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Question Number 155360 by mathlove last updated on 29/Sep/21

Answered by puissant last updated on 30/Sep/21

$$\psi \left(\frac{3}{2}\right)=\psi (1+\frac{1}{2})=\psi \left(\frac{1}{2}\right)+\frac{1}{\frac{1}{2}}=\psi \left(\frac{1}{2}\right)+2$$$$or\psi \left(\frac{1}{2}\right)=-2ln2-\gamma$$$$\Rightarrow \psi \left(\frac{3}{2}\right)=-2ln2-\gamma +2=2+ln\frac{1}{4}-\gamma$$

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