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I am currently reading Titchmarsh's book about the Riemann Zeta function and came across a problem in a proof of the functional equation that I cannot solve.

To be precise, I am referring to this book here at page 15.

I have two problems there:

First:

It states that $$[x]-x+\frac{1}{2}=\sum_{n=1}^{\infty}\frac{sin(2n\pi x)}{n\pi}$$ where $[x]$ is the greatest integer less or equal to $x$.

I know that for $0< y <2\pi $ holds $$\frac{\pi-y}{2}=\sum_{n=1}^{\infty}\frac{sin(ny)}{n}$$ and when I take $y=2\pi x$, I get

$$\sum_{n=1}^{\infty}\frac{sin(2\pi nx)}{n\pi}=\frac{1}{\pi} \cdot \frac{\pi -2\pi x}{2}= \frac{1}{2}-x$$

My question is: what happened to $[x]$? Why does equallity above still hold?

My second question is why we can change series and integral in

\begin{align*} \zeta(s)&= s\int_0^{\infty} \frac{[x]-x+\frac{1}{2}}{x^{s+1}}dx\\ &= s\int_0^{\infty} \frac{\sum_{n=1}^{\infty}\frac{sin(2\pi nx)}{n\pi}}{x^{s+1}}dx\\ &= \frac{s}{\pi}\int_0^{\infty}\sum_{n=1}^{\infty} \frac{sin(2\pi nx)}{n x^{s+1}}dx\\ &=\frac{s}{\pi}\sum_{n=1}^{\infty}\int_0^{\infty} \frac{sin(2\pi nx)}{n x^{s+1}}dx \end{align*}

Titchmarsh states that to justify the term-by-term integration it suffices to show that $$\lim\limits_{\lambda \to \infty} \sum_{n=1}^{\infty} \frac{1}{n}\int_{\lambda}^{\infty} \frac{sin(2\pi nx)}{ x^{s+1}}dx=0 $$

I understand the proof why this limit equals $0$ but I do not understand why this proves that we can integrate term-by-term. I know that if the series converges uniformly or absolute we could do that but I know that it is at least not absolute convergent.

I also don't understand why $$\sum_{n=1}^{\infty}\frac{sin(2n\pi x)}{n\pi} $$ is boundedly convergent.

Thank you very much in advance, i am having a rough time recently because of these questions.

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    $\begingroup$ First question: since $0<y<2\pi$, the formula you write down next is valid only for $0<x<1$. Introducing $[x]$ is a way of extending this formula periodically in $x$. Second question: the definition of an improper integral $\int_0^\infty blah$ is $\lim_{\lambda\to\infty} \int_0^\lambda blah$; I recommend rewriting the computation in this form from the start. $\endgroup$ – Greg Martin May 3 '16 at 22:36
  • $\begingroup$ $\{x\} = x - \lfloor x \rfloor$ is $1$-periodic, and for every $x \not\in \mathbb{Z}$ : $\{x\} = \frac{1}{2}-\sum_{n=1}^\infty \frac{\sin(2 \pi n x)}{\pi n}$, for the proof you can use the Fourier series convergence theorems, or the Taylor series of $-\ln(1-z) = \sum_{n=1}^\infty \frac{z^n}{n}$ for $|z| < 1$, and prove that $|\sum_{n < N} \sin( 2 \pi n x)| < C(x)$ followed by Summation_by_parts $\endgroup$ – reuns May 3 '16 at 23:08
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    $\begingroup$ then you prove that FOR $Re(s) \in ]-1,0[$ : $\zeta(s) = -s \int_0^\infty (\{x\}-1/2) x^{-s-1} dx$, and for the inversion of $\sum$ and $\int$, what you need is proving that $ \int_\lambda^\infty \sum_{n = 1}^\infty \frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx = \sum_{n = 1}^\infty \int_\lambda^\infty \frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx$ and that it $\to 0$ when $\lambda \to \infty$, simply because it is obvious that you can exchange $\int$ and $\sum$ in $ \int_0^\lambda \sum_{n = 1}^N \frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx$, and hence too when $N = \infty$. $\endgroup$ – reuns May 3 '16 at 23:23
  • $\begingroup$ note that by considering $(1-2^s)(1-2^{1-s})$, and grouping the terms by too, you get a full absolutely convergent proof when $Re(s) \in ]0,1[$, detailed here : https://fr.wikipedia.org/wiki/Fonction_zêta_de_Riemann#Relation_fonctionnelle $\endgroup$ – reuns May 3 '16 at 23:25
  • $\begingroup$ Why does it still hold for $ N=\infty $? Don't we need uniform convergence? Sorry but I don't understand on what theorem this argument is based? I think my Problem is that I can't handle two limits simultaneously. $\endgroup$ – user114193 May 4 '16 at 10:42
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in general, for proving from scratch that you can exchange $\sum$ and $\int$ (or any two limits), start with finite bounds. clearly when $\lambda,A,B$ are finite : $$ \int_0^\lambda \sum_{n=A}^B\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx = \sum_{n=A}^B \int_0^\lambda\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx = \sum_{n=A}^B \int_0^{n\lambda} \frac{\sin(2 \pi y)}{\pi n} (y/n)^{-s-1} \frac{dy}{n} $$ $$ = \sum_{n=A}^B n^{s-1} \int_0^{n\lambda}\frac{\sin(2 \pi y)}{\pi } y^{-s-1}dy$$

