Question Number 27613 by abdo imad last updated on 10/Jan/18
![find the value of ∫_0 ^∞ e^(−[x] −x) dx .](Q27613.png)
$${find}\:{the}\:{value}\:{of}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left[{x}\right]\:−{x}} {dx}\:\:. \\ $$
Commented by abdo imad last updated on 12/Jan/18
![I_n = Σ_(k=0) ^n ∫_k ^(k+1) e^(−[x]−x) dx I=lim I_n I_n = Σ_(k=) ^n e^(−k) [e^(−x) ]_k ^(k+1) = Σ_(k=0) ^n e^(−k) ( e^(−k) −e^(−k−1) ) I_n = Σ_(k=0) ^n e^(−2k) −e^(−1) Σ_(k=0) ^n e^(−2k) =(1−e^(−1) ) Σ_(k=0) ^n (e^(−2) )^k ∫_0 ^∞ e^(−[x]−x) dx = lim_(n−>∝) I_n =(1−e^(−1) ).(1/(1−e^(−2) )) = ((e^2 (1−e^(−1) ))/(e^2 −1))= ((e^2 −e)/(e^2 −1))=((e(e−1))/((e+1)(e−1)))= (e/(e+1)) .](Q27661.png)
$${I}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\int_{{k}} ^{{k}+\mathrm{1}} {e}^{−\left[{x}\right]−{x}} {dx}\:\:{I}={lim}\:{I}_{{n}} \\ $$$${I}_{{n}} \:=\:\sum_{{k}=} ^{{n}} {e}^{−{k}} \:\left[{e}^{−{x}} \right]_{{k}} ^{{k}+\mathrm{1}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} {e}^{−{k}} \left(\:{e}^{−{k}} \:−{e}^{−{k}−\mathrm{1}} \right)\: \\ $$$${I}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{e}^{−\mathrm{2}{k}} \:\:−{e}^{−\mathrm{1}} \sum_{{k}=\mathrm{0}} ^{{n}} \:{e}^{−\mathrm{2}{k}} \:\:=\left(\mathrm{1}−{e}^{−\mathrm{1}} \right)\:\sum_{{k}=\mathrm{0}} ^{{n}} \left({e}^{−\mathrm{2}} \right)^{{k}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left[{x}\right]−{x}} {dx}\:=\:{lim}_{{n}−>\propto} \:{I}_{{n}} \:=\left(\mathrm{1}−{e}^{−\mathrm{1}} \right).\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{2}} } \\ $$$$=\:\frac{{e}^{\mathrm{2}} \left(\mathrm{1}−{e}^{−\mathrm{1}} \right)}{{e}^{\mathrm{2}} −\mathrm{1}}=\:\frac{{e}^{\mathrm{2}} \:−{e}}{{e}^{\mathrm{2}} −\mathrm{1}}=\frac{{e}\left({e}−\mathrm{1}\right)}{\left({e}+\mathrm{1}\right)\left({e}−\mathrm{1}\right)}=\:\:\frac{{e}}{{e}+\mathrm{1}}\:. \\ $$
Answered by prakash jain last updated on 12/Jan/18
![=Σ_(i=0) ^∞ e^(−i) ∫_i ^(i+1) e^(−x) dx =Σ_(i=0) ^∞ e^(−i) [−e^(−x) ]_i ^(i+1) =Σ_(i=0) ^∞ (e^(−i) /e^i )−(e^(−i) /e^(i+1) ) =Σ_(i=0) ^∞ (1/e^(2i) )−(1/e^(2i+1) ) =Σ_(i=0) ^∞ (1/e^(2i) )−Σ_(i=0) ^∞ (1/e^(2i+1) ) =(1/(1−(1/e^2 )))−((1/e)/(1−(1/e^2 ))) =((e(e−1))/(e^2 −1))=(e/(e+1))](Q27656.png)
$$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{i}} \int_{{i}} ^{{i}+\mathrm{1}} {e}^{−{x}} {dx} \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{i}} \left[−{e}^{−{x}} \right]_{{i}} ^{{i}+\mathrm{1}} \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{e}^{−{i}} }{{e}^{{i}} }−\frac{{e}^{−{i}} }{{e}^{{i}+\mathrm{1}} } \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{e}^{\mathrm{2}{i}} }−\frac{\mathrm{1}}{{e}^{\mathrm{2}{i}+\mathrm{1}} } \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{e}^{\mathrm{2}{i}} }−\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{e}^{\mathrm{2}{i}+\mathrm{1}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{{e}^{\mathrm{2}} }}−\frac{\mathrm{1}/{e}}{\mathrm{1}−\frac{\mathrm{1}}{{e}^{\mathrm{2}} }} \\ $$$$=\frac{{e}\left({e}−\mathrm{1}\right)}{{e}^{\mathrm{2}} −\mathrm{1}}=\frac{{e}}{{e}+\mathrm{1}} \\ $$