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Question Number 141417 by mnjuly1970 last updated on 18/May/21

$$........advanced.........calculus.......$$$$provethat::$$$$\mathcal{F}:={\int }_{-1}^0\frac{e^x+e^{\frac{1}{x}}-1}{x}dx\stackrel{??}{=}\gamma$$

Answered by mnjuly1970 last updated on 27/May/21

$$:={\int }_{-1}^0\frac{e^{\frac{1}{x}}}{x}dx\stackrel{\frac{1}{x}=-y}{=}{\int }_1^{\mathrm{\infty }}-{ye}^{-y}\frac{dy}{y^2}$$$$:=-{\int }_1^{\mathrm{\infty }}\frac{e^{-y}}{y}dy=-{\int }_1^{\mathrm{\infty }}{e}^{-y}ln\left(y\right)-{\left[ln\left(y\right)e^{-y}\right]}_1^{\mathrm{\infty }}$$$$:=-{\int }_1^{\mathrm{\infty }}{e}^{-y}ln\left(y\right)dy\Rightarrow -{\int }_0^{\mathrm{\infty }}{e}^{-y}ln\left(y\right)dy+{\int }_0^1{e}^{-y}ln\left(y\right)dy$$$${\int }_{-1}^0\frac{e^{\frac{1}{x}}}{x}dx:=\gamma +{\int }_0^1{e}^{-x}ln\left(x\right)dx\left(★\right)$$$$:=\gamma +{\int }_0^{1}ln\left(x\right)d(1-e^{-x})$$$$:=\gamma +{\left[ln\left(x\right)(1-e^{-x})\right]}_0^1-{\int }_0^1\left(\frac{1-e^{-x}}{x}\right)dx$$$$:=\gamma -{\int }_0^1\frac{1-e^{-x}}{x}dx=\gamma +{\int }_0^{-1}\frac{1-e^x}{-x}dx$$$$:=\gamma +{\int }_{-1}^0\frac{1-e^x}{x}dx$$$$\left(★\right)::{\int }_{-1}^0\frac{e^{\frac{1}{x}}+e^x-1}{x}=\gamma .....✓$$$$........\mathcal{F}:={\int }_{-1}^0\frac{e^x+e^{\frac{1}{x}}-1}{x}dx=\gamma ......$$$$$$

Answered by mindispower last updated on 19/May/21

$$x=\frac{1}{t}\Rightarrow \mathcal{F}={\int }_{-1}^0\frac{e^x-1}{x}dx+{\int }_{-1}^0\frac{e^{\frac{1}{x}}}{x}dx=A+B$$$${A=[e^x-1]ln(-x)]}_{-1}^0-{\int }_{-1}^0{e}^{x}ln(-x)$$$$=-{\int }_{-1}^0{\mathit{e}}^{\mathit{x}}\mathit{l}\mathit{n}(-\mathit{x})\mathit{d}\mathit{x}={\int }_1^0{e}^{-t}ln\left(t\right)dt$$$$\mathit{B}={\left[ln(-x)e^{\frac{1}{x}}\right]}_{-1}^0-{\int }_{-1}^0\frac{e^{\frac{1}{x}}}{x^2}ln(-x)dx$$$$={\int }_{-1}^0-\frac{{\mathit{e}}^{\frac{1}{\mathit{x}}}}{{\mathit{x}}^2}\mathit{l}\mathit{n}(-\mathit{x}),t=-\frac{1}{x}\Rightarrow$$$$B=-{\int }_1^{\mathrm{\infty }}{e}^{-t}ln\left(t\right)$$$$\mathcal{F}=A+B=-{\int }_0^{\mathrm{\infty }}ln\left(t\right)e^{-t}dt={\partial }_x(-{\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{x-1}dt){\mid }_{x=1}$$$$={\partial }_x.-\mathrm{\Gamma }(x+1){\mid }_{x=0}=-{\mathrm{\Gamma }}^{\prime }\left(1\right)=-\mathrm{\Gamma }\left(1\right)\mathrm{\Psi }\left(1\right)=-1.-\gamma =\gamma$$$$\Rightarrow {\int }_{-1}^0\frac{e^x+e^{\frac{1}{x}}-1}{x}dx=\gamma$$$$$$$$$$

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