Question-139171

Question Number 139171 by mnjuly1970 last updated on 23/Apr/21

Answered by Dwaipayan Shikari last updated on 23/Apr/21

Γ(1+x)=1−γx+O(x^2 )⇒Γ(x)=(1/x)−γ+O(x) Γ′(1+x)=−γ+O(x)⇒xΓ(x)ψ(x+1)=−γ+O(x) ⇒xΓ(x)ψ(x)+Γ(x)=−γ+O(x)⇒ψ(x)∼−(γ/(xΓ(x)))−(1/x) lim_(x→0) ((ψ(x))/(Γ(x)))=((−(γ/(xΓ(x)))−(1/x))/((1/x)−γ))=((−γx−1)/(1−γx))=−1

$$\Gamma\left(\mathrm{1}+{x}\right)=\mathrm{1}−\gamma{x}+{O}\left({x}^{\mathrm{2}} \right)\Rightarrow\Gamma\left({x}\right)=\frac{\mathrm{1}}{{x}}−\gamma+{O}\left({x}\right) \\ $$$$\Gamma'\left(\mathrm{1}+{x}\right)=−\gamma+{O}\left({x}\right)\Rightarrow{x}\Gamma\left({x}\right)\psi\left({x}+\mathrm{1}\right)=−\gamma+{O}\left({x}\right) \\ $$$$\Rightarrow{x}\Gamma\left({x}\right)\psi\left({x}\right)+\Gamma\left({x}\right)=−\gamma+{O}\left({x}\right)\Rightarrow\psi\left({x}\right)\sim−\frac{\gamma}{{x}\Gamma\left({x}\right)}−\frac{\mathrm{1}}{{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\psi\left({x}\right)}{\Gamma\left({x}\right)}=\frac{−\frac{\gamma}{{x}\Gamma\left({x}\right)}−\frac{\mathrm{1}}{{x}}}{\frac{\mathrm{1}}{{x}}−\gamma}=\frac{−\gamma{x}−\mathrm{1}}{\mathrm{1}−\gamma{x}}=−\mathrm{1} \\ $$

Commented by mnjuly1970 last updated on 24/Apr/21