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Question Number 44706 by maxmathsup by imad last updated on 03/Oct/18
let f_α (x) = ((cos(αx))/(1+x^2 ))  1) calculate f^((n)) (x) and f^((n)) (0)  2) developp f at integr serie  3) give ∫_0 ^x  f_α (t) dt  at form of serie   4) developp  ∫_0 ^∞     f_α (t)dt  at  integr serie .
letfα(x)=cos(αx)1+x21)calculatef(n)(x)andf(n)(0)2)developpfatintegrserie3)give0xfα(t)dtatformofserie4)developp0fα(t)dtatintegrserie.
Commented by maxmathsup by imad last updated on 04/Oct/18
1) we have f_α (x) =((cos(αx))/((x−i)(x+i))) =((cos(αx))/(2i)){(1/(x−i)) −(1/(x+i))}  =(1/(2i)){ ((cos(αx))/(x−i)) −((cos(αx))/(x+i))} ⇒f^((n)) (x) =(1/(2i)){ (((cos(αx))/(x−i)))^((n)) −(((cos(αx))/(x+i)))^((n)) } but  (((cos(αx))/((x−i))))^((n))  =Σ_(k=0) ^n  C_n ^k   (cos(αx)^ )^((k))  ((1/(x−i)))^((n−k))     (leibniz formulae)  =Σ_(k=0) ^n  C_n ^k   cos(αx +((kπ)/2)) (((−1)^(n−k) (n−k)!)/((x−i)^(n−k +1) ))  also  (((cos(αx))/(x+i)))^((n))  =Σ_(k=0) ^n  C_n ^k   cos(αx+(π/2))(((−1)^(n−k) (n−k)!)/((x+i)^(n−k+1) )) ⇒  f^((n)) (x)=(1/(2i)) Σ_(k=0) ^n   C_n ^k   (−1)^(n−k) (n−k)! cos(αx+((kπ)/2)){(1/((x−i)^(n−k+1) )) −(1/((x+i)^(n−k+1) ))}  f^((n)) (0) =(1/(2i)) Σ_(k=0) ^n  (−1)^(n−k)  ((n!)/(k!)) cos(((kπ)/2)){i^(n−k+1)  −(−i)^(n−k+1) }  =Σ_(k=0) ^n  (−1)^(n−k)  ((n!)/(k!)) cos(((kπ)/2))sin(n−k+1)(π/2)
1)wehavefα(x)=cos(αx)(xi)(x+i)=cos(αx)2i{1xi1x+i}=12i{cos(αx)xicos(αx)x+i}f(n)(x)=12i{(cos(αx)xi)(n)(cos(αx)x+i)(n)}but(cos(αx)(xi))(n)=k=0nCnk(cos(αx))(k)(1xi)(nk)(leibnizformulae)=k=0nCnkcos(αx+kπ2)(1)nk(nk)!(xi)nk+1also(cos(αx)x+i)(n)=k=0nCnkcos(αx+π2)(1)nk(nk)!(x+i)nk+1f(n)(x)=12ik=0nCnk(1)nk(nk)!cos(αx+kπ2){1(xi)nk+11(x+i)nk+1}f(n)(0)=12ik=0n(1)nkn!k!cos(kπ2){ink+1(i)nk+1}=k=0n(1)nkn!k!cos(kπ2)sin(nk+1)π2
Commented by maxmathsup by imad last updated on 05/Oct/18
2) we have f(x) =Σ_(n=0) ^∞    ((f^((n)) (0))/(n!)) x^n   =Σ_(n=0) ^∞   ( Σ_(k=0) ^n   (((−1)^(n−k) )/(k!)) cos(((kπ)/2))sin(n−k+1)(π/2))x^n   =Σ_(n=0) ^∞  (Σ_(k=0) ^n   (((−1)^(n−k) )/(k!)) cos(((kπ)/2))cos((n−k)(π/2)) x^n  .
2)wehavef(x)=n=0f(n)(0)n!xn=n=0(k=0n(1)nkk!cos(kπ2)sin(nk+1)π2)xn=n=0(k=0n(1)nkk!cos(kπ2)cos((nk)π2)xn.
Commented by maxmathsup by imad last updated on 05/Oct/18
3) ∫_0 ^x  f_α (t)dt =Σ_(n=0) ^∞ (Σ_(k=0) ^n   (((−1)^(n−k) )/(k!)) cos(((kπ)/2))cos((n−k)(π/2)) (x^(n+1) /(n+1)) .
3)0xfα(t)dt=n=0(k=0n(1)nkk!cos(kπ2)cos((nk)π2)xn+1n+1.
Commented by maxmathsup by imad last updated on 05/Oct/18
4) let calculate first ∫_0 ^∞   f_α (t)dt  =I_α   I_α = ∫_0 ^∞   ((cos(αt))/(1+t^2 ))dt ⇒2I_α  = ∫_(−∞) ^(+∞)    ((cos(αt))/(1+t^2 ))dt =Re(∫_(−∞) ^(+∞)   (e^(iαt) /(1+t^2 ))dt)  let ϕ(z) =(e^(iαz) /(1+z^2 ))  the poles of ϕ are i and −i residus theorem give  ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπRes(ϕ,i) =2iπ  (e^(−α) /(2i)) =π e^(−α)  ⇒ I_α =(π/2) e^(−α)   ⇒∫_0 ^∞   f_α (t)dt =(π/2) e^(−α)  =(π/2) Σ_(n=0) ^∞    (((−α)^n )/(n!)) =(π/2) Σ_(n=0) ^∞  (((−1)^n )/(n!)) α^n  .
4)letcalculatefirst0fα(t)dt=IαIα=0cos(αt)1+t2dt2Iα=+cos(αt)1+t2dt=Re(+eiαt1+t2dt)letφ(z)=eiαz1+z2thepolesofφareiandiresidustheoremgive+φ(z)dz=2iπRes(φ,i)=2iπeα2i=πeαIα=π2eα0fα(t)dt=π2eα=π2n=0(α)nn!=π2n=0(1)nn!αn.

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