【习题4.2 -2】Γ函数

2. 令 ψ(z)=dln⁡Γ(z)dz\begin{aligned}\psi(z) = \frac{d\\ln \\Gamma(z)}{dz}\end{aligned}ψ(z)=dzdlnΓ(z)

① 证明ψ(z+1)=1z+ψ(z)\begin{aligned} \psi(z+1) = \frac{1}{z} + \psi(z)\end{aligned}ψ(z+1)=z1+ψ(z)

证：
ψ(z)=dln⁡Γ(z)dz=Γ(z)′Γ(z)\begin{aligned} \psi(z) = \frac{d\\ln \\Gamma(z)}{dz} = \frac{\\Gamma(z)'}{\Gamma(z)} \end{aligned}ψ(z)=dzdlnΓ(z)=Γ(z)Γ(z)′
ψ(z+1)=Γ(z+1)′Γ(z+1)=zΓ(z)′z⋅Γ(z)=Γ(z)+zΓ(z)′zΓ(z)\begin{aligned} \psi(z+1) = \frac{\\Gamma(z+1)'}{\Gamma(z+1)} = \frac{z\\Gamma(z)'}{z\cdot\Gamma(z)} = \frac{\Gamma(z) + z\\Gamma(z)'}{z\Gamma(z)} \end{aligned}ψ(z+1)=Γ(z+1)Γ(z+1)′=z⋅Γ(z)zΓ(z)′=zΓ(z)Γ(z)+zΓ(z)′
=1z+Γ(z)′Γ(z)=1z+ψ(z)\begin{aligned} = \frac{1}{z} + \frac{\\Gamma(z)'}{\Gamma(z)} = \frac{1}{z} + \psi(z) \end{aligned}=z1+Γ(z)Γ(z)′=z1+ψ(z)

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② 证明 ψ(1−z)=ψ(z)+πcot⁡πz\begin{aligned}\psi(1-z) = \psi(z) + \pi \cot \pi z\end{aligned}ψ(1−z)=ψ(z)+πcotπz

证：
ψ(z)=dln⁡Γ(z)dz=Γ(z)′Γ(z)\begin{aligned} \psi(z) = \frac{d\\ln \\Gamma(z)}{dz} = \frac{\\Gamma(z)'}{\Gamma(z)} \end{aligned}ψ(z)=dzdlnΓ(z)=Γ(z)Γ(z)′
ψ(1−z)=dln⁡Γ(1−z)d(1−z)=−Γ(1−z)′Γ(1−z)\begin{aligned} \psi(1-z) = \frac{d\\ln \\Gamma(1-z)}{d(1-z)} = -\frac{\\Gamma(1-z)'}{\Gamma(1-z)} \end{aligned}ψ(1−z)=d(1−z)dlnΓ(1−z)=−Γ(1−z)Γ(1−z)′
=−πsin⁡πz⋅1Γ(z)′⋅sin⁡πz⋅Γ(z)π\begin{aligned} = -\left \\frac{\\pi}{\\sin \\pi z} \\cdot \\frac{1}{\\Gamma(z)} \\right' \cdot \frac{\sin \pi z \cdot \Gamma(z)}{\pi} \end{aligned}=−sinπzπ⋅Γ(z)1′⋅πsinπz⋅Γ(z)
=−−πsin⁡πz⋅Γ(z)2πcos⁡πz⋅Γ(z)+sin⁡πz⋅Γ′(z)⋅sin⁡πz⋅Γ(z)π\begin{aligned} = -\frac{-\pi}{\\sin \\pi z \\cdot \\Gamma(z)^2} \\pi \\cos \\pi z \\cdot \\Gamma(z) + \\sin \\pi z \\cdot \\Gamma'(z) \cdot \frac{\sin \pi z \cdot \Gamma(z)}{\pi} \end{aligned}=−sinπz⋅Γ(z)2−ππcosπz⋅Γ(z)+sinπz⋅Γ′(z)⋅πsinπz⋅Γ(z)
=1sin⁡πz⋅Γ(z)πcos⁡πz⋅Γ(z)+sin⁡πz⋅Γ′(z)\begin{aligned} = \frac{1}{\sin \pi z \cdot \Gamma(z)} \\pi \\cos \\pi z \\cdot \\Gamma(z) + \\sin \\pi z \\cdot \\Gamma'(z) \end{aligned}=sinπz⋅Γ(z)1πcosπz⋅Γ(z)+sinπz⋅Γ′(z)
=πcot⁡πz+Γ′(z)Γ(z)\begin{aligned} = \pi \cot \pi z + \frac{\Gamma'(z)}{\Gamma(z)} \end{aligned}=πcotπz+Γ(z)Γ′(z)
=πcot⁡πz+ψ(z)\begin{aligned} = \pi \cot \pi z + \psi(z) \end{aligned}=πcotπz+ψ(z)