Riemann $\theta$-function

The Riemann theta function is a number-theoretic functionwhich is only really used inthe derivation of the functional equation for the Riemann xi function.

The Riemann theta function is defined as:

$$\theta (x)=2\omega (x)+1,$$

where $\omega$ is the Riemann omega function.

The domain (http://planetmath.org/Function) of the Riemann theta function is $x>0$.

To give an explicit form for the theta function, note that

$\omega (x)$$=$${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{e}^{-n^2\pi x}$

$=$${\displaystyle \sum _{n=-1}^{-\mathrm{\infty }}}{e}^{-{(-n)}^2\pi x}$

$=$${\displaystyle \sum _{n=-1}^{-\mathrm{\infty }}}{e}^{-n^2\pi x}$

and so

$2\omega (x)+1$$=$${\displaystyle \sum _{n=-1}^{-\mathrm{\infty }}}{e}^{-n^2\pi x}+\omega (x)+1$

$=$${\displaystyle \sum _{n=-1}^{-\mathrm{\infty }}}{e}^{-n^2\pi x}+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{e}^{-n^2\pi x}+e^{-0^2\pi x}$

$=$${\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-n^2\pi x}.$

Thus we have

$$\theta (x)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-n^2\pi x}.$$

Riemann showed that the theta function satisfied a functional equation,which was the key stepin the proof of the analytic continuation for the Riemann xi function.This has direct consequences for the Riemann zeta function.