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In Stein&Shakarchi, Complex Analysis, chapter 6, problem 2-3 (p. 180), they hint at a method to meromorphically continue the zeta function to the entire complex plane. I can see from Abel's summation formula that

$$ \zeta(s) = \frac{s}{s-1} - \frac{1}{2} - s \int_{1}^{\infty} \frac{Q(x)}{x^{s+1}} \, dx \, , $$

where $Q(x) = \{x\} - 1/2$ (with $\{x\}$ being the fractional part of $x$). This is problem 2.

They continue to problem 3. Set $Q_k(x) = B_{k+1}(x)/(k+1)!$, for $0 \leq x \leq 1$, then periodically continue. Here the term $B_{k+1}(x)$ is the $(k+1)$-th Bernoulli polynomial. We have the properties $\frac{d}{dx}Q_{k+1} = Q_k$, and $Q_k(x+1) = Q_k(x)$ and $\int_{0}^{1} Q_k(x) \, dx = 0$. Rewrite

$$ \zeta(s) = \frac{s}{s-1} - \frac{1}{2} - s \int_{1}^{\infty} \left( \frac{d^k}{dx^k} Q_k(x) \right) x^{-s-1} \, dx \, . $$

Now they claim that $k$-fold integration by parts solves the problem for $\text{Re} \, s > -k$. However, I do not quite understand how. Can you provide a hint? This is not homework.

JKoolstra
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  • $Q_k$ is bounded, and $k$-fold integration by parts produces $x^{-s-k-1}$ for the other factor (ignoring parts independent of $x$), so the resulting integral gives a holomorphic function on $\operatorname{Re} s > -k$ (actually, since the integral of $Q_k$ over intervals of integer length vanishes, the integral works for $\operatorname{Re} s > -k-1$, but that takes a bit more work to prove). – Daniel Fischer Jan 24 '18 at 22:47
  • @DanielFischer Yes, I understood that. I think that I've figured my confusion out. I was confused by the fact that, for example, $Q_1(x)x^{-s-1}|_{1}^{\infty}$ is infinity when $\text{Re} , s \leq -1$. But since that is otherwise constant, by analytic continuation we may continue to use that constant, even if the original formula does not hold anymore. Do I say that correctly? – JKoolstra Jan 26 '18 at 10:51
  • @DanielFischer Please, I would love to see how "integer length vanishing" could play a role here. What I am still a bit confused over is the role of the assumption $\int_{0}^{1} Q_k(x) dx = 0$. It is not used I think? One more question: how could I use the above formulation to derive $\zeta(-1) = -1/12$? I know how to do it using the functional equation, but I was looking to avoid it. In (the next) problem 5, Stein and Shakarchi use it to deduce the formula for $\zeta(2k)$, but I'm not seeing that either. Thank you so much in advance for your help! – JKoolstra Jan 26 '18 at 10:56
  • I'm not quite sure what exactly you mean in your first comment, but it looks like you got the principle. We can integrate by parts for $\operatorname{Re} s > -1$, and the boundary term of that integration by parts is a constant. Thus we get a representation $$\zeta(s) = \frac{1}{s-1} + E^{(1)}_k(s) + E^{(2)}_k(s)\cdot I_k(s),$$ where $E^{(j)}_k$ is an entire function, and $I_k$ is an integral that is holomorphic for $\operatorname{Re} s > -k$. This representation is a priori valid for $\operatorname{Re} s > 1$, but by the identity theorem it holds where the right hand side makes sense. – Daniel Fischer Jan 26 '18 at 11:14
  • The assumption $\int_0^1 Q_k(x),dx = 0$ (together with the periodicity) gives the boundedness of $Q_{k+1}$, which we need for the domains of convergence for the integrals to grow. And that boundedness, together with Dirichlet's criterion tells us that $$\int_1^{\infty} \frac{Q_k(x)}{x^{\alpha}},dx$$ converges - as an improper Riemann integral - for $\operatorname{Re} \alpha > 0$, and the convergence is locally uniform (in $\alpha$), so it defines a holomorphic function (of $\alpha$) in the right half-plane. The Lebesgue integral exists only for $\operatorname{Re} \alpha > 1$. – Daniel Fischer Jan 26 '18 at 11:24
  • From our integration by parts, we see that the above integral arises with $\alpha = s + k + 1$, so Lebesgue gives the representation for $\operatorname{Re} s > -k$, and improper Riemann for $\operatorname{Re} s > -k-1$. – Daniel Fischer Jan 26 '18 at 11:28
  • To get $\zeta(-1)$, we need one integration by parts [or two], $$\int_1^{\infty} \frac{Q(x)}{x^{s+1}},dx = \frac{Q_1(x)}{x^{s+1}}\biggr\rvert_1^{\infty} + (s+1)\int_1^{\infty} \frac{Q_1(x)}{x^{s+2}},dx,,$$ which yields $$\zeta(s) = \frac{s}{s-1} -\frac{1}{2} + sQ_1(1) - s(s+1)\int_1^{\infty} \frac{Q_1(x)}{x^{s+2}},dx,.$$ If we use the Lebesgue integral we need another integration by parts, since the integral there exists as a Lebesgue integral only for $\operatorname{Re} s > -1$, but either way, at $s = -1$ that part doesn't contribute because of the factor $s(s+1)$. So that yields – Daniel Fischer Jan 26 '18 at 11:38
  • $$\zeta(-1) = \frac{-1}{-1-1} - \frac{1}{2} -Q_1(1) = -Q_1(1),.$$ – Daniel Fischer Jan 26 '18 at 11:38

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