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Reciprocal times Derivative of Gamma Function 📂Functions

Reciprocal times Derivative of Gamma Function

Definition

The derivative of the logarithm of the Gamma function is called the digamma function.

ψ0(z):=ddzln⁡Γ(z)=Γ′(z)Γ(z) \psi_{0} (z) := \dfrac{d}{dz} \ln \Gamma (z) = \dfrac{\Gamma^{\prime}(z)}{\Gamma (z)}

Theorem

For the Gamma function Γ\Gamma and the Euler-Mascheroni constant γ\gamma, the following holds: Γ′(z)Γ(z)=−γ+∑n=1∞(1n−1z+n−1) {{ \Gamma ' (z) } \over { \Gamma (z) }} = - \gamma + \sum_{n=1}^{\infty} \left( {{ 1 } \over { n }} - {{ 1 } \over { z + n - 1 }} \right)

Proof 1

Weierstrass’s product representation for the Gamma function: For the Gamma function Γ:(0,∞)→R\Gamma : (0, \infty) \to \mathbb{R}, the following holds: 1Γ(x)=xeγxlim⁡n→∞∏k=1n(1+xk)e−xk {1 \over \Gamma (x)} = x e^{\gamma x } \lim_{n \to \infty} \prod_{k=1}^{n} \left( 1 + {x \over k} \right) e^{- {x \over k} }

Taking the reciprocal of the Weierstrass product representation gives: Γ(z)=e−γzz∏n=1nez/n1+z/n \Gamma (z) = {{ e^{-\gamma z} } \over { z }} \prod_{n=1}^{n} {{ e^{z/n} } \over { 1 + z/n }} According to the product rule of differentiation, Γ′(z)=−e−γz(1+γz)z2∏n=1nez/n1+z/n+e−γzz∑n=1∞[zn(z+n)∏k=1∞ez/k1+z/k]=−e−γz(1+γz)z2ze−γzΓ(z)+e−γzz∑n=1∞[zn(z+n)ze−γzΓ(z)]=−1+γzzΓ(z)+Γ(z)∑n=1∞[zn(z+n)] \begin{align*} & \Gamma ' (z) \\ =& - {{ e^{-\gamma z} \left( 1 + \gamma z \right) } \over { z^{2} }} \prod_{n=1}^{n} {{ e^{z/n} } \over { 1 + z/n }} + {{ e^{-\gamma z} } \over { z }} \sum_{n=1}^{\infty} \left[ {{ z } \over { n \left( z + n \right) }} \prod_{k=1}^{\infty} {{ e^{z/k} } \over { 1 + z/k }} \right] \\ =& - {{ e^{-\gamma z} \left( 1 + \gamma z \right) } \over { z^{2} }} {{ z } \over { e^{-\gamma z} }} \Gamma (z) + {{ e^{-\gamma z} } \over { z }} \sum_{n=1}^{\infty} \left[ {{ z } \over { n \left( z + n \right) }} {{ z } \over { e^{-\gamma z} }} \Gamma (z) \right] \\ =& - {{ 1 + \gamma z } \over { z }} \Gamma (z) + \Gamma (z) \sum_{n=1}^{\infty} \left[ {{ z } \over { n \left( z + n \right) }} \right] \end{align*} and dividing both sides by Γ(z)\Gamma (z) gives: Γ′(z)Γ(z)=−1+γzz+∑n=1∞[zn(z+n)]=−γ−1z+∑n=1∞[1n−1z+n]=−γ+∑n=1∞[1n−1z+n−1] \begin{align*} {{ \Gamma ' (z) } \over { \Gamma (z) }} =& - {{ 1 + \gamma z } \over { z }} + \sum_{n=1}^{\infty} \left[ {{ z } \over { n \left( z + n \right) }} \right] \\ =& - \gamma - {{ 1 } \over { z }} + \sum_{n=1}^{\infty} \left[ {{ 1 } \over { n }} - {{ 1 } \over { z + n }} \right] \\ =& - \gamma + \sum_{n=1}^{\infty} \left[ {{ 1 } \over { n }} - {{ 1 } \over { z + n - 1 }} \right] \end{align*}

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  1. https://proofwiki.org/wiki/Reciprocal_times_Derivative_of_Gamma_Function ↩︎