Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I try to understand the proof of Chap. VI, n° 3.1, Prop. 10 in Serre's "A course in arithmetic" (page 70). The goal is to prove that zeta-function can be written as \begin{align*} \zeta(s)=\frac{1}{s-1}+\phi(s) \end{align*} with an holomorphic $\phi$ for any $s\in\mathbb{C}$ with $\mathfrak{Re}(s)>0$.

I understand the proof except for one detail: $\phi$ is given explicitly by $\phi=\sum_{n=1}^\infty \phi_n$ with \begin{align*} \phi_n(s)=\int_n^{n+1} \left(n^{-s}-t^{-s}\right) dt. \end{align*} To prove the convergence of $\phi$, we want to show that \begin{align*} \left|\phi_n(s)\right|\leq\frac{|s|}{n^{x+1}} \quad \text{with}\quad x=\mathfrak{Re}(s)\quad(*) \end{align*} For this Serre first notes that \begin{align*} \left|\phi_n(s)\right| \leq \sup_{n\leq t\leq n+1}\left| n^{-s}-t^{-s}\right| \end{align*} Which is clear since the range of integration is $(n+1)-n=1$. Then he sais that the derivative of $n^{-s}-t^{-s}$ is equal to $\frac{s}{t^{s+1}}$. And from this somehow follows (*). My question is: How exactly does this follow?

I noted the following statements for $s=x+iy$ where $x=\mathfrak{Re}(s)$ and $y=\mathfrak{Im}(s)$ and $f(t)=n^{-s}-t^{-s}$

  1. Because $|t^{iy}|=|e^{iy\ln(t)}|=1$ (since it is on the unit circle) and $|t^{x+1}|=t^{x+1}$ (since $t$ and $x$ are real) we see that \begin{align*} \left|f'(t)\right|=\left|\frac{s}{t^{x+1+iy}}\right|=\frac{|s|}{|t^{x+1}| |t^{iy}|}=\frac{|s|}{t^{x+1}} \end{align*}
  2. Because $f(n)=0$ we can calculate \begin{align*} |f(n+1)|=|f(n+1)-f(n)|=\left|\int_n^{n+1}f'(t)dt\right|\leq\sup_{n\leq t\leq n+1}\left|f'(t)\right|=\sup_{n\leq t\leq n+1}\frac{|s|}{t^{x+1}}=\frac{|s|}{n^{x+1}} \end{align*}

Am I close? :-)

share|cite|improve this question
Looks exactly right, you should have more confidence :). The key idea that $|\int f| \le \int|f|$ is pervasive and often considered a form of the triangle inequality. Great choice of book, by the way. – Erick Wong Feb 21 '13 at 15:09
I don't see how this implies $|\phi_n(s)|\leq\frac{|s|}{n^{x+1}}$ and what it has to do with the derivative. – born Feb 21 '13 at 15:38
Oh, I must have taken this for granted. In the first equation after (*) you cite that $|\phi_n(s)|$ is bounded by $\sup |f(t)|$. Although point $2$ only refers to $f(n+1)$, use the same argument to get a bound for $f(t)$. – Erick Wong Feb 21 '13 at 15:48
The key idea is that the derivative controls how quickly $f$ can vary. Since $f'$ is small, $f(n+\epsilon)$ cannot stray too far from $f(n)$, which is $0$. – Erick Wong Feb 21 '13 at 15:49
up vote 4 down vote accepted

By the definition of $\phi_n$ and the Fundamental theorem of calculus we have

$$|\phi_n(s)| = \left| \int_n^{n+1} f(t) \, dt \right| \stackrel{f(n)=0}{=} \left| \int_n^{n+1} \left(\int_n^t f'(y) \, dy \right) \, dt \right| \leq \int_n^{n+1} |f'(y)| \, dy$$

Use your calculation from 1. (and note that $y \in [n,n+1]$, hence $y\geq n$).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.