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Question Number 126631 by mnjuly1970 last updated on 23/Dec/20
...calculus... prove that:: :: Ω=^(??) ∫_0 ^( 1) x^(n−1) ln^2 (1−x)dx = (2/n)Σ_(k=1) ^n ((H_k /k)) ✓
…calculus…provethat::::Ω=??∫01xn−1ln2(1−x)dx=2n∑nk=1(Hkk)✓
Answered by mindispower last updated on 23/Dec/20
β(x,y)=∫_0 ^1 t^(x−1) (1−t)^(y−1) dt (∂^2 β/∂y^2 )=∫_0 ^1 t^(x−1) ln^2 (1−t)y^(t−1) dt ((∂^2 β(x,y))/∂y^2 )∣_((x,y)=(n,1)) =∫_0 ^1 t^(n−1) ln^2 (1−t)dt (∂β/∂y)=(Ψ(y)−Ψ(x+y))β(x,y) (∂^2 β/∂y^2 )=(Ψ^1 (y)−Ψ^1 (x+y))β(x,y)+(Ψ(y)−Ψ(x+y))^2 β(x,y) Ω=(Ψ^1 (1)−Ψ^1 (1+n))β(1,n)+(Ψ(1)−Ψ(1+n))^2 β(1,n) Ω=(Σ_(k=1) ^n (1/k^2 ))(1/n)+(Σ_(k=1) ^n (1/k))^2 .(1/n) H_n =Σ_(k≤n) (1/k),H_n ^((2)) =Σ_(k≤n) (1/k^2 ) =((H_n ^((2)) +H_n )/n)
β(x,y)=∫01tx−1(1−t)y−1dt∂2β∂y2=∫01tx−1ln2(1−t)yt−1dt∂2β(x,y)∂y2∣(x,y)=(n,1)=∫01tn−1ln2(1−t)dt∂β∂y=(Ψ(y)−Ψ(x+y))β(x,y)∂2β∂y2=(Ψ1(y)−Ψ1(x+y))β(x,y)+(Ψ(y)−Ψ(x+y))2β(x,y)Ω=(Ψ1(1)−Ψ1(1+n))β(1,n)+(Ψ(1)−Ψ(1+n))2β(1,n)Ω=(∑nk=11k2)1n+(∑nk=11k)2.1nHn=∑k⩽n1k,Hn(2)=∑k⩽n1k2=Hn(2)+Hnn
Commented by mnjuly1970 last updated on 23/Dec/20
very nice..thanking...
verynice..thanking…
Answered by mnjuly1970 last updated on 23/Dec/20
solution: consider: I_n =∫_0 ^( 1) x^(n−1) ln(1−x)dx we can write: I_n =(1/n)∫(x^n −1) ^′ ln(1−x)dx =(1/n){[(x^n −1)ln(1−x)]_0 ^1 +∫_0 ^( 1) ((x^n −1)/(1−x))dx} =((−H_n )/n) .... (1) put:: Ω=∫_0 ^( 1) x^(n−1) ln^2 (1−x)dx =(1/n)∫_0 ^( 1) (x^n −1) ^′ ln^2 (1−x)dx =(1/n){[(x^n −1)ln^2 (1−x)]_0 ^1 +2∫_0 ^( 1) ((x^n −1)/(1−x))ln(1−x)dx =(2/n) ∫_0 ^( 1) (−Σ_(k=1) ^n x^(k−1) )ln(1−x)dx =((−2)/n^2 )Σ_(k=1) ^n ∫_0 ^( 1) x^(k−1) ln(1−x)dx=(2/n)Σ_(k=1) ^n (H_k /k) ✓
solution:consider:In=∫01xn−1ln(1−x)dxwecanwrite:In=1n∫(xn−1)′ln(1−x)dx=1n{[(xn−1)ln(1−x)]01+∫01xn−11−xdx}=−Hnn….(1)put::Ω=∫01xn−1ln2(1−x)dx=1n∫01(xn−1)′ln2(1−x)dx=1n{[(xn−1)ln2(1−x)]01+2∫01xn−11−xln(1−x)dx=2n∫01(−∑nk=1xk−1)ln(1−x)dx=−2n2∑nk=1∫01xk−1ln(1−x)dx=2n∑nk=1Hkk✓

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