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Question Number 39711 by math khazana by abdo last updated on 10/Jul/18
calculate ∫_(−∞) ^(+∞) (x^n /((1+x^2 )^n )) dx with n natral integr
$${calculate}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}^{{n}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }\:{dx}\:{with}\:{n}\:{natral}\:{integr} \\ $$
Commented by maxmathsup by imad last updated on 11/Jul/18
let A_n = ∫_(−∞) ^(+∞) ((x^n dx)/((1+x^2 )^n )) and let ϕ(z)=(z^n /((1+z^2 )^n )) ϕ(z) = (z^n /((z−i)^n (z+i)^n )) ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ Res(ϕ,i) but Res(ϕ,i)=lim_(z→i) (1/((n−1)!)){(z−i)^n ϕ(z)}^((n−1)) =lim_(z→i) (1/((n−1)!)) { x^n (z+i)^(−n) }^((n−1)) and { x^n (z+i)^(−n) }^((n−1)) =Σ_(k=0) ^(n−1) C_(n−1) ^k {(z+i)^(−n) }^((k)) (x^n )^((n−1−k)) but (z+i)^(−n) }^((k)) =(−1)^k n(n+1)...(n+k−1)(z+i)^(−n−k) (x^n )^((p)) =n(n−1)...(n−p+1)x^(n−p) ⇒(x^n )^((n−1−k)) =n(n−1)...(k+2)x^(k+1) ⇒ Res(ϕ,i) = (1/((n−1)!)) Σ_(k=0) ^(n−1) C_(n−1) ^k (−1)^k n(n+1)...(n+k−1)(2i)^(−n−k) (n−1)...(k+2)i^(k+1) A_n =2iπ Res(ϕ,i)
$${let}\:{A}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{x}^{{n}} {dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }\:\:\:\:{and}\:{let}\:\varphi\left({z}\right)=\frac{{z}^{{n}} }{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{{n}} } \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}^{{n}} }{\left({z}−{i}\right)^{{n}} \left({z}+{i}\right)^{{n}} } \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)\:\:{but} \\ $$$${Res}\left(\varphi,{i}\right)={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{{n}} \varphi\left({z}\right)\right\}^{\left({n}−\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\:\:\left\{\:\:{x}^{{n}} \left({z}+{i}\right)^{−{n}} \right\}^{\left({n}−\mathrm{1}\right)} \:\:{and} \\ $$$$\left\{\:{x}^{{n}} \left({z}+{i}\right)^{−{n}} \right\}^{\left({n}−\mathrm{1}\right)} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:{C}_{{n}−\mathrm{1}} ^{{k}} \:\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left({k}\right)} \:\:\left({x}^{{n}} \right)^{\left({n}−\mathrm{1}−{k}\right)} \:\:{but} \\ $$$$\left.\left({z}+{i}\right)^{−{n}} \right\}^{\left({k}\right)} =\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)…\left({n}+{k}−\mathrm{1}\right)\left({z}+{i}\right)^{−{n}−{k}} \\ $$$$\left({x}^{{n}} \right)^{\left({p}\right)} ={n}\left({n}−\mathrm{1}\right)…\left({n}−{p}+\mathrm{1}\right){x}^{{n}−{p}} \Rightarrow\left({x}^{{n}} \right)^{\left({n}−\mathrm{1}−{k}\right)} ={n}\left({n}−\mathrm{1}\right)…\left({k}+\mathrm{2}\right){x}^{{k}+\mathrm{1}} \:\Rightarrow \\ $$$${Res}\left(\varphi,{i}\right)\:=\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:{C}_{{n}−\mathrm{1}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)…\left({n}+{k}−\mathrm{1}\right)\left(\mathrm{2}{i}\right)^{−{n}−{k}} \:\left({n}−\mathrm{1}\right)…\left({k}+\mathrm{2}\right){i}^{{k}+\mathrm{1}} \\ $$$${A}_{{n}} =\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$

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