Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 134469 by SEKRET last updated on 04/Mar/21

∫ ((x^m +1)/(x^n +1)) dx =? n>m m;n∈Z^+

$$\:\:\:\int\:\:\frac{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{m}}} +\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}} +\mathrm{1}}\:\boldsymbol{\mathrm{dx}}\:\:=?\:\:\:\:\:\boldsymbol{\mathrm{n}}>\boldsymbol{\mathrm{m}}\:\:\:\:\:\:\boldsymbol{\mathrm{m}};\boldsymbol{\mathrm{n}}\in\boldsymbol{\mathrm{Z}}^{+} \\ $$

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

$${x}^{{n}} +\mathrm{1}\:=\:\mathrm{0}\:\Leftrightarrow\:{x}\:=\:{e}^{{i}\frac{\pi}{{n}}\left(\mathrm{1}+\mathrm{2}{k}\right)} ,\:{k}\in\left\{\mathrm{0},\:\mathrm{1},...,\:{n}−\mathrm{1}\right\} \\ $$ $$\mathrm{Let}\:{x}_{{k}} \:=\:{e}^{{i}\frac{\pi}{{n}}\left(\mathrm{1}+\mathrm{2}{k}\right)} ,\:{k}\in\left\{\mathrm{0},\:\mathrm{1},...,\:{n}−\mathrm{1}\right\} \\ $$ $$\mathrm{R}_{{m},{n}} \left({x}\right)\:=\:\frac{{x}^{{m}} +\mathrm{1}}{{x}^{{n}} +\mathrm{1}},\:{n}>{m} \\ $$ $$\mathrm{R}_{{m},{n}} \left({x}\right)\:=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{A}_{{k}} }{{x}−{x}_{{k}} } \\ $$ $$\mathrm{A}_{{k}} \:=\:\frac{{x}_{{k}} ^{{m}} +\mathrm{1}}{\underset{\underset{{j}\neq{k}} {{j}=\mathrm{0}}} {\overset{{n}−\mathrm{1}} {\prod}}\left({x}_{{k}} −{x}_{{j}} \right)} \\ $$ $$\mathrm{I}_{{m},{n}} \left({x}\right)\:=\:\int\mathrm{R}_{{m},{n}} \left({x}\right){dx} \\ $$ $$\mathrm{I}_{{m},{n}} \left({x}\right)\:=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{A}_{{k}} \mathrm{ln}\mid{x}−{x}_{{k}} \mid+\mathrm{C} \\ $$

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) )

$$\int\frac{{x}^{{m}} +\mathrm{1}}{{x}^{{n}} +\mathrm{1}}{dx} \\ $$ $$=\int\left({x}^{{m}} +\mathrm{1}\right)\left(\mathrm{1}−{x}^{{n}} \right)\frac{\mathrm{1}}{\left(\mathrm{1}−{x}^{\mathrm{2}{n}} \right)}{dx} \\ $$ $$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\int\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} }{{k}!}\left({x}^{{m}} +\mathrm{1}\right)\left(\mathrm{1}−{x}^{{n}} \right){x}^{\mathrm{2}{nk}} {dx} \\ $$ $$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} }{{k}!}\left(\frac{{x}^{\mathrm{2}{nk}+{m}+\mathrm{1}} }{\mathrm{2}{nk}+{m}+\mathrm{1}}+\frac{{x}^{\mathrm{2}{nk}+\mathrm{1}} }{\mathrm{2}{nk}+\mathrm{1}}−\frac{{x}^{{m}+{n}+\mathrm{2}{nk}+\mathrm{1}} }{{m}+{n}+\mathrm{2}{nk}+\mathrm{1}}−\frac{{x}^{{n}+\mathrm{2}{nk}+\mathrm{1}} }{{n}+\mathrm{1}+\mathrm{2}{nk}}\right) \\ $$ $$=\frac{{x}}{\mathrm{2}{n}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} {x}^{\mathrm{2}{nk}+{m}} }{{k}!\left({k}+\frac{{m}+\mathrm{1}}{\mathrm{2}{n}}\right)}+\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} {x}^{\mathrm{2}{nk}} }{{k}!\left({k}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)}−\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} {x}^{{m}+{n}+\mathrm{2}{nk}} }{{k}!\left({k}+\frac{{m}+{n}+\mathrm{1}}{\mathrm{2}{n}}\right)}−\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} {x}^{\mathrm{2}{nk}+{n}} }{{k}!\left({k}+\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\right)} \\ $$ $$=\frac{{x}}{\mathrm{2}{n}}\left(\Psi+\Phi−\Lambda−\varphi\right) \\ $$ $$\Psi=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} \Gamma\left({k}+\frac{{m}+\mathrm{1}}{\mathrm{2}{n}}\right)}{{k}!\Gamma\left({k}+\frac{{m}+\mathrm{1}}{\mathrm{2}{n}}+\mathrm{1}\right)}{x}^{\mathrm{2}{nk}+{m}} =\left(\mathrm{1}+\frac{\mathrm{2}{n}}{{m}+\mathrm{1}}\right)\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} \left(\frac{{m}+\mathrm{1}}{\mathrm{2}{n}}\right)_{{k}} }{{k}!\left(\frac{{m}+\mathrm{1}}{\mathrm{2}{n}}+\mathrm{1}\right)_{{k}} }{x}^{\mathrm{2}{nk}+{m}} \\ $$ $$={x}^{{m}} \left(\mathrm{1}+\frac{\mathrm{2}{n}}{{m}+\mathrm{1}}\right)\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{{m}+\mathrm{1}}{\mathrm{2}{n}};\frac{{m}+\mathrm{1}+\mathrm{2}{n}}{\mathrm{2}{n}}\:;{x}^{\mathrm{2}{n}} \right) \\ $$ $$\Phi=\left(\mathrm{1}+\mathrm{2}{n}\right)\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}{n}};\frac{\mathrm{1}+\mathrm{2}{n}}{\mathrm{2}{n}};{x}^{\mathrm{2}{n}} \right) \\ $$ $$\Lambda={x}^{{m}+{n}} \left(\mathrm{1}+\frac{\mathrm{2}{n}}{{m}+{n}+\mathrm{1}}\right)\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{{m}+{n}+\mathrm{1}}{\mathrm{2}{n}},\frac{{m}+\mathrm{3}{n}+\mathrm{1}}{\mathrm{2}{n}};{x}^{\mathrm{2}{n}} \right)\: \\ $$ $$\varphi={x}^{{n}} \left(\mathrm{1}+\frac{\mathrm{2}{n}}{{n}+\mathrm{1}}\right)\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{{n}+\mathrm{1}}{\mathrm{2}{n}};\frac{\mathrm{3}{n}+\mathrm{1}}{\mathrm{2}{n}};{x}^{\mathrm{2}{n}} \right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com