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I-0-x-n-1-x-1-x-n-3-1-dx-

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Question Number 141133 by mnjuly1970 last updated on 16/May/21
I:=∫_0 ^( ∞) (((x^n −1)(x−1))/(x^(n+3) −1))dx=??
I:=∫0∞(xn−1)(x−1)xn+3−1dx=??
Answered by mathmax by abdo last updated on 17/May/21
at form of series Ψ=∫_0 ^1 ((x^(n+1) −x^n −x+1)/(x^(n+3) −1))dx +∫_1 ^∞ ((x^(n+1) −x^n −x+1)/(x^(n+3) −1))dx =Ψ_1 +Ψ_2 Ψ_1 =−∫_0 ^1 (x^(n+1) −x^n −x+1)Σ_(k=0) ^∞ x^(k(n+3)) =−Σ_(k=0) ^∞ ∫_0 ^1 (x^(n+1 +k(n+3)) −x^(k(n+3)+n) +x^(k(n+3)) dx =−Σ_(k=0) ^∞ [(1/(n+2+(n+3)k))x^(n+2+(n+3)k) −(1/(n+1+(n+3)k))x^(n+1+(n+3)k) +(1/(1+(n+3)k))x^(1+(n+3)k) ]_0 ^1 =−Σ_(k=0) ^∞ ((1/(n+2+(n+3)k))−(1/(n+1+(n+3)k))+(1/(1+(n+3)k))) Ψ_2 =_(x=(1/t)) −∫_0 ^1 (((1/t^(n+1) )−(1/t^n )−(1/t)+1)/((1/t^(n+3) )−1))×(−(dt/t^2 )) =∫_0 ^1 t^(n+1) ((((1/t^(n+1) )−(1/t^n )−(1/t)+1)/(1−t^(n+3) )))dt =∫_0 ^1 ((1−t−t^n +t^(n+1) )/(1−t^(n+3) ))dt =∫_0 ^1 ((−t^(n+1) +t^n +t−1)/(t^(n+3) −1))dt ⇒ Ψ_1 +Ψ_2 =∫_0 ^1 ((x^(n+1) −x^n −x+1−x^(n+1) +x^n +x−1)/(x^(n+3) −1))dx =0 ⇒Ψ=0 !
atformofseriesΨ=∫01xn+1−xn−x+1xn+3−1dx+∫1∞xn+1−xn−x+1xn+3−1dx=Ψ1+Ψ2Ψ1=−∫01(xn+1−xn−x+1)∑k=0∞xk(n+3)=−∑k=0∞∫01(xn+1+k(n+3)−xk(n+3)+n+xk(n+3)dx=−∑k=0∞[1n+2+(n+3)kxn+2+(n+3)k−1n+1+(n+3)kxn+1+(n+3)k+11+(n+3)kx1+(n+3)k]01=−∑k=0∞(1n+2+(n+3)k−1n+1+(n+3)k+11+(n+3)k)Ψ2=x=1t−∫011tn+1−1tn−1t+11tn+3−1×(−dtt2)=∫01tn+1(1tn+1−1tn−1t+11−tn+3)dt=∫011−t−tn+tn+11−tn+3dt=∫01−tn+1+tn+t−1tn+3−1dt⇒Ψ1+Ψ2=∫01xn+1−xn−x+1−xn+1+xn+x−1xn+3−1dx=0⇒Ψ=0!

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