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calculateA-n-0-dx-x-2-n-x-2-2n-with-n-integr-natural-1-

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Question Number 109214 by mathmax by abdo last updated on 22/Aug/20
calculateA_n = ∫_0 ^∞ (dx/((x^2 +n)(x^2 +2n))) with n integr natural≥1
$$\mathrm{calculateA}_{\mathrm{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{n}\right)\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{2n}\right)}\:\:\mathrm{with}\:\mathrm{n}\:\mathrm{integr}\:\mathrm{natural}\geqslant\mathrm{1} \\ $$
Answered by mathmax by abdo last updated on 22/Aug/20
2A_n =∫_(−∞) ^(+∞) (dx/((x^2 +n)(x^2 +2n))) let ϕ(z) =(1/((z^2 +n)(z^2 +2n))) ⇒ ϕ(z) =(1/((z−i(√n))(z+i(√n))(z−i(√(2n)))(z+i(√(2n))))) residus theorem ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ {Res(ϕ,i(√n)) +Res(ϕ,−i(√n))} Res(ϕ,i(√n)) =(1/(2i(√n)(−n+2n))) =(1/(2in(√n))) Res(ϕ,i(√(2n))) =(1/(2i(√(2n))(−2n+n))) =−(1/(2in(√(2n)))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{(1/(2in(√n)))−(1/(2in(√(2n))))} =(π/(n(√n))) −(π/(n(√2)(√n))) =(π/(n(√n))){1−(1/( (√2)))} =((π((√2)−1))/(n(√(2n)))) ⇒ A_n =((π((√2)−1))/(2n(√(2n))))
$$\mathrm{2A}_{\mathrm{n}} =\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{n}\right)\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{2n}\right)}\:\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{n}\right)\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{2n}\right)}\:\Rightarrow \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{z}−\mathrm{i}\sqrt{\mathrm{n}}\right)\left(\mathrm{z}+\mathrm{i}\sqrt{\mathrm{n}}\right)\left(\mathrm{z}−\mathrm{i}\sqrt{\mathrm{2n}}\right)\left(\mathrm{z}+\mathrm{i}\sqrt{\mathrm{2n}}\right)}\:\mathrm{residus}\:\mathrm{theorem}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\left\{\mathrm{Res}\left(\varphi,\mathrm{i}\sqrt{\mathrm{n}}\right)\:+\mathrm{Res}\left(\varphi,−\mathrm{i}\sqrt{\mathrm{n}}\right)\right\} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{i}\sqrt{\mathrm{n}}\right)\:=\frac{\mathrm{1}}{\mathrm{2i}\sqrt{\mathrm{n}}\left(−\mathrm{n}+\mathrm{2n}\right)}\:=\frac{\mathrm{1}}{\mathrm{2in}\sqrt{\mathrm{n}}} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{i}\sqrt{\mathrm{2n}}\right)\:=\frac{\mathrm{1}}{\mathrm{2i}\sqrt{\mathrm{2n}}\left(−\mathrm{2n}+\mathrm{n}\right)}\:=−\frac{\mathrm{1}}{\mathrm{2in}\sqrt{\mathrm{2n}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\frac{\mathrm{1}}{\mathrm{2in}\sqrt{\mathrm{n}}}−\frac{\mathrm{1}}{\mathrm{2in}\sqrt{\mathrm{2n}}}\right\}\:=\frac{\pi}{\mathrm{n}\sqrt{\mathrm{n}}}\:−\frac{\pi}{\mathrm{n}\sqrt{\mathrm{2}}\sqrt{\mathrm{n}}} \\ $$$$=\frac{\pi}{\mathrm{n}\sqrt{\mathrm{n}}}\left\{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right\}\:=\frac{\pi\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{n}\sqrt{\mathrm{2n}}}\:\Rightarrow\:\mathrm{A}_{\mathrm{n}} =\frac{\pi\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2n}\sqrt{\mathrm{2n}}} \\ $$

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