Question Number 124548 by mnjuly1970 last updated on 04/Dec/20
![...nice calculus... evaluate ::: lim_(x→0) {(1/x)[((ln(Γ(1+x))/x)−ψ(x+1)]}=?](Q124548.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:\:\:\:\:{evaluate}\:::: \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \left\{\frac{\mathrm{1}}{{x}}\left[\frac{{ln}\left(\Gamma\left(\mathrm{1}+{x}\right)\right.}{{x}}−\psi\left({x}+\mathrm{1}\right)\right]\right\}=? \\ $$$$ \\ $$
Answered by mindispower last updated on 04/Dec/20
![ln(Γ(1+x))=f(x) f(x)=f(0)+f′(0)x+((f′′(0)x^2 )/2)+o(x^2 ) f(0)=0,f′(0)=Ψ(1),f′′(0)=Ψ^1 (1) Ψ(x+1)=Ψ(1)+Ψ^1 (1)x+o(x) ((ln(Γ(1+x)))/x)−Ψ(x+1)=((xΨ(1)+(x^2 /2)Ψ^1 (1)+o(x^2 ))/x)−Ψ(1)−Ψ^1 (1)x+o(x) =−((Ψ^1 (1))/2)x+o(x) lim_(x→0) {(1/x)[((ln(Γ(1+x)))/x)−Ψ(1+x)]}=lim_(x→0) (1/x).((−Ψ(1))/2)x+o(x) =lim_(x→0) −((Ψ^1 (1))/2)+o(1)=−((Ψ^1 (1))/2) Ψ^1 (z)=Σ_(j≥0) (1/((j+z)^2 ))⇒Ψ^1 (1)=Σ_(j≥0) (1/((1+j)^2 ))=(π^2 /6) we get −(π^2 /(12))](Q124563.png)
$${ln}\left(\Gamma\left(\mathrm{1}+{x}\right)\right)={f}\left({x}\right) \\ $$$${f}\left({x}\right)={f}\left(\mathrm{0}\right)+{f}'\left(\mathrm{0}\right){x}+\frac{{f}''\left(\mathrm{0}\right){x}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0},{f}'\left(\mathrm{0}\right)=\Psi\left(\mathrm{1}\right),{f}''\left(\mathrm{0}\right)=\Psi^{\mathrm{1}} \left(\mathrm{1}\right) \\ $$$$\Psi\left({x}+\mathrm{1}\right)=\Psi\left(\mathrm{1}\right)+\Psi^{\mathrm{1}} \left(\mathrm{1}\right){x}+{o}\left({x}\right) \\ $$$$\frac{{ln}\left(\Gamma\left(\mathrm{1}+{x}\right)\right)}{{x}}−\Psi\left({x}+\mathrm{1}\right)=\frac{{x}\Psi\left(\mathrm{1}\right)+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\Psi^{\mathrm{1}} \left(\mathrm{1}\right)+{o}\left({x}^{\mathrm{2}} \right)}{{x}}−\Psi\left(\mathrm{1}\right)−\Psi^{\mathrm{1}} \left(\mathrm{1}\right){x}+{o}\left({x}\right) \\ $$$$=−\frac{\Psi^{\mathrm{1}} \left(\mathrm{1}\right)}{\mathrm{2}}{x}+{o}\left({x}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{x}}\left[\frac{{ln}\left(\Gamma\left(\mathrm{1}+{x}\right)\right)}{{x}}−\Psi\left(\mathrm{1}+{x}\right)\right]\right\}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}.\frac{−\Psi\left(\mathrm{1}\right)}{\mathrm{2}}{x}+{o}\left({x}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\frac{\Psi^{\mathrm{1}} \left(\mathrm{1}\right)}{\mathrm{2}}+{o}\left(\mathrm{1}\right)=−\frac{\Psi^{\mathrm{1}} \left(\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Psi^{\mathrm{1}} \left({z}\right)=\underset{{j}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({j}+{z}\right)^{\mathrm{2}} }\Rightarrow\Psi^{\mathrm{1}} \left(\mathrm{1}\right)=\underset{{j}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{1}+{j}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${we}\:{get}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$
Commented by mnjuly1970 last updated on 04/Dec/20
$${bravo}\:{mr}\:{mindspower} \\ $$$$\:{maclauren}\:{expansion}\:{of}\:\:\Gamma\left({x}+\mathrm{1}\right) \\ $$$${grateful}... \\ $$