Question Number 155360 by mathlove last updated on 29/Sep/21
Answered by puissant last updated on 30/Sep/21
$$\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\psi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}=\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{2} \\ $$$${or}\:\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{2}{ln}\mathrm{2}−\gamma \\ $$$$\Rightarrow\:\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\:−\mathrm{2}{ln}\mathrm{2}−\gamma+\mathrm{2}\:=\:\mathrm{2}+{ln}\frac{\mathrm{1}}{\mathrm{4}}−\gamma \\ $$