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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)

findthevalueof∫0∞e−[x]sinxdxinthat[x]=E(x)

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

letputI=∫0∝e−[x]sinxdxIn=∫0ne−[x]sinxdxwehaveI=limn−>∝InbutIn=∑k=0n−1∫kk+1e−[x]sinxdx=∑k=0k=n−1e−k∫kk+1sinxdxIn=∑k=0k=n−1e−k[−cosx]kk+1=∑k=0k=n−1e−k(cosk−cos(k+1))=∑k=0k=n−1e−kcosk−∑k=0k=n−1e−kcos(k+1)=∑k=0k=n−1e−kcosk−∑k=1ne−(k−1)cosk=(1−e)∑k=0n−1e−kcoskbut∑k=0n−1e−kcosk=Re(∑k=0n−1e(−1+i)k)and∑k=0k=n−1e(−1+i)k=1−e(−1+i)n1−e−1+ilimn−>∝In=(1−e)∑k=0∝e−kcosk=(1−e)Re(11−e−1+i)but11−e−1+i=11−e−1(cos(1)+isin(1)=1−e−1cos(1)+ie−1sin(1)(1−e−1cos(1))2+e−2sin2(1)⇒limn−>∝In=1−e−1cos(1)(1−e−1cos(1))2+e−2sin2(1)=∫0∞e−[x]sinxdx

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