Question and Answers Forum

All Questions      Topic List

Logarithms Questions

Previous in All Question      Next in All Question      

Previous in Logarithms      Next in Logarithms      

Question Number 117826 by snipers237 last updated on 13/Oct/20

$$ProvethattheEulerConstantisqlsoequalto$$$$\underset{x\to -1}{\lim }\mathrm{\Gamma }\left(x\right)-\frac{1}{x(x+1)}$$

Answered by mnjuly1970 last updated on 14/Oct/20

$$solution::$$$${lim}_{x\to -1}[\mathrm{\Gamma }\left(x\right)-\frac{1}{x(x+1)}]$$$$={lim}_{x\to -1}\left(\frac{\mathrm{\Gamma }(x+2)-1}{x(x+1)}\right)$$$$\stackrel{l^{\prime }hospital}{=}{lim}_{x\to -1}\left(\frac{\mathrm{\Gamma }{}^{\prime }(x+2)}{2x+1}\right)$$$$$$$$=\frac{{\mathrm{\Gamma }}^{\prime }\left(1\right)}{-1}=-{\mathrm{\Gamma }}^{\prime }\left(1\right)=-\psi \left(1\right)\mathrm{\Gamma }\left(1\right)=\gamma$$$$where::\gamma is\mathcal{E}uler-$$$$\mathcal{M}asheroniconstant$$$$...m.n.1970...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com