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Question Number 90113 by M±th+et£s last updated on 21/Apr/20

$${\int }_0^{\mathrm{\infty }}{e}^{-x}(\frac{1}{1-e^{-x}}-\frac{1}{x})dx$$

Answered by maths mind last updated on 21/Apr/20

$$\mathrm{\Psi }\left(z\right)={\int }_0^{+\mathrm{\infty }}(\frac{e^{-t}}{t}-\frac{e^{-zt}}{1-e^{-t}})dt$$$${\int }_0^{+\mathrm{\infty }}{e}^{-x}(\frac{1}{1-e^{-x}}-\frac{1}{x})dx={\int }_0^{\mathrm{\infty }}(\frac{e^{-x}}{1-e^{-x}}-\frac{e^{-x}}{x})dx$$$$=-{\int }_0^{\mathrm{\infty }}(\frac{e^{-x}}{x}-\frac{e^{-1x}}{1-e^{-x}})dx=-\mathrm{\Psi }\left(1\right)=\gamma$$

Commented by M±th+et£s last updated on 21/Apr/20

$$wownicedigamma!$$

Commented by maths mind last updated on 21/Apr/20

$$thanxsir$$$$$$

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