Relationship between the Gamma Function and the Riemann Zeta Function and the Dirichlet Eta Function
정리
If $\operatorname{Re} (s) > 1$ then $$ \zeta (s) \Gamma (s) = \mathcal{M} \left[ {{ 1 } \over { e^{x} - 1 }} \right] (s) = \int_{0}^{\infty} {{ x^{s-1} } \over { e^{x} - 1 }} dx \\ \eta (s) \Gamma (s) = \mathcal{M} \left[ {{ 1 } \over { e^{x} + 1 }} \right] (s) = \int_{0}^{\infty} {{ x^{s-1} } \over { e^{x} + 1 }} dx $$
- $\mathcal{M}$ is a Mellin transform.
- $\operatorname{Re} (s)$ represents the real part of the complex number $s$.
설명
The Dirichlet eta function $\eta (s)$ is not only mathematically interesting due to its relationship with the Riemann zeta function $\zeta (s)$ as an alternating series, but it can also be neatly summarized through intermediation by the Gamma function $\Gamma (S)$ and Mellin transform $\mathcal{M}$ as shown above.
증명
Strategy: Expanding $f(x) = \left( e^{x} - 1 \right)^{-1}$ and $g(x) = \left( e^{x} + 1 \right)^{-1}$ into series and using substitution integration within the definite integral to derive $n$. If $x > 0$ then by the geometric series $$ {{ 1 } \over { 1 - e^{-x} }} = 1 + e^{-x} + e^{-2x} + \cdots $$ Rearranging the term $1$ on the right side gives $$ \begin{align*} & e^{-x} + e^{-2x} + \cdots \\ =& {{ 1 } \over { 1 - e^{-x} }} - 1 \\ =& {{ e^{-x} } \over { 1 - e^{-x} }} \\ =& {{ 1 } \over { e^{x} -1 }} \end{align*} $$ Let’s define the function $f,g$ and the sequence of functions $\left\{ f_{N} \right\}_{N \in \mathbb{N}}, \left\{ g_{N} \right\}_{N \in \mathbb{N}}$ as follows: $$ f(x) := {{ 1 } \over { e^{x} - 1 }} = e^{-x} + e^{-2x} + \cdots \\ f_{N}(x) := \sum_{n=1}^{N} e^{-nx} \\ g(x) := {{ 1 } \over { e^{x} + 1 }} = e^{-x} - e^{-2x} + \cdots \\ g_{N}(x) := \sum_{n=1}^{N} (-1)^{n-1} e^{-nx} $$ Then, when $N \to \infty$, $$ f_{N} \to f \\ g_{N} \to g $$ thus, allowing the use of the Dominated Convergence Theorem for integration.
In the Mellin transform of $f$, substituting as in $z := nx$ gives $\displaystyle {{ 1 } \over { n }} dz =dx$, and by the Dominated Convergence Theorem (DCT), $$ \begin{align*} \mathcal{M} \left[ {{ 1 } \over { e^{x} - 1 }} \right] (s) =& \int_{0}^{\infty} x^{s-1} {{ 1 } \over { e^{x} - 1 }} dx \\ \overset{\text{DCT}}{=}& \lim_{N \to \infty} \int_{0}^{\infty} x^{s-1} \sum_{n=1}^{N} e^{-nx} dx \\ =& \lim_{n \to \infty} \sum_{n=1}^{N} \int_{0}^{\infty} \left( {{ z } \over { n }} \right)^{s-1} e^{-z} {{ 1 } \over { n }} dz \\ =& \sum_{n=1}^{\infty} {{ 1 } \over { n^{s} }} \int_{0}^{\infty} z^{s-1} e^{-z} dz \\ =& \zeta (s) \Gamma (s) \end{align*} $$ Similarly, $$ \begin{align*} \mathcal{M} \left[ {{ 1 } \over { e^{x} + 1 }} \right] (s) =& \int_{0}^{\infty} x^{s-1} {{ 1 } \over { e^{x} + 1 }} dx \\ \overset{\text{DCT}}{=}& \lim_{N \to \infty} \int_{0}^{\infty} x^{s-1} \sum_{n=1}^{N} (-1)^{n-1} e^{-nx} dx \\ =& \lim_{N \to \infty} \sum_{n=1}^{N} (-1)^{n-1} \int_{0}^{\infty} \left( {{ z } \over { n }} \right)^{s-1} e^{-z} {{ 1 } \over { n }} dz \\ =& \sum_{n=1}^{\infty} {{ (-1)^{n-1} } \over { n^{s} }} \int_{0}^{\infty} z^{s-1} e^{-z} dz \\ =& \eta (s) \Gamma (s) \end{align*} $$
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