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let-P-n-x-1-x-2-1-x-4-1-x-2-n-calculate-lim-n-0-x-P-n-t-dt-with-0-lt-x-lt-1-

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Question Number 33744 by prof Abdo imad last updated on 23/Apr/18
let P_n (x)=(1+x^2 )(1+x^4 )....(1+x^2^n ) calculate lim_(n→+∞) ∫_0 ^x P_n (t)dt with 0<x<1 .
letPn(x)=(1+x2)(1+x4)….(1+x2n)calculatelimn→+∞∫0xPn(t)dtwith0<x<1.
Commented by prof Abdo imad last updated on 25/Apr/18
we have proved that P_n (t) = ((1−t^2^(n+1) )/(1−t^2 )) ⇒ ∫_0 ^x P_n (t)dt = ∫_0 ^x ((1−t^2^(n+1) )/(1−t^2 ))dt = ∫_0 ^x (dt/(1−t^2 )) −∫_0 ^x (t^2^(n+1) /(1−t^2 )) dt but lim_(n→+∞) ∫_0 ^x (t^2^(n+1) /(1−t^2 )) dt =0 because 0≤ ⇒t≤x<1 ⇒ lim_(n→+∞) ∫_0 ^x P_n (t)dt = ∫_0 ^x (dt/(1−t^2 )) =(1/2) ∫_0 ^x ( (1/(1−t)) +(1/(1+t)))dt =[(1/2)ln∣((1+t)/(1−t))∣]_0 ^c =(1/2) ln∣ ((1+x)/(1−x))∣ .
wehaveprovedthatPn(t)=1−t2n+11−t2⇒∫0xPn(t)dt=∫0x1−t2n+11−t2dt=∫0xdt1−t2−∫0xt2n+11−t2dtbutlimn→+∞∫0xt2n+11−t2dt=0because0⩽⇒t⩽x<1⇒limn→+∞∫0xPn(t)dt=∫0xdt1−t2=12∫0x(11−t+11+t)dt=[12ln∣1+t1−t∣]0c=12ln∣1+x1−x∣.

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