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x-m-1-x-n-1-dx-n-gt-m-m-n-Z-

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Question Number 134469 by SEKRET last updated on 04/Mar/21
∫ ((x^m +1)/(x^n +1)) dx =? n>m m;n∈Z^+
∫xm+1xn+1dx=?n>mm;n∈Z+
Answered by Olaf last updated on 04/Mar/21
x^n +1 = 0 ⇔ x = e^(i(π/n)(1+2k)) , k∈{0, 1,..., n−1} Let x_k = e^(i(π/n)(1+2k)) , k∈{0, 1,..., n−1} R_(m,n) (x) = ((x^m +1)/(x^n +1)), n>m R_(m,n) (x) = Σ_(k=0) ^(n−1) (A_k /(x−x_k )) A_k = ((x_k ^m +1)/(Π_(j=0_(j≠k) ) ^(n−1) (x_k −x_j ))) I_(m,n) (x) = ∫R_(m,n) (x)dx I_(m,n) (x) = Σ_(k=0) ^(n−1) A_k ln∣x−x_k ∣+C
xn+1=0⇔x=eiπn(1+2k),k∈{0,1,…,n−1}Letxk=eiπn(1+2k),k∈{0,1,…,n−1}Rm,n(x)=xm+1xn+1,n>mRm,n(x)=∑n−1k=0Akx−xkAk=xkm+1∏n−1j=0j≠k(xk−xj)Im,n(x)=∫Rm,n(x)dxIm,n(x)=∑n−1k=0Akln∣x−xk∣+C
Answered by Dwaipayan Shikari last updated on 04/Mar/21
∫((x^m +1)/(x^n +1))dx =∫(x^m +1)(1−x^n )(1/((1−x^(2n) )))dx =Σ_(k=0) ^∞ ∫((((1/2))_k )/(k!))(x^m +1)(1−x^n )x^(2nk) dx =Σ_(k=0) ^∞ ((((1/2))_k )/(k!))((x^(2nk+m+1) /(2nk+m+1))+(x^(2nk+1) /(2nk+1))−(x^(m+n+2nk+1) /(m+n+2nk+1))−(x^(n+2nk+1) /(n+1+2nk))) =(x/(2n))Σ_(k=0) ^∞ ((((1/2))_k x^(2nk+m) )/(k!(k+((m+1)/(2n)))))+((((1/2))_k x^(2nk) )/(k!(k+(1/(2n)))))−((((1/2))_k x^(m+n+2nk) )/(k!(k+((m+n+1)/(2n)))))−((((1/2))_k x^(2nk+n) )/(k!(k+((n+1)/(2n))))) =(x/(2n))(Ψ+Φ−Λ−ϕ) Ψ=Σ_(k=0) ^∞ ((((1/2))_k Γ(k+((m+1)/(2n))))/(k!Γ(k+((m+1)/(2n))+1)))x^(2nk+m) =(1+((2n)/(m+1)))Σ_(k=0) ^∞ ((((1/2))_k (((m+1)/(2n)))_k )/(k!(((m+1)/(2n))+1)_k ))x^(2nk+m) =x^m (1+((2n)/(m+1))) _2 F_1 ((1/2),((m+1)/(2n));((m+1+2n)/(2n)) ;x^(2n) ) Φ=(1+2n) _2 F_1 ((1/2),(1/(2n));((1+2n)/(2n));x^(2n) ) Λ=x^(m+n) (1+((2n)/(m+n+1))) _2 F_1 ((1/2),((m+n+1)/(2n)),((m+3n+1)/(2n));x^(2n) ) ϕ=x^n (1+((2n)/(n+1))) _2 F_1 ((1/2),((n+1)/(2n));((3n+1)/(2n));x^(2n) )
∫xm+1xn+1dx=∫(xm+1)(1−xn)1(1−x2n)dx=∑∞k=0∫(12)kk!(xm+1)(1−xn)x2nkdx=∑∞k=0(12)kk!(x2nk+m+12nk+m+1+x2nk+12nk+1−xm+n+2nk+1m+n+2nk+1−xn+2nk+1n+1+2nk)=x2n∑∞k=0(12)kx2nk+mk!(k+m+12n)+(12)kx2nkk!(k+12n)−(12)kxm+n+2nkk!(k+m+n+12n)−(12)kx2nk+nk!(k+n+12n)=x2n(Ψ+Φ−Λ−φ)Ψ=∑∞k=0(12)kΓ(k+m+12n)k!Γ(k+m+12n+1)x2nk+m=(1+2nm+1)∑∞k=0(12)k(m+12n)kk!(m+12n+1)kx2nk+m=xm(1+2nm+1)2F1(12,m+12n;m+1+2n2n;x2n)Φ=(1+2n)2F1(12,12n;1+2n2n;x2n)Λ=xm+n(1+2nm+n+1)2F1(12,m+n+12n,m+3n+12n;x2n)φ=xn(1+2nn+1)2F1(12,n+12n;3n+12n;x2n)

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