I am having trouble filling the blanks in a proof in Edwards book about the Riemann Zeta-Function, pages 253/254 to be precise.

The problem is the following:

Let $1/2 \leq \sigma \leq 1$ and $\alpha_k$ such that $\sum_{k=1}^{\infty} \frac{\alpha_k} {k^{s}}=\zeta(s)^{-1/2}$ for $\text{Re } s>1$.

Using Perron's Formula I could verify that

$$\sum_{k\leq Y} \frac{\alpha_k}{k^{\sigma}} \log\frac{Y}{k}= \frac{1}{2\pi i} \int_{a-i\infty}^{a+i\infty} \frac{1}{(\zeta(s+\sigma))^{1/2}} \prod_{p|\rho} \left(1-\frac{1}{p^{s+\sigma}} \right)^{-1/2} Y^s \frac{ds}{s^2} \tag{1} $$

for any $a$ such that $a>1-\sigma$ and $k$ which are relatively prime to given $\rho$. The argument is as follows:

Let $\rho$ be given. Then by the definition of $\alpha_k$ we get for $\sigma>1$

$$ \sum_{k}\frac{\alpha_k}{k^{\sigma}}= \prod_{p\, \nmid\, \rho} \left(1- \frac{1}{p^\sigma} \right)^{1/2}= \frac{1}{(\zeta(\sigma))^{1/2}} \prod_{p\,\mid \,\rho}\left(1-\frac{1}{p^{\sigma}}\right)^{-1/2} $$

where $k$ runs over positive integers relatively prime to $\rho$.

Using integration by parts and Perron's Formula we get

$$ \frac{1}{2\pi i} \int_{a-i\infty}^{a+i\infty} x^s \frac{ds}{s^2} = \frac{1}{2\pi i} \int_{a-i\infty}^{a+i\infty} (\log x) x^s \frac{ds}{s}= \begin{cases} 0, & 0<x\leq 1,\\ \log x, &1\leq x < \infty\end{cases}. $$

Using $x= Y/k$ we have $0<\frac{Y}{k}\leq 1$ for $k>Y$ and $1 \leq \frac{Y}{k}< \infty$ for $k\leq Y$. Thus, multiplying with $\frac{\alpha_k}{k^{s+\sigma}}$ and summing over all $k$ we get

$$\sum_{k\leq Y} \alpha_k \left(\log \frac{Y}{n} \right) n^{-\sigma}= \frac{1}{2\pi i}\sum_{k=1}^{\infty} \int_{a-i\infty}^{a+i \infty} \frac{\alpha_k}{n^{\sigma}} \left(\frac{Y}{n} \right)^s \frac{ds}{s^2}. $$

If we split $(Y/n)^s$ to $Y^s \cdot n^{-s}$ and interchange the sum with the integral, which is justified by absolute convergence of the series in $a>1-\sigma$ the equation $(1)$ follows.

What Edwards now tries is to proof that $(1)$ still holds for $a=1-\sigma$.

Now he looks at two cases and I have problems with both of them:

First Case: $\sigma\neq 1$.

He claims that using the inequality

$$\frac{1}{|\zeta(s)|} \leq A |s-1| \hspace{1cm} \text{ for }\operatorname{Re}s \geq 1 \tag{2} $$

it is easy to show that $(1)$ still holds with $a=1-\sigma$.

I think I did show that the integral converges but I couldn't show that the equality still holds. For this I think I need to verify that I can interchange sum and Integral for $a=1-\sigma$ but I don't know how to do that.

Second Case: $\sigma=1$.

Because of the $s^2$ in the denominator the line of integration cannot be moved to $a=1-\sigma$ but instead we take the path from $-i\infty$ to $-iB$, the semicircle in Re $s\geq 0$ from $-iB$ to $iB$ and the line from $iB$ to $i\infty$ where $B$ will later be set to $B=(\log Y)^{-1} $ for $Y>1$.

For this to show I think I need to use Cauchy's Theorem, the estimate $(2)$ and show that

$$ \int_{0+iT}^{a+iT} \frac{1}{(\zeta(s+1))^{1/2}} \prod_{p|\rho}\left(1-\frac{1}{p^{s+1}} \right)^{-1/2} Y^s \frac{ds}{s^2} \tag{3} $$

tends to $0$ as $T \to \infty$ but my estimations only led to $(3) \leq \infty$ as $T\to \infty$.

Another argument he makes and which I didn't grasp at all is the following:

Assuming everything above was proved, we get that

$$ \sum_{k\leq Y} \frac{\alpha_k}{k^{\sigma}} \log\frac{Y}{k} \leq \text{const} (\log Y)^{1/2} \prod_{p|\rho} \left(1+\frac{1}{p} \right)^{1/2} $$

for $\sigma=1$. This far it is all fine but now he claims that, I quote,

" This estimate extends from $\sigma=1$ to $1-\frac{1}{2}(\log Y)^{-1} \leq \sigma \leq 1$ if the path of integration is taken to be the line Re $s=1-\sigma$ with a detour around the right side of the circle $|s|=(\log Y)^{-1}$ where it intercepts the line. The result is again that the quantity $(1)$ to be estimated has modulus at most a constant times $(\log Y)^{1/2} \prod_{p|\rho} (1+p^{-1})^{1/2}$. But for $ \frac{1}{2} \leq \sigma \leq 1-\frac{1}{2} (\log Y)^{-1} $ the previous estimate shows it has modulus at most a constant times $\prod_{p|\rho } (1+p^{-1})^{1/2} Y^{1-\sigma} (\log Y)^{1/2}$."

