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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)

$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left[{x}\right]} {sinxdx}\:\:\:{in}\:{that}\:\left[{x}\right]={E}\left({x}\right) \\ $$

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

$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\propto} \:{e}^{−\left[{x}\right]} {sinx}\:{dx}\:\:\:{I}_{{n}} \:\:=\:\:\int_{\mathrm{0}} ^{{n}} \:\:{e}^{−\left[{x}\right]} {sinx}\:{dx} \\ $$$${we}\:{have}\:{I}=\:{lim}_{{n}−>\propto} \:\:\:{I}_{{n}} \:\:\:{but} \\ $$$${I}_{{n}} =\:\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{{k}} ^{{k}+\mathrm{1}} {e}^{−\left[{x}\right]} \:{sinxdx}=\:\sum_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \:{e}^{−{k}} \:\int_{{k}} ^{{k}+\mathrm{1}} {sinx}\:{dx} \\ $$$${I}_{{n}} \:\:=\:\sum_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \:{e}^{−{k}} \left[\:\:−{cosx}\right]_{{k}} ^{{k}+\mathrm{1}} \:\:=\:\:\sum_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \:{e}^{−{k}} \left(\:{cosk}−{cos}\left({k}+\mathrm{1}\right)\right) \\ $$$$=\:\:\sum_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \:{e}^{−{k}} \:{cosk}\:−\:\sum_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \:{e}^{−{k}} {cos}\left({k}+\mathrm{1}\right) \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} {e}^{−{k}} \:{cosk}\:\:−\:\sum_{{k}=\mathrm{1}} ^{{n}} {e}^{−\left({k}−\mathrm{1}\right)} {cosk} \\ $$$$=\:\left(\mathrm{1}−{e}\right)\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{−{k}} \:{cosk} \\ $$$${but}\:\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{−{k}} \:{cos}\:{k}=\:\:{Re}\left(\:\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:{e}^{\left(−\mathrm{1}+{i}\right){k}} \right) \\ $$$${and}\:\:\:\:\sum_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \:{e}^{\left(−\mathrm{1}+{i}\right){k}} \:\:=\:\frac{\mathrm{1}\:−\:{e}^{\left(−\mathrm{1}+{i}\right){n}} }{\mathrm{1}−{e}^{−\mathrm{1}+{i}} } \\ $$$${lim}\:_{{n}−>\propto} {I}_{{n}} \:=\:\:\left(\mathrm{1}−{e}\right)\:\sum_{{k}=\mathrm{0}} ^{\propto} \:{e}^{−{k}} {cosk}\:\:=\left(\mathrm{1}−{e}\right)\:{Re}\left(\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}+{i}} }\right) \\ $$$${but}\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}+{i}} }\:\:=\:\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} \left(\:{cos}\left(\mathrm{1}\right)\:+{isin}\left(\mathrm{1}\right)\right.} \\ $$$$=\:\frac{\mathrm{1}−{e}^{−\mathrm{1}} {cos}\left(\mathrm{1}\right)\:+{i}\:{e}^{−\mathrm{1}} {sin}\left(\mathrm{1}\right)}{\left(\mathrm{1}−{e}^{−\mathrm{1}} {cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} +{e}^{−\mathrm{2}} {sin}^{\mathrm{2}} \left(\mathrm{1}\right)} \\ $$$$\Rightarrow{lim}_{{n}−>\propto} \:{I}_{{n}} \:=\:\:\frac{\mathrm{1}−\:{e}^{−\mathrm{1}} {cos}\left(\mathrm{1}\right)}{\left(\mathrm{1}−\:{e}^{−\mathrm{1}} {cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{e}^{−\mathrm{2}} {sin}^{\mathrm{2}} \left(\mathrm{1}\right)}=\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left[{x}\right]} {sinxdx} \\ $$$$ \\ $$

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