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Question Number 192668 by mathocean1 last updated on 24/May/23

$${\int }_0^4\frac{1}{x}{e}^{x}dx-{\int }_0^4\frac{1}{x}{e}^{\frac{1}{2}x}dx=?$$

Answered by witcher3 last updated on 24/May/23

$$\mathrm{fals}$$$${\int }_0^1\frac{{\mathrm{e}}^{\mathrm{kx}}}{\mathrm{x}}\mathrm{dx}\mathrm{not}\mathrm{rieman}\mathrm{integrabl}\mathrm{but}\mathrm{lesbegue}\mathrm{integrable}$$$$\mathrm{cause}\mathrm{\forall }\mathrm{n}\in {\mathbb{Z}}_{+}\mu \left([0,\frac{1}{\mathrm{n}}]\right)=\frac{1}{\mathrm{n}}\to 0\mathrm{for}\mathrm{n}\to \mathrm{\infty }$$$$\mathrm{\forall }\mathrm{n}\in \mathbb{N}$$$${\int }_{\frac{1}{\mathrm{n}}}^4\frac{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{x}}\mathrm{dx}={\int }_0^4\frac{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{x}}{1}_{[\frac{1}{\mathrm{n}},4]}..\mathrm{exist}\mathrm{for}\mathrm{all}\mathrm{n}\in \mathbb{N}$$$$$$$$$$$$$$

Answered by Frix last updated on 24/May/23

$$\int \frac{{\mathrm{e}}^x}{x}dx=\mathrm{Ei}\left(x\right)+C$$$$\int \frac{e^{\frac{x}{2}}}{x}dx=\mathrm{Ei}\left(\frac{x}{2}\right)+C$$$$\mathrm{You}\mathrm{need}\mathrm{to}\mathrm{find}\mathrm{a}\mathrm{table}\mathrm{for}\mathrm{values}\mathrm{of}\mathrm{Ei}\left(x\right)$$$$\mathrm{The}\mathrm{answer}\mathrm{is}\approx .747750730933$$

Commented by mathocean1 last updated on 25/May/23

$$okthanks$$

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