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Question Number 99234 by abdomathmax last updated on 19/Jun/20

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}\mathrm{lnx}\mathrm{dx}$$

Answered by maths mind last updated on 19/Jun/20

$$\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{x-1}dt$$$${\mathrm{\Gamma }}^{\prime }\left(x\right)={\int }_0^{+\mathrm{\infty }}{e}^{-t}ln\left(t\right)t^{x-1}dt$$$${\mathrm{\Gamma }}^{\prime }\left(1\right)={\int }_0^{+\mathrm{\infty }}{e}^{-t}ln\left(t\right)dt=-\gamma$$

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