Question Number 87013 by mathmax by abdo last updated on 01/Apr/20
![calculate ∫_0 ^∞ (e^(−[x]) /(x+1))dx](Q87013.png)
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\left[{x}\right]} }{{x}+\mathrm{1}}{dx} \\ $$
Commented by mathmax by abdo last updated on 02/Apr/20
![I =∫_1 ^∞ (e^(−[x]) /(x+1))dx ⇒ I =Σ_(n=1) ^∞ ∫_n ^(n+1) (e^(−n) /(x+1))dx =Σ_(n=1) ^∞ e^(−n) [ln(x+1)]_n ^(n+1) =Σ_(n=1) ^∞ e^(−n) {ln(n+1)−ln(n)} =Σ_(n=1) ^∞ e^(−n) ln(((n+1)/n)) =Σ_(n=1) ^∞ e^(−n) ln(1+(1/n)) and this serie is convergent.](Q87082.png)
$${I}\:=\int_{\mathrm{1}} ^{\infty} \:\:\frac{{e}^{−\left[{x}\right]} }{{x}+\mathrm{1}}{dx}\:\Rightarrow\:{I}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\int_{{n}} ^{{n}+\mathrm{1}} \:\frac{{e}^{−{n}} }{{x}+\mathrm{1}}{dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{−{n}} \left[{ln}\left({x}+\mathrm{1}\right)\right]_{{n}} ^{{n}+\mathrm{1}} \:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{−{n}} \left\{{ln}\left({n}+\mathrm{1}\right)−{ln}\left({n}\right)\right\} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{−{n}} {ln}\left(\frac{{n}+\mathrm{1}}{{n}}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{−{n}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\:\:{and}\:{this}\:{serie}\:{is} \\ $$$${convergent}. \\ $$
Commented by mathmax by abdo last updated on 02/Apr/20
![forgive the Q is calculate ∫_1 ^(+∞) (e^(−[x]) /(x+1))dx](Q87083.png)
$${forgive}\:{the}\:{Q}\:{is}\:{calculate}\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{e}^{−\left[{x}\right]} }{{x}+\mathrm{1}}{dx} \\ $$