Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function

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Theorem

Let $\zeta$ be the Riemann zeta function.

Let $s \in \C$ be a complex number with real part $\sigma>1$.


Then:

$\displaystyle \pi^{-s / 2} \Gamma \left({\frac s 2}\right) \zeta \left({s}\right) = - \frac 1 {s \left({1 - s}\right)} + \int_1^\infty \left({x^{s / 2 - 1} + x^{- \left({s + 1}\right) / 2} }\right) \omega \left({x}\right) \rd x$

where:

$\Gamma$ is the gamma function
$\displaystyle \omega \left({x}\right) = \sum_{n \mathop = 1}^\infty e^{- \pi n^2 x}$


Proof

By definition of the Gamma function:

$\displaystyle \Gamma \left({\frac s 2}\right) = \int_0^\infty t^{s / 2 - 1} e^{-t} \rd t$

First substitute $t = \pi n^2 x$ to give:

$\displaystyle \pi^{-s / 2} \Gamma \left({\frac s 2}\right) n^{-s} = \int_0^\infty x^{s / 2 - 1} e^{-\pi n^2 x} \rd x$

By definition, on $\operatorname{Re} \left({s}\right) > 1$:

$\displaystyle \zeta \left({s}\right) = \sum_{n \mathop = 1}^\infty n^{-s}$

Therefore summing over $n \ge 1$:

$\displaystyle \pi^{-s / 2} \Gamma \left({\frac s 2}\right) \zeta \left({s}\right) = \int_0^\infty x^{s / 2 - 1} \omega \left({x}\right) \rd x$

where:

$\displaystyle \omega \left({x}\right) = \sum_{n \mathop = 1}^\infty e^{-\pi n^2 x}$

Next the integral is split at $x = 1$ as follows:

\(\displaystyle \pi^{-s / 2}\Gamma \left({\frac s 2}\right) \zeta \left({s}\right)\) \(=\) \(\displaystyle \int_0^1 x^{s / 2 - 1} \omega \left({x}\right) \rd x + \int_1^\infty x^{s / 2 - 1} \omega \left({x}\right) \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \int_1^\infty x^{-s / 2 - 1} \omega \left({\frac 1 x}\right) \rd x + \int_1^\infty x^{s / 2 - 1} \omega \left({x}\right) \rd x\) substituting $x \mapsto \dfrac 1 x$ in the first integral.

Recall the Jacobi theta function:

\(\displaystyle \theta \left({x}\right)\) \(=\) \(\displaystyle \sum_{n \mathop \in \Z} e^{- \pi n^2 x}\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \omega \left({x}\right) + 1\)

Since $e^{-x^2}$ is a fixed point of the Fourier transform, by Properties of the Fourier Transform:

$\mathcal F \left[{e^{-\pi t^2 x} }\right] \left({u}\right) = x^{-1 / 2} e^{-\pi u^2 / x}$

where $\mathcal F$ denotes the Fourier transform.

Therefore, by the Poisson Summation Formula:

$\theta \left({x}\right) = \dfrac 1 {\sqrt x} \theta \left({\dfrac 1 x}\right)$

whence:

$\omega \left({\dfrac 1 x}\right) = -\dfrac 1 2 + \dfrac 1 2 \sqrt x + \sqrt x \omega \left({x}\right)$

Therefore:

\(\displaystyle \int_1^\infty x^{-s / 2 - 1} \omega \left({\frac 1 x}\right) \rd x\) \(=\) \(\displaystyle \int_1^\infty x^{-s / 2 - 1} \left({-\frac 1 2 + \frac 1 2 \sqrt x + \sqrt x \omega \left({x}\right)}\right) \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle - \frac 1 s + \frac 1 {s - 1} + \int_1^\infty x^{-\left({s + 1}\right)/2} \omega \left({x}\right) \rd x\)

So:

$\displaystyle \pi^{-s / 2} \Gamma \left({\frac s 2}\right) \zeta \left({s}\right) = -\frac 1 {s \left({1 - s}\right)} + \int_1^\infty \left[{x^{s / 2} + x^{\left({1 - s}\right) / 2} }\right] \omega \left({x}\right) \frac {\rd x} x$

as required.

$\blacksquare$


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