Question Number 65332 by aliesam last updated on 28/Jul/19
Commented by mathmax by abdo last updated on 29/Jul/19
![let suppose n inter let decompose the fraction F(x) =(x/(x^n +1)) z^n +1 =0 ⇒z^n =−1 ⇒r^n e^(inθ) =e^((2k+1)π) (z=re^(iθ) ) ⇒r=1 andθ θ =(((2k+1)π)/n) so the roots are Z_k =e^(i(((2k+1)π)/n)) k∈[[0,n−1]] F(x) =Σ_(i=0) ^(n−1) (λ_i /(x−Z_i )) with λ_i =(Z_i /(nZ_i ^(n−1) )) =(Z_i ^2 /(−n)) ⇒ F(x) =−(1/n)Σ_(i=0) ^(n−1) (Z_i ^2 /(x−Z_i )) ⇒ ∫ F(x)dx =−(1/n)Σ_(i=0) ^(n−1) Z_i ^2 ∫ (dx/(x−Z_i )) rest to determine ∫ (dx/(x−Z_i )) be continued...](https://www.tinkutara.com/question/Q65434.png)
$${let}\:{suppose}\:{n}\:{inter}\:{let}\:{decompose}\:{the}\:{fraction} \\ $$$${F}\left({x}\right)\:=\frac{{x}}{{x}^{{n}} \:+\mathrm{1}} \\ $$$${z}^{{n}} \:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow{z}^{{n}} \:=−\mathrm{1}\:\:\Rightarrow{r}^{{n}} \:{e}^{{in}\theta} \:={e}^{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:\:\:\:\left({z}={re}^{{i}\theta} \right)\:\Rightarrow{r}=\mathrm{1}\:{and}\theta \\ $$$$\theta\:=\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}}\:{so}\:{the}\:{roots}\:{are}\:{Z}_{{k}} ={e}^{{i}\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}}} \:\:\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right] \\ $$$${F}\left({x}\right)\:=\sum_{{i}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{\lambda_{{i}} }{{x}−{Z}_{{i}} }\:\:\:\:\:\:{with}\:\lambda_{{i}} =\frac{{Z}_{{i}} }{{nZ}_{{i}} ^{{n}−\mathrm{1}} }\:=\frac{{Z}_{{i}} ^{\mathrm{2}} }{−{n}}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=−\frac{\mathrm{1}}{{n}}\sum_{{i}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\frac{{Z}_{{i}} ^{\mathrm{2}} }{{x}−{Z}_{{i}} }\:\Rightarrow\:\int\:{F}\left({x}\right){dx}\:=−\frac{\mathrm{1}}{{n}}\sum_{{i}=\mathrm{0}} ^{{n}−\mathrm{1}} {Z}_{{i}} ^{\mathrm{2}} \:\int\:\:\:\frac{{dx}}{{x}−{Z}_{{i}} } \\ $$$${rest}\:{to}\:{determine}\:\int\:\:\frac{{dx}}{{x}−{Z}_{{i}} }\:\:\:{be}\:{continued}… \\ $$