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Question Number 140399 by mnjuly1970 last updated on 07/May/21

$$$$$$\mathit{\xi }:={\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}-e^{-x}}{x}dx=k.\gamma$$$$find{}^{″}k{}^{″}...$$$$\gamma :=\mathcal{E}ulerconstant....$$

Answered by qaz last updated on 07/May/21

$$-\gamma ={\int }_0^{\mathrm{\infty }}(e^{-x}-\frac{1}{1+x})\frac{dx}{x}$$$$-\gamma ={\int }_0^{\mathrm{\infty }}(e^{-x^2}-\frac{1}{1+x^2})\frac{d\left(x^2\right)}{x^2}=2{\int }_0^{\mathrm{\infty }}(e^{-x^2}-\frac{1}{1+x^2})\frac{dx}{x}$$$$\Rightarrow -\frac{\gamma }{2}={\int }_0^{\mathrm{\infty }}(e^{-x^2}-\frac{1}{1+x^2})\frac{dx}{x}$$$$\xi ={\int }_0^{\mathrm{\infty }}(e^{-x^2}-e^{-x})\frac{dx}{x}$$$$={\int }_0^{\mathrm{\infty }}\{(e^{-x^2}-\frac{1}{1+x^2})-(e^{-x}-\frac{1}{1+x})+(\frac{1}{1+x^2}-\frac{1}{1+x})\}\frac{dx}{x}$$$$=-\frac{\gamma }{2}+\gamma +{\int }_0^{\mathrm{\infty }}(\frac{1}{1+x^2}-\frac{1}{1+x})\frac{dx}{x}$$$$=\frac{\gamma }{2}+{\int }_0^{\mathrm{\infty }}(\frac{1}{1+x}-\frac{x}{1+x^2})dx$$$$=\frac{\gamma }{2}+ln\frac{1+x}{\sqrt{1+x^2}}{\mid }_0^{\mathrm{\infty }}$$$$=\frac{\gamma }{2}$$$$\Rightarrow k=\frac{1}{2}$$

Commented by mnjuly1970 last updated on 07/May/21

$$bravo...mrpayan...$$

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