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let-u-n-0-1-x-n-sin-pix-dx-1-prove-that-u-n-converges-2-prove-that-u-n-0-pi-sint-t-dt-




Question Number 32042 by abdo imad last updated on 18/Mar/18
let u_n   =∫_0 ^1   x^n  sin(πx)dx  1) prove that Σ u_n  converges  2) prove that Σ u_n = ∫_0 ^π   ((sint)/t)dt .
letun=01xnsin(πx)dx1)provethatΣunconverges2)provethatΣun=0πsinttdt.
Commented by abdo imad last updated on 20/Mar/18
let put S_n =Σ_(k=0) ^n  ∫_0 ^1  x^k  sin(πx)dx  S_n = ∫_0 ^1 (Σ_(k=0) ^n  x^k )sin(πx)dx =∫_0 ^1  ((1−x^(n+1) )/(1−x)) sin(πx)dx  ⇒ S_(n )  − ∫_0 ^1   ((sin(πx))/(1−x))dx = −∫_0 ^1   (x^(n+1) /(1−x)) sin(πx)dx and ∃m>0/  ∣ S_n  − ∫_0 ^1   ((sin(πx))/(1−x))dx∣  ≤ m ∫_0 ^1  x^(n+1) dx =(m/(n+2)) →_(n→∞) 0  ⇒ S_n converged and lim S_n  = ∫_0 ^1   ((sin(πx))/(1−x)) dx  2) ch .1−x=t give ∫_0 ^1   ((sin(πx))/(1−x)) dx =∫_0 ^1   ((sin(π(1−t)))/t)= dt  = ∫_0 ^1   ((sin(πt))/t) dt after ch. πt =u give  lim _(n→∞)  S_n =  ∫_0 ^π   ((sin(u))/(u/π)) (du/π) =∫_0 ^π    ((sinu)/u) du . so  Σ_(n=0) ^∞  u_n = ∫_0 ^π   ((sinu)/u) du  .
letputSn=k=0n01xksin(πx)dxSn=01(k=0nxk)sin(πx)dx=011xn+11xsin(πx)dxSn01sin(πx)1xdx=01xn+11xsin(πx)dxandm>0/Sn01sin(πx)1xdxm01xn+1dx=mn+2n0SnconvergedandlimSn=01sin(πx)1xdx2)ch.1x=tgive01sin(πx)1xdx=01sin(π(1t))t=dt=01sin(πt)tdtafterch.πt=ugivelimnSn=0πsin(u)uπduπ=0πsinuudu.son=0un=0πsinuudu.

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