1. ## Euler Product

Show $\displaystyle{\prod_{p\equiv1\;(4)}\frac{p^3+1}{p^ 3-1} = \frac{105}{4\pi^3}\zeta(3)}$ and $\displaystyle{\prod_{p\equiv3\;(4)}\frac{p^3+1}{p^ 3-1} = \frac{28}{\pi^3}\zeta(3)}$.

2. Originally Posted by chiph588@
Show $\prod_{p\equiv1\;(4)}\frac{p^3+1}{p^3-1} = \frac{105}{4\pi^3}\zeta(3)$ and $\prod_{p\equiv3\;(4)}\frac{p^3+1}{p^3-1} = \frac{28}{\pi^3}\zeta(3)$.

I consider

$\ F(s) = \big{ \frac{1}{1^s} - \frac{1}{3^s} + \frac{1}{5^s} - ... } = \displaystyle{\sum_{k=0}^{\infty} } \frac{(-1)^k}{(2k+1)^s}$ which is similar to zeta function .

I find that it can be ' factorised ' in this form :

$\left( \displaystyle{\prod_{p\equiv 1(4)}} ( 1 - \frac{1}{p^s} ) \displaystyle{\prod_{p\equiv 3(4)}} ( 1 + \frac{1}{p^s} ) \right)^{-1} ~~~~~ (1)$

The reason is when we expand the product from RHS we obtain something like :

$\displaystyle{ \prod_{1(4)}} ( 1 + \frac{1}{p^s } + ... ) \prod_{3(4)} ( 1 - \frac{1}{p^s } + ... )$

For a number $n$ if it is $1 \bmod{4}$ , no matter how many primes $3\bmod{4}$ it contains , the number must be even , and for $n \equiv 3 \bmod{4}$ , we conclude that the number of primes $3\bmod{4}$ as factors must be odd , so we have this formula (1) .

Together with $\zeta(s) = \left( \displaystyle{\prod_p} ( 1 - \frac{1}{p^s} ) \right) ^{-1}$

$= \frac{1}{1 - \frac{1}{2^s} } \left( \displaystyle{\prod_{p\equiv 1(4)}} ( 1 - \frac{1}{p^s} ) \displaystyle{ \prod_{p \equiv 3(4) }} ( 1 - \frac{1}{p^s} ) \right) ^{-1}$

We obtain , by division :

$\boxed{ \prod_{p \equiv 3(4) } \frac{p^s+1 }{ p^s - 1 } = ( 1 - \frac{1}{2^s} ) ~ \frac{ \zeta(s) }{ F(s) } }$

Also , we have

$\left( \frac{ 2^s + 1 }{2^s - 1 } \right) \displaystyle{ \prod_{p \equiv 1(4) }} \frac{ 1 + \frac{1}{p^s} }{ 1 - \frac{1}{p^s} } \displaystyle{\prod_{p \equiv 3(4) }} \frac{ 1 + \frac{1}{p^s} }{ 1 - \frac{1}{p^s} }$

$= \frac{ \zeta^2(s) }{\zeta(2s ) }$ so we obtain

$\boxed{ \prod_{p \equiv 1(4) } \frac{ p^s + 1 }{ p^s - 1 } = \frac{1}{ 1 + \frac{1}{2^s } } ~ \frac{ \zeta(s) F(s) }{ \zeta(2s) }}$

I don't know the value of $F(3)$ but from your writing , it should be $\frac{\pi^3}{32}$ and if I remember correctly , $\zeta(6) = \frac{\pi^6}{945}$ we should have

$\displaystyle{ \prod_{p \equiv 1(4) }} \frac{ p^3 + 1 }{ p^3 - 1 } } = \frac{8}{9} ~ \frac{\pi^3}{32 } ~ \frac{945}{\pi^6} ~ \zeta(3)$

$= \frac{ 105}{4 \pi^3 } ~ \zeta(3)$ , it looks ok to me ...

Remarks : I've read a book by Euler , called Introduction to Analysis of the Infinite , in this book he really introduced a method of evaluating $F(2n-1)$ ( of course , in the mean time , he evaluated $\zeta(2n)$ , too ) but i forgot this method , recently i got another method using Fourier Analysis , I guess you experts are using this skilfully .

3. Very good!

It can be shown $F(2k+1)=\displaystyle{{{{({-1})^k}{E_{2k}}{\pi^{2k+1}} \over {4^{k+1}}(2k!)}}}$, where $E_{n}$ are the Euler Numbers.

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Also if anyone didn't follow the first step, $F(s)=\displaystyle{\sum_{k=0}^{\infty} } \frac{(-1)^k}{(2k+1)^s} = \displaystyle{\prod_p} \left(1+(-1)^{(p-1)/2}p^{-s}\right)^{-1} = \left( \displaystyle{\prod_{p\equiv 1(4)}} ( 1 - p^{-s}) \displaystyle{\prod_{p\equiv 3(4)}} ( 1 + p^{-s} ) \right)^{-1}$.

The second equality comes from the fact that $F(s) = \displaystyle{\sum_{n=0}^{\infty} } \frac{\chi(n)}{n^s}$, where $\chi(n)=\begin{cases} (1)^{(n-1)/2} ,\;\;\text{if }n\text{ is odd}\\0,\;\;\text{if }n\text{ is even}\end{cases}$

Now observe $\chi(n)$ is completely multiplicative so we can transform $F(s)$ into its Euler Product representaion.

4. Oh , is $\chi(n)$ the Dirichlet function ? I first saw it in the later part of a book about elementary number theory .

5. Originally Posted by simplependulum
Oh , is $\chi(n)$ the Dirichlet function ? I first saw it in the later part of a book about elementary number theory .
Sort of... it's not the but rather a Dirichlet character.

The link I provided goes in depth and has tables for you to see, but this particular $\chi(n)$ is the only non-principle character modulo $4$.