1. ## Riemann Zeta Function

For $1 define $\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$. Prove that $\zeta(s) = s \int_{1}^{\infty} \frac{[x]}{x^{s+1}} \ dx$. So consider an interval $[1,N]$. Let $g(s) = s \int_{1}^{N} \frac{[x]}{x^{s+1}} \ dx - \sum_{n=1}^{N} \frac{1}{n^s}$. Now we want to show that $g(s) = 0$. Then we take $N \to \infty$ and conclude that $g(s) = 0$ on $[1, \infty)$. The second series is convergent.

2. Hello,
Originally Posted by manjohn12
For $1 define $\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$. Prove that $\zeta(s) = s \int_{1}^{\infty} \frac{[x]}{x^{s+1}} \ dx$. So consider an interval $[1,N]$. Let $g(s) = s \int_{1}^{N} \frac{[x]}{x^{s+1}} \ dx - \sum_{n=1}^{N} \frac{1}{n^s}$. Now we want to show that $g(s) = 0$. Then we take $N \to \infty$ and conclude that $g(s) = 0$ on $[1, \infty)$. The second series is convergent.
Ha ! I think I found it \o/

But no, g(s) is not necessarily equal to 0. I'm not too sure (I hate those limit/convergence theorems), but proving that $\lim_{N\to \infty} g(s)=0$ may be enough (because both terms of the difference converge (*))

Consider the integral $I(s)=s \int_1^N \frac{\lfloor x \rfloor}{x^{s+1}} ~dx$
Then divide the integral on intervals which boundaries are integers :

$I(s)=s \sum_{k=1}^{N-1} \int_k^{k+1} \frac{\lfloor x \rfloor}{x^{s+1}} ~dx$

But $x \in [k,k+1) \Rightarrow \lfloor x\rfloor =k$

So :
\begin{aligned}
I(s)
&=\sum_{k=1}^{N-1} sk \int_k^{k+1} \frac{1}{x^{s+1}} ~dx \\
&=\sum_{k=1}^{N-1} sk \cdot \left. \frac{1}{x^s \cdot (-s)}\right|_k^{k+1} \\
&=\sum_{k=1}^{N-1} k \cdot \left(\frac{1}{k^s}-\frac{1}{(k+1)^s}\right)
\end{aligned}

Now :
\begin{aligned}
g(s) &=\sum_{k=1}^{N-1} k \cdot \left(\frac{1}{k^s}-\frac{1}{(k+1)^s}\right)-\sum_{k=1}^N \frac{1}{k^s} \\
&=-\frac{1}{N^s}+\sum_{k=1}^{N-1} k \cdot \left(\frac{1}{k^s}-\frac{1}{(k+1)^s}\right)-\frac{1}{k^s} \quad \text{(because these are finite sums)} \\
&=-\frac{1}{N^s}+\sum_{k=1}^{N-1} \frac{k-1}{k^s}-\frac{k}{(k+1)^s}
\end{aligned}

But this is a telescoping sum !
So we have :
$g(s)=-\frac{1}{N^s}+\frac{1-1}{1^s}-\frac{N-1}{N^s}=-\frac{N}{N^s}$

And this obviously goes to 0 as N goes to infinity (because $s>1$)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
(*)
$\int_1^\infty \frac{\lfloor x \rfloor}{x^{s+1}} ~dx \leq \int_1^\infty \frac{x+1}{x^{s+1}} ~dx=\int_1^\infty \frac{1}{x^s}+\frac{1}{x^{s+1}} ~dx$
and this converges

3. Originally Posted by Moo
Hello,

Ha ! I think I found it \o/

But no, g(s) is not necessarily equal to 0. I'm not too sure (I hate those limit/convergence theorems), but proving that $\lim_{N\to \infty} g(s)=0$ may be enough (because both terms of the difference converge (*))

Consider the integral $I(s)=s \int_1^N \frac{\lfloor x \rfloor}{x^{s+1}} ~dx$
Then divide the integral on intervals which boundaries are integers :

