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Question Number 71663 by mathmax by abdo last updated on 18/Oct/19

calculate A_n =∫_0 ^∞ (dx/((x^2 +1)(x^2 +2)....(x^2 +n))) with n integr and n≥1

$${calculate}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{2}\right)....\left({x}^{\mathrm{2}} \:+{n}\right)} \\ $$$${with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{1} \\ $$

Commented by mathmax by abdo last updated on 18/Oct/19

we have 2A_n =∫_(−∞) ^(+∞) (dx/((x^2 +1)(x^2 +2)..(x^2 +n))) let ϕ(z)=(1/((z^2 +1)(z^2 +2)...(z^2 +n))) =(1/(Π_(k=1) ^n (z^2 +k))) =(1/(Π_(k=1) ^n (z−i(√k))(z+i(√k)))) so the poles of ϕ are +^− i(√k) and ∫_(−∞) ^(+∞) ϕ(z)dz =2iπΣ_(k=1) ^n Res(ϕ,i(√k)) Res(ϕ,i(√k)) =lim_(z→i(√k)) (z−i(√k))ϕ(z) = (1/(Π_(k=1) ^n (2i(√k))Π_(j=1and j≠k) ^n (i(√k)−i(√j)))) =(1/((2i)^n (√(n!))Π_(j=1 and j≠k) ^n (i(√k)−i(√j)))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =((2iπ)/((2i)^n (√(n!))Π_(j=1) ^n (i(√k)−i(√j)))) ⇒ A_n =Σ_(k=1) ^n ((iπ)/((2i)^n (√(n!))Π_(j=1and j≠k) ^n ( i(√k)−i(√j))))

$${we}\:{have}\:\mathrm{2}{A}_{{n}} =\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{2}\right)..\left({x}^{\mathrm{2}} +{n}\right)}\:{let}\: \\ $$$$\varphi\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left({z}^{\mathrm{2}} +\mathrm{2}\right)...\left({z}^{\mathrm{2}} \:+{n}\right)}\:=\frac{\mathrm{1}}{\prod_{{k}=\mathrm{1}} ^{{n}} \left({z}^{\mathrm{2}} +{k}\right)} \\ $$$$=\frac{\mathrm{1}}{\prod_{{k}=\mathrm{1}} ^{{n}} \left({z}−{i}\sqrt{{k}}\right)\left({z}+{i}\sqrt{{k}}\right)}\:\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}{i}\sqrt{{k}} \\ $$$${and}\:\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\sum_{{k}=\mathrm{1}} ^{{n}} \:{Res}\left(\varphi,{i}\sqrt{{k}}\right) \\ $$$${Res}\left(\varphi,{i}\sqrt{{k}}\right)\:={lim}_{{z}\rightarrow{i}\sqrt{{k}}} \:\:\:\left({z}−{i}\sqrt{{k}}\right)\varphi\left({z}\right) \\ $$$$=\:\:\:\frac{\mathrm{1}}{\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{2}{i}\sqrt{{k}}\right)\prod_{{j}=\mathrm{1}{and}\:{j}\neq{k}} ^{{n}} \left({i}\sqrt{{k}}−{i}\sqrt{{j}}\right)}\:=\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{{n}} \sqrt{{n}!}\prod_{{j}=\mathrm{1}\:{and}\:{j}\neq{k}} ^{{n}} \left({i}\sqrt{{k}}−{i}\sqrt{{j}}\right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\frac{\mathrm{2}{i}\pi}{\left(\mathrm{2}{i}\right)^{{n}} \sqrt{{n}!}\prod_{{j}=\mathrm{1}} ^{{n}} \left({i}\sqrt{{k}}−{i}\sqrt{{j}}\right)}\:\Rightarrow \\ $$$${A}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \frac{{i}\pi}{\left(\mathrm{2}{i}\right)^{{n}} \sqrt{{n}!}\prod_{{j}=\mathrm{1}{and}\:{j}\neq{k}} ^{{n}} \left(\:{i}\sqrt{{k}}−{i}\sqrt{{j}}\right)} \\ $$

