Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function
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Theorem
Let $s \in \C: \map \Re s > 1$.
Let $x \in \R_{>0}$.
Then:
- $\ds \pi^{-s / 2} \map \Gamma {\frac s 2} \map \zeta s = -\frac 1 {s \paren {1 - s} } + \dfrac 1 2 \int_1^\infty \paren {x^{s / 2 - 1} + x^{-\paren {s + 1} / 2} } \paren {\map {\vartheta_3} {0, e^{-\pi x} } - 1} \rd x$
where:
- $\map \Gamma s$ is the gamma function
- $\map \zeta s$ is the Riemann zeta function
- $\ds \map {\vartheta_3} {0, e^{-\pi x} }$ is the Jacobi theta function of the third type.
Proof
Lemma 1
- $\ds \pi^{-s / 2} \map \Gamma {\frac s 2} \map \zeta s = \int_0^1 x^{s / 2 - 1} \sum_{n \mathop = 1}^\infty e^{-\pi n^2 x } \rd x + \int_1^\infty x^{s / 2 - 1} \sum_{n \mathop = 1}^\infty e^{-\pi n^2 x } \rd x$
$\Box$
Lemma 2
- $\ds \int_0^1 x^{s / 2 - 1} \sum_{n \mathop = 1}^\infty e^{-\pi n^2 x} \rd x = -\frac 1 {s \paren {1 - s} } + \int_1^\infty x^{-\paren {s + 1} / 2} \sum_{n \mathop = 1}^\infty e^{-\paren {\pi n^2 x} } \rd x$
$\Box$
Lemma 3
- $\ds \sum_{n \mathop = 1}^\infty e^{-\pi n^2 x} = \dfrac 1 2 \paren {\map {\vartheta_3} {0, e^{-\pi x} } - 1}$
$\Box$
\(\ds \pi^{-s / 2} \map \Gamma {\frac s 2} \map \zeta s\) | \(=\) | \(\ds \int_0^1 x^{s / 2 - 1} \sum_{n \mathop = 1}^\infty e^{-\pi n^2 x} \rd x + \int_1^\infty x^{s / 2 - 1} \sum_{n \mathop = 1}^\infty e^{-\pi n^2 x} \rd x\) | Lemma $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 {s \paren {1 - s} } + \int_1^\infty x^{-\paren {s + 1} / 2} \sum_{n \mathop = 1}^\infty e^{-\paren {\pi n^2 x} } \rd x + \int_1^\infty x^{s / 2 - 1} \sum_{n \mathop = 1}^\infty e^{-\pi n^2 x} \rd x\) | Lemma $2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 {s \paren {1 - s} } + \int_1^\infty \paren {x^{s / 2 - 1} + x^{-s / 2 - 1 / 2} } \sum_{n \mathop = 1}^\infty e^{-\paren {\pi n^2 x} } \rd x\) | Linear Combination of Definite Integrals | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 {s \paren {1 - s} } + \frac 1 2 \int_1^\infty \paren {x^{s / 2 - 1} + x^{-s / 2 - 1 / 2} } \paren {\map {\vartheta_3} {0, e^{-\pi x} } - 1} \rd x\) | Lemma $3$ |
$\blacksquare$