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Question Number 119965 by sdfg last updated on 28/Oct/20

Answered by Dwaipayan Shikari last updated on 28/Oct/20

$$\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt$$$$\frac{d\mathrm{\Gamma }\left(x\right)}{dx}={\int }_0^{\mathrm{\infty }}\frac{\partial }{\partial x}\left(t^{x-1}{e}^{-t}\right)dt$$$$\frac{d\mathrm{\Gamma }\left(x\right)}{dx}={\int }_0^{\mathrm{\infty }}logt{t}^{x-1}{e}^{-t}dt$$$$So$$$$Generally$$$$\frac{d^n\mathrm{\Gamma }\left(x\right)}{{dx}^n}={\int }_0^{\mathrm{\infty }}{log}^{n}t{t}^{x-1}{e}^{-t}dt$$$$$$$$$$

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