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Question Number 186223 by SANOGO last updated on 02/Feb/23

$${\int }_o^{+oo}{e}^{-E\left(x\right)dx}$$

Answered by Mathspace last updated on 02/Feb/23

$$I=\sum _{n=0}^{\mathrm{\infty }}{\int }_n^{n+1}{e}^{-n}dx$$$$=\sum _{n=0}^{\mathrm{\infty }}{e}^{-n}=\sum _{n=0}^{\mathrm{\infty }}{\left(e^{-1}\right)}^n$$$$=\frac{1}{1-e^{-1}}=\frac{1}{1-\frac{1}{e}}=\frac{e}{e-1}$$$$\mathrm{\$}I=\frac{e}{e-1}\mathrm{\$}$$

Commented by SANOGO last updated on 02/Feb/23

$$mercibien$$

Commented by Mathspace last updated on 03/Feb/23

$$youarewelcome$$

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