Question Number 30580 by abdo imad last updated on 23/Feb/18
![decompose inside C[x] F= (x^n /(x^m +1)) with m≥n+2 then find ∫_0 ^∞ (x^n /(x^m +1))dx.](Q30580.png)
$${decompose}\:{inside}\:{C}\left[{x}\right]\:{F}=\:\frac{{x}^{{n}} }{{x}^{{m}} \:+\mathrm{1}}\:{with}\:{m}\geqslant{n}+\mathrm{2} \\ $$$${then}\:{find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{{x}^{{m}} \:+\mathrm{1}}{dx}. \\ $$
Commented by prof Abdo imad last updated on 24/Feb/18
![z^m +1=0⇔z^m =e^(i(2k+1)π) so the roots of this equation are z_k = e^(i(2k+1)(π/(m ))) /k∈[[0,m−1]]⇒ F(x)= Σ_(k=0) ^(m−1) (λ_k /(x−z_k )) we have λ_k = (z_k ^n /(m z_k ^(m−1) )) ⇒ λ_k = −(1/m) z_k ^(n+1) ⇒F(x)= −(1/m)Σ_(k=0) ^(m−1) (z_k ^(n+1) /(x−z_k )) . 2) the ch. x^m =t give ∫_0 ^∞ (x^n /(x^m +1))dx = ∫_0 ^∞ (t^(n/m) /(1+t)) (1/m)t^((1/m)−1) dt =∫_0 ^∞ (1/m) (t^(((n+1)/m)−1) /(1+t))dt = (1/m) (π/(sin(π((n+1)/m)))) we have used the result ∫_0 ^∞ (t^(a−1) /(1+t))dt= (π/(sin(πa))) if 0<a<1.](Q30716.png)
$${z}^{{m}} \:+\mathrm{1}=\mathrm{0}\Leftrightarrow{z}^{{m}} ={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:{so}\:{the}\:{roots}\:{of}\:{this} \\ $$$${equation}\:{are}\:\:{z}_{{k}} =\:{e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{{m}\:}} \:\:/{k}\in\left[\left[\mathrm{0},{m}−\mathrm{1}\right]\right]\Rightarrow \\ $$$${F}\left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{m}−\mathrm{1}} \:\:\frac{\lambda_{{k}} }{{x}−{z}_{{k}} }\:\:{we}\:{have}\:\lambda_{{k}} =\:\:\frac{{z}_{{k}} ^{{n}} }{{m}\:{z}_{{k}} ^{{m}−\mathrm{1}} }\:\Rightarrow \\ $$$$\lambda_{{k}} =\:−\frac{\mathrm{1}}{{m}}\:{z}_{{k}} ^{{n}+\mathrm{1}} \:\Rightarrow{F}\left({x}\right)=\:−\frac{\mathrm{1}}{{m}}\sum_{{k}=\mathrm{0}} ^{{m}−\mathrm{1}} \:\:\frac{{z}_{{k}} ^{{n}+\mathrm{1}} }{{x}−{z}_{{k}} }\:\:. \\ $$$$\left.\mathrm{2}\right)\:{the}\:{ch}.\:{x}^{{m}} ={t}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{{n}} }{{x}^{{m}} +\mathrm{1}}{dx}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}^{\frac{{n}}{{m}}} }{\mathrm{1}+{t}}\:\:\frac{\mathrm{1}}{{m}}{t}^{\frac{\mathrm{1}}{{m}}−\mathrm{1}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{{m}}\:\:\frac{{t}^{\frac{{n}+\mathrm{1}}{{m}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\:\frac{\mathrm{1}}{{m}}\:\frac{\pi}{{sin}\left(\pi\frac{{n}+\mathrm{1}}{{m}}\right)}\:{we}\:{have} \\ $$$${used}\:{the}\:{result}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}=\:\frac{\pi}{{sin}\left(\pi{a}\right)}\:\:{if}\:\mathrm{0}<{a}<\mathrm{1}. \\ $$