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Question Number 145197 by Gbenga last updated on 03/Jul/21

$$d/dxofx!=?$$

Commented by Dwaipayan Shikari last updated on 03/Jul/21

$$x!=\mathrm{\Gamma }(x+1)$$$$\frac{dx!}{dx}={\mathrm{\Gamma }}^{\prime }(x+1)=\mathrm{\Gamma }(x+1)\psi (x+1)As\frac{{\mathrm{\Gamma }}^{\prime }(x+1)}{\mathrm{\Gamma }(x+1)}=\psi (x+1)$$

Answered by puissant last updated on 03/Jul/21

$$\mathrm{\Gamma }\left(\mathrm{x}\right)=(\mathrm{x}-1)!={\int }_0^{+\mathrm{\infty }}{\mathrm{t}}^{\mathrm{x}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}$$$$\frac{1}{\mathrm{\Gamma }\left(\mathrm{x}\right)}={\mathrm{xe}}^{\gamma \mathrm{x}}\underset{\mathrm{k}=1}{\prod ^{\mathrm{\infty }}}(1+\frac{\mathrm{x}}{\mathrm{k}}){\mathrm{e}}^{-\frac{\mathrm{x}}{\mathrm{k}}}$$$$\ln \left(\frac{1}{\mathrm{\Gamma }\left(\mathrm{x}\right)}\right)=-\ln \left(\mathrm{\Gamma }\left(\mathrm{x}\right)\right)$$$$=\ln \left(\mathrm{x}\right)+\gamma \mathrm{x}+\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}(1+\frac{\mathrm{x}}{\mathrm{k}})-\frac{\mathrm{x}}{\mathrm{k}}$$$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\ln \left(\frac{1}{\mathrm{\Gamma }\left(\mathrm{x}\right)}\right)\right)=-{\left(\ln \left(\mathrm{\Gamma }\left(\mathrm{x}\right)\right)\right)}^{\prime }=-\frac{{\mathrm{\Gamma }}^{\prime }\left(\mathrm{x}\right)}{\mathrm{\Gamma }\left(\mathrm{x}\right)}$$$$=\frac{1}{\mathrm{x}}+\gamma +\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{1+\frac{\mathrm{x}}{\mathrm{k}}}\times \frac{1}{\mathrm{k}}-\frac{1}{\mathrm{k}}$$$$=-\frac{1}{\mathrm{x}}-\gamma +\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{k}}-\frac{1}{\mathrm{x}+\mathrm{k}}$$$$\lambda \left(\mathrm{x}\right)=-\gamma +\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{k}}-\frac{1}{\mathrm{k}-1+\mathrm{x}}$$$$\Rightarrow \lambda (\mathrm{x}+1)=-\gamma +\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{k}}-\frac{1}{\mathrm{k}+\mathrm{x}}$$$$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}!)=\frac{{\mathrm{\Gamma }}^{\prime }(\mathrm{x}+1)}{\mathrm{\Gamma }(\mathrm{x}+1)}....$$

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