3.\(^\ast\) If \(Q ( x ) = \{ x \} - 1 / 2\) , then we can write the expression in the previous problem as
\[ \zeta (s) = \frac {s}{s - 1} - \frac {1}{2} - s \int_ {1} ^ {\infty} \frac {Q (x)}{x ^ {s + 1}} d x. \]
Let us construct \(Q _ { k } ( x )\) recursively so that
\[ \int_ {0} ^ {1} Q _ {k} (x) d x = 0, \quad \frac {d Q _ {k + 1}}{d x} = Q _ {k} (x), \quad Q _ {0} (x) = Q (x) \quad \text { and } \quad Q _ {k} (x + 1) = Q _ {k} (x). \]
Then we can write
\[ \zeta (s) = \frac {s}{s - 1} - \frac {1}{2} - s \int_ {1} ^ {\infty} \left(\frac {d ^ {k}}{d x ^ {k}} Q _ {k} (x)\right) x ^ {- s - 1} d x, \]
and a k-fold integration by parts gives the analytic continuation for \(\zeta ( s )\) when \(\operatorname { R e } ( s ) > - k\)
Solution
For \(j\ge0\) and \(\Re s>-j\) put \(I_j(s)\da\int_1^\infty Q_j(x)x^{-s-1-j}\,dx\).
1. Each \(Q_j\) is continuous for \(j\ge1\), periodic, and bounded.
\(Q_0\) is bounded and periodic. If \(Q_j\) is bounded and periodic with \(\int_0^1Q_j=0\), then \(Q_{j+1}(x)=Q_{j+1}(0)+\int_0^xQ_j\) is continuous and satisfies \(Q_{j+1}(x+1)-Q_{j+1}(x)=\int_x^{x+1}Q_j=0\), so it is periodic, hence bounded; the constant \(Q_{j+1}(0)\) is fixed by \(\int_0^1Q_{j+1}=0\).
2. \(I_j\) is holomorphic on \(\Re s>-j\).
By step 1, \(\abs{Q_j(x)x^{-s-1-j}}\le Cx^{-\sigma-1-j}\) with \(\sigma=\Re s\), which is integrable on \([1,\infty)\) uniformly for \(\sigma\ge-j+\eps\). So the integral converges locally uniformly on \(\Re s>-j\) and defines a holomorphic function.
3. \(I_j(s)=-Q_{j+1}(1)+(s+j+1)I_{j+1}(s)\) for \(\Re s>-j\).
Integrate by parts on \([1,R]\) with \(Q_j=Q_{j+1}'\): \(\int_1^RQ_jx^{-s-1-j}\,dx=\bigl[Q_{j+1}x^{-s-1-j}\bigr]_1^R+(s+j+1)\int_1^RQ_{j+1}x^{-s-2-j}\,dx\). As \(R\to\infty\) the boundary term at \(R\) tends to \(0\), since \(Q_{j+1}\) is bounded and \(\Re s>-j-1\).
4. Q.E.D.
For \(\Re s>1\), the formula of the previous problem gives \(\zeta(s)=\frac{s}{s-1}-\frac12-sI_0(s)\). Applying step 3 \(k\) times, \[I_0(s)=P_k(s)+(s+1)(s+2)\cdots(s+k)\,I_k(s),\] where \(P_k\) is a polynomial in \(s\) built from the constants \(Q_1(1),\ldots,Q_k(1)\). The right side is holomorphic on \(\Re s>-k\) by step 2, so \(\frac{s}{s-1}-\frac12-s\bigl(P_k(s)+(s+1)\cdots(s+k)I_k(s)\bigr)\) is meromorphic on \(\Re s>-k\) with only a simple pole at \(s=1\), and it agrees with \(\zeta\) on \(\Re s>1\). This is the analytic continuation of \(\zeta\) to \(\Re s>-k\).