Quote Originally Posted by EinStone View Post
Define $\displaystyle L : \{s \in \mathbb{C}| Re(s) > 0\} \rightarrow \mathbb{C}$ by $\displaystyle L(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$.

Note that $\displaystyle L(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}}{e^t+e^{-t}}dt$

Show that L can be extended to an entire function $\displaystyle L : \mathbb{C} \rightarrow \mathbb{C}$
It turns out that $\displaystyle L(s) $ is continued by the following:

$\displaystyle L(1-s) = \left(\frac{2}{\pi}\right)^s\sin\left(\frac{\pi s}{2}\right)\Gamma(s)L(s) $