Question Number 124548 by mnjuly1970 last updated on 04/Dec/20
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:{calculus}… \\ $$$$\:\:\:\:\:\:\:{evaluate}\:::: \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \left\{\frac{\mathrm{1}}{{x}}\left[\frac{{ln}\left(\Gamma\left(\mathrm{1}+{x}\right)\right.}{{x}}−\psi\left({x}+\mathrm{1}\right)\right]\right\}=? \\ $$$$ \\ $$
Answered by mindispower last updated on 04/Dec/20
$${ln}\left(\Gamma\left(\mathrm{1}+{x}\right)\right)={f}\left({x}\right) \\ $$$${f}\left({x}\right)={f}\left(\mathrm{0}\right)+{f}'\left(\mathrm{0}\right){x}+\frac{{f}''\left(\mathrm{0}\right){x}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0},{f}'\left(\mathrm{0}\right)=\Psi\left(\mathrm{1}\right),{f}''\left(\mathrm{0}\right)=\Psi^{\mathrm{1}} \left(\mathrm{1}\right) \\ $$$$\Psi\left({x}+\mathrm{1}\right)=\Psi\left(\mathrm{1}\right)+\Psi^{\mathrm{1}} \left(\mathrm{1}\right){x}+{o}\left({x}\right) \\ $$$$\frac{{ln}\left(\Gamma\left(\mathrm{1}+{x}\right)\right)}{{x}}−\Psi\left({x}+\mathrm{1}\right)=\frac{{x}\Psi\left(\mathrm{1}\right)+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\Psi^{\mathrm{1}} \left(\mathrm{1}\right)+{o}\left({x}^{\mathrm{2}} \right)}{{x}}−\Psi\left(\mathrm{1}\right)−\Psi^{\mathrm{1}} \left(\mathrm{1}\right){x}+{o}\left({x}\right) \\ $$$$=−\frac{\Psi^{\mathrm{1}} \left(\mathrm{1}\right)}{\mathrm{2}}{x}+{o}\left({x}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{x}}\left[\frac{{ln}\left(\Gamma\left(\mathrm{1}+{x}\right)\right)}{{x}}−\Psi\left(\mathrm{1}+{x}\right)\right]\right\}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}.\frac{−\Psi\left(\mathrm{1}\right)}{\mathrm{2}}{x}+{o}\left({x}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\frac{\Psi^{\mathrm{1}} \left(\mathrm{1}\right)}{\mathrm{2}}+{o}\left(\mathrm{1}\right)=−\frac{\Psi^{\mathrm{1}} \left(\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Psi^{\mathrm{1}} \left({z}\right)=\underset{{j}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({j}+{z}\right)^{\mathrm{2}} }\Rightarrow\Psi^{\mathrm{1}} \left(\mathrm{1}\right)=\underset{{j}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{1}+{j}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${we}\:{get}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$
Commented by mnjuly1970 last updated on 04/Dec/20
$${bravo}\:{mr}\:{mindspower} \\ $$$$\:{maclauren}\:{expansion}\:{of}\:\:\Gamma\left({x}+\mathrm{1}\right) \\ $$$${grateful}… \\ $$