Approximaths

Zeta summation IV

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In the previous post we derived the functional equation of the theta function

\displaystyle \vartheta(t) = t^{-1/2} \vartheta(1/t), \qquad t>0.

With t > 0. We observe that:

\displaystyle \vartheta(t) \leq C t^{-1/2} \text{ as } t \to 0

note that

\displaystyle \sum_{n \geq 1}e^{-\pi n^2 t} \leq \sum_{n \geq 1}e^{-\pi n t} \leq Ce^{-\pi t}, t \geq 1

Remember the definition of the Gamma function. For s >0 the Gamma function is defined by:

\displaystyle \Gamma(s) = \int_{0}^{\infty} e^{-t} t^{s-1}\,dt
\displaystyle \Gamma(s/2) = \int_{0}^{\infty} e^{-t} t^{\frac{s}{2}-1}\,dt

Define u:= \frac{t}{\pi n^2} \implies du = \frac{1}{\pi n^2}\,dt and dt = \pi n^2 \,du

\displaystyle \Gamma(s/2) = \int_{0}^{\infty} e^{-\pi n^2 u} (\pi n^2 u)^{\frac{s}{2}-1} \pi n^2\,du
\displaystyle \Gamma(s/2) = \pi^{s/2} n^{s}\int_{0}^{\infty} e^{-\pi n^2 u} u^{\frac{s}{2}-1}\,du
\displaystyle \pi^{-s/2} \Gamma(s/2) n^{-s} = \int_{0}^{\infty} e^{-\pi n^2 u} u^{\frac{s}{2}-1}\,du \qquad (n \geq 1)

Observe that for the \vartheta(t) defined above:

\displaystyle \vartheta(u) = \sum_{n \in \mathbb{Z}} e^{-\pi n^2 u}
\displaystyle = 2 \sum_{n = 1}^{\infty} e^{-\pi n^2 u} + 1
\displaystyle \frac{\vartheta(u) -1}{2} = \sum_{n = 1}^{\infty} e^{-\pi n^2 u}

It follows that:

\displaystyle \int_{0}^{\infty} u^{\frac{s}{2}-1} \biggl(\frac{\vartheta(u) -1}{2}\biggr) \,du = \int_{0}^{\infty} u^{\frac{s}{2}-1} \sum_{n = 1}^{ \infty} e^{-\pi n^2 u} \,du
\displaystyle = \sum_{n = 1}^{\infty} \int_{0}^{\infty} u^{\frac{s}{2}-1} e^{-\pi n^2 u}\,du
\displaystyle = \sum_{n = 1}^{\infty} \int_{0}^{\infty} e^{-\pi n^2 u} u^{\frac{s}{2}-1} \,du
\displaystyle = \sum_{n = 1}^{\infty} \pi^{-s/2} \Gamma(s/2) n^{-s}
\displaystyle = \pi^{-s/2} \Gamma(s/2) \sum_{n = 1}^{\infty} n^{-s}
\displaystyle = \pi^{-s/2} \Gamma(s/2) \zeta(s)

Therefore:

\displaystyle \zeta(s) = \frac{\int_{0}^{\infty} u^{\frac{s}{2}-1} \biggl(\frac{\vartheta(u) -1}{2}\biggr) \,du}{\pi^{-s/2} \Gamma(s/2) }

The define the Xi function as :

\displaystyle \xi(s) := \int_{0}^{\infty} u^{\frac{s}{2}-1} \biggl(\frac{\vartheta(u) -1}{2}\biggr) \,du

We finally have

\boxed{\displaystyle \zeta(s) = \pi^{s/2}\frac{\xi(s)}{\Gamma(s/2)}}

which is the meromorphic continuation into the complex plane of the Zeta function with a simple pole at s = 1.

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