$I(s)=s \sum_{k=1}^{N-1} \int_k^{k+1} \frac{\lfloor x \rfloor}{x^{s+1}} ~dx$

But $x \in [k,k+1) \Rightarrow \lfloor x\rfloor =k$

So :
\begin{aligned}
I(s)
&=\sum_{k=1}^{N-1} sk \int_k^{k+1} \frac{1}{x^{s+1}} ~dx \\
&=\sum_{k=1}^{N-1} sk \cdot \left. \frac{1}{x^s \cdot (-s)}\right|_k^{k+1} \\
&=\sum_{k=1}^{N-1} k \cdot \left(\frac{1}{k^s}-\frac{1}{(k+1)^s}\right)
\end{aligned}

Now :
\begin{aligned}
g(s) &=\sum_{k=1}^{N-1} k \cdot \left(\frac{1}{k^s}-\frac{1}{(k+1)^s}\right)-\sum_{k=1}^N \frac{1}{k^s} \\
&=-\frac{1}{N^s}+\sum_{k=1}^{N-1} k \cdot \left(\frac{1}{k^s}-\frac{1}{(k+1)^s}\right)-\frac{1}{k^s} \quad \text{(because these are finite sums)} \\
&=-\frac{1}{N^s}+\sum_{k=1}^{N-1} \frac{k-1}{k^s}-\frac{k}{(k+1)^s}
\end{aligned}

But this is a telescoping sum !
So we have :
$g(s)=-\frac{1}{N^s}+\frac{1-1}{1^s}-\frac{N-1}{N^s}=-\frac{N}{N^s}$

And this obviously goes to 0 as N goes to infinity (because $s>1$)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
(*)
$\int_1^\infty \frac{\lfloor x \rfloor}{x^{s+1}} ~dx \leq \int_1^\infty \frac{x+1}{x^{s+1}} ~dx=\int_1^\infty \frac{1}{x^s}+\frac{1}{x^{s+1}} ~dx$
and this converges
Why is $g(s)$ not necessarily equal to 0?

4. Originally Posted by manjohn12
Why is $g(s)$ not necessarily equal to 0?
Because you want to prove that $\lim_{n \to \infty} I(s)=\zeta(s)=\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k^s}$

That is to say :
$\lim_{n \to \infty} I(s)-\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k^s}=0$

Since the two converge, that's :
$\lim_{n \to \infty} \left[I(s)-\sum_{k=1}^n \frac{1}{k^s}\right]=0$

$\lim_{n \to \infty} g(s)=0$

That's the limit as n goes to infinity, not the value for any n.
You can check it if you take n=1. It's not equal to 0.

5. According to the defintion reported in...

Floor Function -- from Wolfram MathWorld

... the 'fractional part of x' is defined as...

$\{x\}= x - \lfloor x \rfloor$ (1)

... and it is the function you see in figure...

In the interval $n\le x< n+1$ is...

$\{x\}= x-n$ (2)

According to (1) and remembering that...

$\int \frac{dx}{x^{s}}= -\frac{1}{(s-1)\cdot x^{s-1}} + c$ (3)

... we obtain...

$\int_{n}^{n+1} \frac{\lfloor x \rfloor}{x^{s+1}}\cdot dx= \frac{n}{s}\cdot \{\frac{1}{n^{s}}-\frac{1}{(n+1)^{s}}\}$ (4)

... so that...

$\int_{1}^{n} \frac{\lfloor x \rfloor}{x^{s+1}}\cdot dx= \frac{1}{s}\cdot (\frac{1}{1^{s}}- \frac{1}{2^{s}}) + \frac{2}{s}\cdot (\frac{1}{2^{s}}- \frac{1}{3^{s}}) +\dots + \frac{n-1}{s}\cdot \{\frac{1}{(n-1)^{s}}- \frac{1}{n^{s}}\} =$

$= \frac{1}{s}\cdot \{\frac{1}{1^{s}} + \frac{1}{2^{s}} + \dots + \frac{1}{(n-1)^{s}} - \frac{n-1}{n^{s}}\}$ (5)

From (5) it derives that for $Re(s)>1$ ...

$s\cdot \int_{1}^{\infty} \frac{\lfloor x \rfloor}{x^{s+1}}\cdot dx= \sum_{n=1}^{\infty}\frac{1}{n^{s}} = \zeta(s)$ (6)

$g(s)= s\cdot \int_{1}^{n} \frac{\lfloor x \rfloor}{x^{s+1}}\cdot dx - \sum_{k=1}^{n}\frac{1}{k^{s}} = - \frac{n}{n^{s}}$ (7)

Kind regards

$\chi$ $\sigma$