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o-oo-e-E-x-dx-




Question Number 186223 by SANOGO last updated on 02/Feb/23
∫_o ^(+oo) e^(−E(x)dx)
$$\int_{{o}} ^{+{oo}} {e}^{−{E}\left({x}\right){dx}} \\ $$
Answered by Mathspace last updated on 02/Feb/23
I=Σ_(n=0) ^∞ ∫_n ^(n+1) e^(−n) dx  =Σ_(n=0) ^∞ e^(−n)  =Σ_(n=0) ^∞ (e^(−1) )^n   =(1/(1−e^(−1) ))=(1/(1−(1/e)))=(e/(e−1))  $I=(e/(e−1))$
$${I}=\sum_{{n}=\mathrm{0}} ^{\infty} \int_{{n}} ^{{n}+\mathrm{1}} {e}^{−{n}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} {e}^{−{n}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \left({e}^{−\mathrm{1}} \right)^{{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{{e}}}=\frac{{e}}{{e}−\mathrm{1}} \\ $$$$\${I}=\frac{{e}}{{e}−\mathrm{1}}\$ \\ $$
Commented by SANOGO last updated on 02/Feb/23
merci bien
$${merci}\:{bien} \\ $$
Commented by Mathspace last updated on 03/Feb/23
you are welcome
$${you}\:{are}\:{welcome} \\ $$

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