I don't understand why this splitting of $\sigma$ at $1-\frac{1}{2}(\log Y)^{-1}$ is necessary or even why it is useful and why the estimates really hold.

I am sorry that these are so many questions but I really want to understand the arguments he made because I feel they are not that complicated and maybe even standard calculations but I somehow fail to really understand what is happening and doing the necessary calculations properly.

I am thankful for any tips and hints.


1 Answer 1


There is the same idea as when proving the prime number theorem.

  • To make things simple set $\alpha_n$ such that for $\Re(s) > 1$ $$\sum_{n=1}^\infty \alpha_n n^{-s} = \frac{1}{\zeta(s)^{1/2}}$$

    Then for $\Re(s)+\sigma > 1$ $$\frac{1}{\zeta(s+\sigma)^{1/2}} = s\int_1^\infty (\sum_{n \le x} \alpha_n n^{-\sigma})x^{-s-1}dx,\\ \frac{d}{d\sigma}\frac{-1}{s\zeta(s+\sigma)^{1/2}}=\int_1^\infty (\sum_{n \le x} \alpha_n n^{-\sigma}\log n)x^{-s-1}dx, \\ \frac{d}{ds}\frac{-1}{s\zeta(s+\sigma)^{1/2}}=\int_1^\infty (\sum_{n \le x} \alpha_n n^{-\sigma})(\log x )x^{-s-1}dx$$

$$\int_1^\infty (\sum_{n \le x} \alpha_n n^{-\sigma}(\log x - \log n))x^{-s-1}dx=\frac{1}{s^2\zeta(s+\sigma)^{1/2}}$$

  • Thus for $a +\sigma > 1$ by inverse Fourier/Laplace/Mellin transform

$$\sum_{n \le x} \alpha_n n^{-\sigma}\log (x/ n) = \frac{1}{2i\pi} \int_{a-i\infty}^{a+i\infty} \frac{1}{\zeta(s+\sigma)^{1/2}} x^s\frac{ds}{s^2}\\=\frac{1}{2i\pi} \int_{a-iT}^{a+iT} \frac{1}{\zeta(s+\sigma)^{1/2}} x^s\frac{ds}{s^2}+ \mathcal{O}(T^{-1/2})$$

Where $\mathcal{O}(T^{-1/2})$ is because for a fixed $a+\sigma > 1$, $\frac{1}{\zeta(a+\sigma+it)} = \mathcal{O}(t)$.

  • But we have the zero-free region $\ \ \sigma > 1-\frac{c}{\log t}, t > 1$ where again $\frac{1}{\zeta(\sigma+it)} = \mathcal{O}(t)$ uniformly in $\sigma$ this time.

    Then to treat the branch point at $s+\sigma=1$ let $f_\sigma(x) = \frac{1}{2i\pi}\int_{|s+\sigma|=1+\epsilon} \frac{1}{\zeta(\sigma+s)^{1/2}} x^s \frac{ds}{s^2}$ we find

$$\sum_{n \le x} \alpha_n n^{-\sigma}\log (x/ n) = \frac{1}{2i\pi} \int_{a-i\infty}^{a+i\infty} \frac{1}{\zeta(s+\sigma)^{1/2}}x^s\frac{ds}{s^2}$$ $$ = \frac{1}{2i\pi} \int_{a-iT}^{a+iT} \frac{1}{\zeta(s+\sigma)^{1/2}} x^s\frac{ds}{s^2}+ \mathcal{O}(T^{-1/2}) $$ $$ =f_\sigma(x)+ \frac{1}{2i\pi} \int_{1-\sigma-\frac{c}{\log T}-iT}^{1-\sigma-\frac{c}{\log T}+iT}\!\!\!\!+\int_{1-\sigma-\frac{c}{\log T}+iT}^{a+iT}\!\!\!\!+\int_{a-iT}^{1-\sigma-\frac{c}{\log T}-iT} \frac{1}{\zeta(s+\sigma)^{1/2}} x^s\frac{ds}{s^2}+ \mathcal{O}(T^{-1/2})$$

$$= f_\sigma(x)+\mathcal{O}(x^{1-\sigma-\frac{c}{\log T}})+\mathcal{O}(T^{-1/2})$$

  • $\begingroup$ Sorry, but I don't understand what exactly you are answering. Alas, I am not familiar with Fourier/Laplace/Mellin transformation and the zero-free region you mentioned. Besides that, I really do not understand how this answers any of my questions. I think you argue a little bit too quick for my understanding - sorry for that. Could you explain what exactly you are answering and why your argument answers that question? $\endgroup$
    – user114193
    Sep 17, 2017 at 20:12
  • $\begingroup$ @Tilman I'm explaining what you didn't understand : we shift the line of integration $\int_{a-i\infty}^{a+i\infty}$ on the left of $a+\sigma=1$ as usual in the theory of $\zeta(s)$, using the zero-free region that your book mentioned when proving the PNT.. For $a+\sigma > 1$ inverse Fourier/Laplace/Mellin transform is the same as Perron's formula, easy to prove because everything converges absolutely. $\endgroup$
    – reuns
    Sep 17, 2017 at 20:15
  • $\begingroup$ I see now, I think I need some time to fully understand your answer and using it to answer my question. $\endgroup$
    – user114193
    Sep 17, 2017 at 20:26
  • $\begingroup$ do you have an explanation without using the zero-free region you mentioned? I think Edwards thinks of another argument for if he actually needed that theorem he would mention it. $\endgroup$
    – user114193
    Sep 18, 2017 at 6:35
  • $\begingroup$ @Tilman Edwards does exactly the same $\endgroup$
    – reuns
    Sep 18, 2017 at 17:40

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