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Relationship between the Gamma Function and the Riemann Zeta Function and the Dirichlet Eta Function 📂Functions

Relationship between the Gamma Function and the Riemann Zeta Function and the Dirichlet Eta Function

정리

If Re⁡(s)>1\operatorname{Re} (s) > 1 then ζ(s)Γ(s)=M[1ex−1](s)=∫0∞xs−1ex−1dxη(s)Γ(s)=M[1ex+1](s)=∫0∞xs−1ex+1dx \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

설명

The Dirichlet eta function η(s)\eta (s) is not only mathematically interesting due to its relationship with the Riemann zeta function ζ(s)\zeta (s) as an alternating series, but it can also be neatly summarized through intermediation by the Gamma function Γ(S)\Gamma (S) and Mellin transform M\mathcal{M} as shown above.

증명

Strategy: Expanding f(x)=(ex−1)−1f(x) = \left( e^{x} - 1 \right)^{-1} and g(x)=(ex+1)−1g(x) = \left( e^{x} + 1 \right)^{-1} into series and using substitution integration within the definite integral to derive nn. If x>0x > 0 then by the geometric series 11−e−x=1+e−x+e−2x+⋯ {{ 1 } \over { 1 - e^{-x} }} = 1 + e^{-x} + e^{-2x} + \cdots Rearranging the term 11 on the right side gives e−x+e−2x+⋯=11−e−x−1=e−x1−e−x=1ex−1 \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,gf,g and the sequence of functions {fN}N∈N,{gN}N∈N\left\{ f_{N} \right\}_{N \in \mathbb{N}}, \left\{ g_{N} \right\}_{N \in \mathbb{N}} as follows: f(x):=1ex−1=e−x+e−2x+⋯fN(x):=∑n=1Ne−nxg(x):=1ex+1=e−x−e−2x+⋯gN(x):=∑n=1N(−1)n−1e−nx 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→∞N \to \infty, fN→fgN→g f_{N} \to f \\ g_{N} \to g thus, allowing the use of the Dominated Convergence Theorem for integration.

In the Mellin transform of ff, substituting as in z:=nxz := nx gives 1ndz=dx\displaystyle {{ 1 } \over { n }} dz =dx, and by the Dominated Convergence Theorem (DCT), M[1ex−1](s)=∫0∞xs−11ex−1dx=DCTlim⁡N→∞∫0∞xs−1∑n=1Ne−nxdx=lim⁡n→∞∑n=1N∫0∞(zn)s−1e−z1ndz=∑n=1∞1ns∫0∞zs−1e−zdz=ζ(s)Γ(s) \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, M[1ex+1](s)=∫0∞xs−11ex+1dx=DCTlim⁡N→∞∫0∞xs−1∑n=1N(−1)n−1e−nxdx=lim⁡N→∞∑n=1N(−1)n−1∫0∞(zn)s−1e−z1ndz=∑n=1∞(−1)n−1ns∫0∞zs−1e−zdz=η(s)Γ(s) \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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