is continuable in the complex plane with only singularity a simple pole at \(s = 1\)
[Hint: Write
\[ \zeta (s) = \frac {1}{\Gamma (s)} \int_ {0} ^ {1} \frac {x ^ {s - 1}}{e ^ {x} - 1} d x + \frac {1}{\Gamma (s)} \int_ {1} ^ {\infty} \frac {x ^ {s - 1}}{e ^ {x} - 1} d x. \]
The second integral defines an entire function, while
\[ \int_ {0} ^ {1} \frac {x ^ {s - 1}}{e ^ {x} - 1} d x = \sum_ {m = 0} ^ {\infty} \frac {B _ {m}}{m ! (s + m - 1)}, \]
where \(B _ { m }\) denotes the \(m ^ { \mathrm { t h } }\) Bernoulli number defined by
\[ {\frac {x}{e ^ {x} - 1}} = \sum_ {m = 0} ^ {\infty} {\frac {B _ {m}}{m !}} x ^ {m}. \]
Then \(B _ { 0 } = 1\) , and since \(z / ( e ^ { z } - 1 )\) is holomorphic for \(| z | < 2 \pi\) , we must have lim s \(\begin{array} { r } { \limsup _ { m \to \infty } | B _ { m } / m ! | ^ { 1 / m } = 1 / 2 \pi . ] } \end{array}\)
Solution
Split \[ \Gamma(s)\zeta(s) =\int_0^1\frac{x^{s-1}}{e^x-1}\,dx +\int_1^\infty\frac{x^{s-1}}{e^x-1}\,dx. \tag{1} \] The second integral defines an entire function of \(s\), because on every compact set in the \(s\)-plane its integrand is dominated by \(C x^N e^{-x}\) for large \(x\).
Write \[ h(x)=\frac{x}{e^x-1}=\sum_{m=0}^\infty \frac{B_m}{m!}x^m, \qquad |x|<2\pi. \] Fix \(N\ge0\). Then near \(0\), \[ h(x)=\sum_{m=0}^N\frac{B_m}{m!}x^m+x^{N+1}r_N(x), \] where \(r_N\) is holomorphic near \([0,1]\). Hence \[ \int_0^1\frac{x^{s-1}}{e^x-1}\,dx =\int_0^1 x^{s-2}h(x)\,dx =\sum_{m=0}^N\frac{B_m}{m!(s+m-1)}+H_N(s), \tag{2} \] where \(H_N\) is holomorphic for \(\Re s>-N\).
Multiplying (1) and (2) by \(1/\Gamma(s)\) therefore extends \(\zeta(s)\) meromorphically to \(\Re s>-N\). Since \(N\) is arbitrary, this gives a meromorphic continuation to all of \(\mathbb C\).
The possible poles arising from (2) occur at \[ s=1-m\qquad(m=0,1,2,\ldots). \] For \(m\ge1\), these are nonpositive integers, and \(1/\Gamma(s)\) has a simple zero there, so those poles cancel. At \(s=1\), the \(m=0\) term is \(1/(s-1)\) because \(B_0=1\), and \(1/\Gamma(1)=1\), so the pole survives and is simple, with residue \(1\).
Thus \(\zeta\) continues meromorphically to \(\mathbb C\) with exactly one singularity, a simple pole at \(s=1\).