# How does one compute the Euler product for the Dirichlet Beta function?

In this post, the author derives the Euler product for Dirichlet Beta function, defined as

$$\beta(s) = \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}$$

for $$\Re(s)>0$$ and obtains

$$\beta(s) = \prod_p \bigg(\frac {p^s}{p^s-\sin\left(\frac{p \,\pi}{2}\right)} \bigg).$$

I would like to know how one arrives to the Euler product, can someone show me a proof?

As the author of this post did, using Dirichlet character $$\chi:=\chi_4$$, we can write $$\displaystyle \beta(s)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}=\sum_{n=1}^\infty \frac{\chi(n)}{n^s}$$ where $$\chi(n)=\begin{cases}1\ \ \ \text{ for }n\equiv 1\ \text{mod }4\\-1\ \ \ \text{ for }n\equiv 3\ \text{mod }4\\0\ \ \ \text{ otherwise}\end{cases}.$$ We can see that $$\chi(nm)=\chi(n)\chi(m)$$ for all $$n,m\ge 1$$, hence \begin{align*} \beta(s)&=\sum_{n=1}^\infty \frac{\chi(n)}{n^s}\\&=\sum_{r_1,r_2,\cdots\ge 0} \frac{\prod_j \chi(p_j)^{r_j}}{\prod_j p_j^{sr_j}} \\&=\prod_j \left(1+\frac{\chi(p_j)}{p_j^s} +\frac{\chi(p_j)^2}{p_j^{2s}}+\cdots\right)\\&=\prod_p \frac1{1-p^{-s}\chi(p)}\\&=\prod_p \frac{p^s}{p^s-\chi(p)}=\prod_p \frac{p^s}{p^s-\sin\left(\frac{p\pi}2\right)}, \end{align*} for all $$\Re(s)>1$$.