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Question-62455




Question Number 62455 by aliesam last updated on 21/Jun/19
Commented by mathmax by abdo last updated on 22/Jun/19
S =Σ_(n=1) ^∞  A_n    with A_n =∫_(n−(π/2)) ^(n+(π/2))  e^(−x) ∣cos(x)∣dx   changement x =n−(π/2) +t give  A_n = ∫_0 ^π    e^(−(n−(π/2)+t)))  ∣cos((π/2)−(n+t))∣dt               = e^((π/2)−n)  ∫_0 ^π   e^(−t)   ∣sin(n+t)∣ dt      if we suppose sin(n+t)≥0 on[0,π]( that need  a proof) ⇒A_n = e^((π/2)−n)  ∫_0 ^π (sin(n)cos(t) +cos(n)sin(t))dt  =sin(n) e^((π/2)−n)  ∫_0 ^π  cost dt+cos(n) e^((π/2)−n)  ∫_0 ^π  sint dt  =0 +cos(n) e^((π/2)−n)  [−cost]_0 ^π  =2 cos(n) e^((π/2)−n)  ⇒  S =2Σ_(n=1) ^∞   cos(n)e^((π/2)−n)  =2 e^(π/2)  Σ_(n=1) ^∞  e^(−n)  cos(n)  =2 e^(π/2)  (Σ_(n=0) ^∞   e^(−n)  cos(n)−1)  but  Σ_(n=0) ^∞  e^(−n)  cosn =Re( Σ_(n=0) ^∞  e^(−n+in) ) =Re(Σ_(n=0) ^∞  (e^(−1+i) )^n )  we have ∣e^(−1 +i) ∣ =(1/e)<1 ⇒  Σ_(n=0) ^∞ (e^(−1+i) )^n  =(1/(1−e^(−1+i) )) =(1/(1−e^(−1) (cos(1) +isin(1))))  =(1/(1−e^(−1) cos(1)−ie^(−1) sin(1))) =((1−e^(−1) cos(1)+e^(−1)  sin(1))/((1−e^(−1) cos(1))^2  +e^(−2)  sin^2 (1))) ⇒  Σ_(n=0) ^∞  e^(−n)  cos(n) =((1−e^(−1)  cos(1))/((1−e^(−1)  cos(1))^2  +e^(−2) sin^2 (1))) ⇒  S =2 e^(π/2) { ((1−e^(−1)  cos(1))/(1−2e^(−1)  cos(1) +e^(−2) )) −1} .
S=n=1AnwithAn=nπ2n+π2excos(x)dxchangementx=nπ2+tgiveAn=0πe(nπ2+t))cos(π2(n+t))dt=eπ2n0πetsin(n+t)dtifwesupposesin(n+t)0on[0,π](thatneedaproof)An=eπ2n0π(sin(n)cos(t)+cos(n)sin(t))dt=sin(n)eπ2n0πcostdt+cos(n)eπ2n0πsintdt=0+cos(n)eπ2n[cost]0π=2cos(n)eπ2nS=2n=1cos(n)eπ2n=2eπ2n=1encos(n)=2eπ2(n=0encos(n)1)butn=0encosn=Re(n=0en+in)=Re(n=0(e1+i)n)wehavee1+i=1e<1n=0(e1+i)n=11e1+i=11e1(cos(1)+isin(1))=11e1cos(1)ie1sin(1)=1e1cos(1)+e1sin(1)(1e1cos(1))2+e2sin2(1)n=0encos(n)=1e1cos(1)(1e1cos(1))2+e2sin2(1)S=2eπ2{1e1cos(1)12e1cos(1)+e21}.
Commented by aliesam last updated on 22/Jun/19
thank you sir brilliant solution
thankyousirbrilliantsolution
Commented by mathmax by abdo last updated on 22/Jun/19
you are welcome sir.
youarewelcomesir.

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