now all you need is noticing that when $Re(s) \in ]-1,0[$ : $\displaystyle\sum_{n} n^{s-1}$ is absolutely convergent, and $\displaystyle\left|\int_0^{n\lambda}\frac{\sin(2 \pi y)}{\pi } y^{-s-1}dy\right| <C \left|\int_0^\infty\frac{\sin(2 \pi y)}{\pi } y^{-s-1}dy\right|$ (you can use integration by part for proving that the integral converges)

hence, when $Re(s) \in ]-1,0[$ the series $$\sum_{n= A}^\infty n^{s-1} \int_0^{n\lambda}\frac{\sin(2 \pi y)}{\pi } y^{-s-1}dy$$ converges absolutely (and hence it $\to 0$ when $A \to \infty$)

finally, prove that when $A \to \infty$ : $$ \int_0^\lambda \sum_{n=A}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx \to 0$$ (again you can use integration by part)

and from all this you get that :

$$ \int_0^\lambda \sum_{n=1}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx = \int_0^\lambda \sum_{n=1}^A\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx + \int_0^\lambda \sum_{n=A+1}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx $$

$$ = \sum_{n=1}^A \int_0^\lambda \frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx + \int_0^\lambda \sum_{n=A+1}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx $$

$$= \sum_{n=1}^\infty \int_0^\lambda \frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx - \sum_{n=A+1}^\infty \int_0^\lambda \frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx+\int_0^\lambda \sum_{n=A+1}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx$$ $$ = \sum_{n=1}^\infty \int_0^\lambda \frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx\qquad\qquad(\text{ letting } \ A \to \infty)$$

(since the two other terms $\to 0$)

and from what Titchmarsh prove, that

$$\lim\limits_{\lambda \to \infty} \sum_{n=1}^{\infty} \frac{1}{n}\int_{\lambda}^{\infty} \frac{sin(2\pi nx)}{ x^{s+1}}dx=0$$

you get that when $Re(s) \in ]-1,0[$ :

$$ \int_0^\infty \sum_{n=1}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx = \sum_{n=1}^\infty\int_0^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx$$

which (with the functional equations for the $\Gamma$ function) proves the functional equation for $\zeta(s)$.

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  • $\begingroup$ Thank you very much! That helped a lot. I finally understand the argument - there is just one problem left: you say i can prove $\int_0^{\lambda} \sum_{n=A}^{\infty} \frac {sin (2n\pi x)}{n\pi } x^{-s-1} dx \rightarrow 0$ using integration by parts. My problem is that for integration by parts I already need to interchange series and integral. Am I missing something? $\endgroup$ – user114193 May 10 '16 at 0:06
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    $\begingroup$ @Tilman : no you don't need, $\sum_{n=A}^\infty \frac{\sin(2 \pi n x)}{n\pi} = 1/2+ \lfloor x \rfloor - x - \sum_{n=1}^{A-1} \frac{\sin(2 \pi n x)}{n\pi} $ $\endgroup$ – reuns May 10 '16 at 0:21

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