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Question Number 131546 by mathmax by abdo last updated on 05/Feb/21

$$\mathrm{prove}\mathrm{that}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}\ln \left(\mathrm{x}\right)\mathrm{dx}=-\gamma$$

Answered by Dwaipayan Shikari last updated on 06/Feb/21

$$I\left(a\right)={\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{a}dx=\mathrm{\Gamma }(a+1)$$$$I^{\prime }\left(a\right)={\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{a}log\left(x\right)dx={\mathrm{\Gamma }}^{\prime }(a+1)$$$$I^{\prime }\left(0\right)={\int }_0^{\mathrm{\infty }}{e}^{-x}log\left(x\right)dx={\mathrm{\Gamma }}^{\prime }\left(1\right)=\mathrm{\Gamma }\left(1\right)\psi \left(1\right)=-\gamma$$

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