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Question Number 44318 by abdo.msup.com last updated on 26/Sep/18

$$letf\left(x\right)={\int }_x^{+\mathrm{\infty }}\frac{e^{-t}}{t}dt$$$$1)calculate{f}^{{}^{\prime }}\left(x\right)$$$$2)findaequivalentoff\left(x\right)when$$$$x\to +\mathrm{\infty }.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Sep/18

Commented by maxmathsup by imad last updated on 27/Sep/18

$$wehavef\left(x\right)={\int }_x^1\frac{e^{-t}}{t}dt+{\int }_1^{+\mathrm{\infty }}\frac{e^{-t}}{t}dt={\int }_1^{+\mathrm{\infty }}\frac{e^{-t}}{t}dt-{\int }_1^x\frac{e^{-t}}{t}dt\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=-\frac{e^{-x}}{x}(x>0)$$$$2)letintegratebyparts$$$$f\left(x\right)={[-\frac{e^{-t}}{t}]}_x^{+\mathrm{\infty }}-{\int }_x^{+\mathrm{\infty }}\frac{e^{-t}}{t^2}dt=\frac{e^{-x}}{x}-{\int }_x^{+\mathrm{\infty }}\frac{e^{-t}}{t^2}dt\Rightarrow$$$$\mid f\left(x\right)-\frac{e^{-x}}{x}\mid =\mid {\int }_x^{+\mathrm{\infty }}\frac{e^{-t}}{t^2}dt\mid ⩽{\int }_x^{+\mathrm{\infty }}\frac{dt}{t^2}={[-\frac{1}{t}]}_x^{+\mathrm{\infty }}=\frac{1}{x}\to 0(x\to +\mathrm{\infty })\Rightarrow$$$$f\left(x\right)\sim \frac{e^{-x}}{x}(x\to +\mathrm{\infty }).$$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Sep/18

$$\frac{df\left(x\right)}{dx}=-\frac{e^{-x}}{x}\times \frac{dx}{dx}$$$$f^{{}^{\prime }}\left(x\right)=-\frac{e^{-x}}{x}$$$$$$

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