Question Number 26575 by abdo imad last updated on 26/Dec/17
![find the value of ∫_0 ^∞ e^(−[x]) sinxdx in that [x]=E(x)](https://www.tinkutara.com/question/Q26575.png)
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left[{x}\right]} {sinxdx}\:\:\:{in}\:{that}\:\left[{x}\right]={E}\left({x}\right) \\ $$
Commented by abdo imad last updated on 29/Dec/17
![let put I= ∫_0 ^∝ e^(−[x]) sinx dx I_n = ∫_0 ^n e^(−[x]) sinx dx we have I= lim_(n−>∝) I_n but I_n = Σ_(k=0) ^(n−1) ∫_k ^(k+1) e^(−[x]) sinxdx= Σ_(k=0) ^(k=n−1) e^(−k) ∫_k ^(k+1) sinx dx I_n = Σ_(k=0) ^(k=n−1) e^(−k) [ −cosx]_k ^(k+1) = Σ_(k=0) ^(k=n−1) e^(−k) ( cosk−cos(k+1)) = Σ_(k=0) ^(k=n−1) e^(−k) cosk − Σ_(k=0) ^(k=n−1) e^(−k) cos(k+1) = Σ_(k=0) ^(k=n−1) e^(−k) cosk − Σ_(k=1) ^n e^(−(k−1)) cosk = (1−e) Σ_(k=0) ^(n−1) e^(−k) cosk but Σ_(k=0) ^(n−1) e^(−k) cos k= Re( Σ_(k=0) ^(n−1) e^((−1+i)k) ) and Σ_(k=0) ^(k=n−1) e^((−1+i)k) = ((1 − e^((−1+i)n) )/(1−e^(−1+i) )) lim _(n−>∝) I_n = (1−e) Σ_(k=0) ^∝ e^(−k) cosk =(1−e) Re( (1/(1−e^(−1+i) ))) but (1/(1−e^(−1+i) )) = (1/(1−e^(−1) ( cos(1) +isin(1))) = ((1−e^(−1) cos(1) +i e^(−1) sin(1))/((1−e^(−1) cos(1))^2 +e^(−2) sin^2 (1))) ⇒lim_(n−>∝) I_n = ((1− e^(−1) cos(1))/((1− e^(−1) cos(1))^2 +e^(−2) sin^2 (1)))= ∫_0 ^∞ e^(−[x]) sinxdx](https://www.tinkutara.com/question/Q26813.png)
$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\propto} \:{e}^{−\left[{x}\right]} {sinx}\:{dx}\:\:\:{I}_{{n}} \:\:=\:\:\int_{\mathrm{0}} ^{{n}} \:\:{e}^{−\left[{x}\right]} {sinx}\:{dx} \\ $$$${we}\:{have}\:{I}=\:{lim}_{{n}−>\propto} \:\:\:{I}_{{n}} \:\:\:{but} \\ $$$${I}_{{n}} =\:\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{{k}} ^{{k}+\mathrm{1}} {e}^{−\left[{x}\right]} \:{sinxdx}=\:\sum_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \:{e}^{−{k}} \:\int_{{k}} ^{{k}+\mathrm{1}} {sinx}\:{dx} \\ $$$${I}_{{n}} \:\:=\:\sum_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \:{e}^{−{k}} \left[\:\:−{cosx}\right]_{{k}} ^{{k}+\mathrm{1}} \:\:=\:\:\sum_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \:{e}^{−{k}} \left(\:{cosk}−{cos}\left({k}+\mathrm{1}\right)\right) \\ $$$$=\:\:\sum_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \:{e}^{−{k}} \:{cosk}\:−\:\sum_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \:{e}^{−{k}} {cos}\left({k}+\mathrm{1}\right) \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} {e}^{−{k}} \:{cosk}\:\:−\:\sum_{{k}=\mathrm{1}} ^{{n}} {e}^{−\left({k}−\mathrm{1}\right)} {cosk} \\ $$$$=\:\left(\mathrm{1}−{e}\right)\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{−{k}} \:{cosk} \\ $$$${but}\:\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{−{k}} \:{cos}\:{k}=\:\:{Re}\left(\:\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:{e}^{\left(−\mathrm{1}+{i}\right){k}} \right) \\ $$$${and}\:\:\:\:\sum_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \:{e}^{\left(−\mathrm{1}+{i}\right){k}} \:\:=\:\frac{\mathrm{1}\:−\:{e}^{\left(−\mathrm{1}+{i}\right){n}} }{\mathrm{1}−{e}^{−\mathrm{1}+{i}} } \\ $$$${lim}\:_{{n}−>\propto} {I}_{{n}} \:=\:\:\left(\mathrm{1}−{e}\right)\:\sum_{{k}=\mathrm{0}} ^{\propto} \:{e}^{−{k}} {cosk}\:\:=\left(\mathrm{1}−{e}\right)\:{Re}\left(\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}+{i}} }\right) \\ $$$${but}\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}+{i}} }\:\:=\:\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} \left(\:{cos}\left(\mathrm{1}\right)\:+{isin}\left(\mathrm{1}\right)\right.} \\ $$$$=\:\frac{\mathrm{1}−{e}^{−\mathrm{1}} {cos}\left(\mathrm{1}\right)\:+{i}\:{e}^{−\mathrm{1}} {sin}\left(\mathrm{1}\right)}{\left(\mathrm{1}−{e}^{−\mathrm{1}} {cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} +{e}^{−\mathrm{2}} {sin}^{\mathrm{2}} \left(\mathrm{1}\right)} \\ $$$$\Rightarrow{lim}_{{n}−>\propto} \:{I}_{{n}} \:=\:\:\frac{\mathrm{1}−\:{e}^{−\mathrm{1}} {cos}\left(\mathrm{1}\right)}{\left(\mathrm{1}−\:{e}^{−\mathrm{1}} {cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{e}^{−\mathrm{2}} {sin}^{\mathrm{2}} \left(\mathrm{1}\right)}=\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left[{x}\right]} {sinxdx} \\ $$$$ \\ $$