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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.

decomposeinsideC[x]F=xnxm+1withm⩾n+2thenfind∫0∞xnxm+1dx.

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.

zm+1=0⇔zm=ei(2k+1)πsotherootsofthisequationarezk=ei(2k+1)πm/k∈[[0,m−1]]⇒F(x)=∑k=0m−1λkx−zkwehaveλk=zknmzkm−1⇒λk=−1mzkn+1⇒F(x)=−1m∑k=0m−1zkn+1x−zk.2)thech.xm=tgive∫0∞xnxm+1dx=∫0∞tnm1+t1mt1m−1dt=∫0∞1mtn+1m−11+tdt=1mπsin(πn+1m)wehaveusedtheresult∫0∞ta−11+tdt=πsin(πa)if0<a<1.

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