Answered by mind is power last updated on 18/Oct/19

(1/((x^2 +1).....(x^2 +n)))=Σ_(k=1) ^n ((a_k /((x−i(√k))))+(b_k /((x+i(√k)))) a_k =lim_ (x−i(√k)).(1/((x^2 +1)......(x^2 +n))) =Π_(l=1,l≠k) ^n (1/(2i(√k).(l−k))) b_k =Π_(l=1,l≠k) ^n (1/(−2i(√k)(l−k))) (1/((x^2 +1).....(x^2 +n)))=Σ_(k=1) ^n (Π_(l=1,l≠k) ^n (1/(2i(√k)(l−k)))(.(1/(x−i(√k)))−(1/(x+i(√k)))) =Σ_(k=1) ^n (Π_(l=1,k≠l) ^n (1/((l−k)(x^2 +k)))) A_n =Σ_(k=1) ^n Π_(l=1,l≠k) ^n (1/((l−k)))∫_0 ^(+∞) (dx/(x^2 +k)) ∫(dx/(x^2 +k))=(1/(√k)).tan^(−1) ((x/(√k)))+c ∫_0 ^(+∞) (dx/(x^2 +k))=(π/(2(√k))) A_n =Σ_(k=1) ^n Π_(l=1,l≠k) ^n (π/(2(√k).(l−k)))

$$\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right).....\left(\mathrm{x}^{\mathrm{2}} +\mathrm{n}\right)}=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\frac{\mathrm{a}_{\mathrm{k}} }{\left(\mathrm{x}−\mathrm{i}\sqrt{\mathrm{k}}\right)}+\frac{\mathrm{b}_{\mathrm{k}} }{\left(\mathrm{x}+\mathrm{i}\sqrt{\mathrm{k}}\right.}\right) \\ $$$$\mathrm{a}_{\mathrm{k}} =\mathrm{lim}_{} \left(\mathrm{x}−\mathrm{i}\sqrt{\mathrm{k}}\right).\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)......\left(\mathrm{x}^{\mathrm{2}} +\mathrm{n}\right)} \\ $$$$=\underset{\mathrm{l}=\mathrm{1},\mathrm{l}\neq\mathrm{k}} {\overset{\mathrm{n}} {\prod}}\frac{\mathrm{1}}{\mathrm{2i}\sqrt{\mathrm{k}}.\left(\mathrm{l}−\mathrm{k}\right)} \\ $$$$\mathrm{b}_{\mathrm{k}} =\underset{\mathrm{l}=\mathrm{1},\mathrm{l}\neq\mathrm{k}} {\overset{\mathrm{n}} {\prod}}\frac{\mathrm{1}}{−\mathrm{2i}\sqrt{\mathrm{k}}\left(\mathrm{l}−\mathrm{k}\right)} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right).....\left(\mathrm{x}^{\mathrm{2}} +\mathrm{n}\right)}=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\underset{\mathrm{l}=\mathrm{1},\mathrm{l}\neq\mathrm{k}} {\overset{\mathrm{n}} {\prod}}\frac{\mathrm{1}}{\mathrm{2i}\sqrt{\mathrm{k}}\left(\mathrm{l}−\mathrm{k}\right)}\left(.\frac{\mathrm{1}}{\mathrm{x}−\mathrm{i}\sqrt{\mathrm{k}}}−\frac{\mathrm{1}}{\mathrm{x}+\mathrm{i}\sqrt{\mathrm{k}}}\right)\right. \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\underset{\mathrm{l}=\mathrm{1},\mathrm{k}\neq\mathrm{l}} {\overset{\mathrm{n}} {\prod}}\frac{\mathrm{1}}{\left(\mathrm{l}−\mathrm{k}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{k}\right)}\right) \\ $$$$\mathrm{A}_{\mathrm{n}} =\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\underset{\mathrm{l}=\mathrm{1},\mathrm{l}\neq\mathrm{k}} {\overset{\mathrm{n}} {\prod}}\frac{\mathrm{1}}{\left(\mathrm{l}−\mathrm{k}\right)}\underset{\mathrm{0}} {\overset{+\infty} {\int}}\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{k}} \\ $$$$\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{k}}=\frac{\mathrm{1}}{\sqrt{\mathrm{k}}}.\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\sqrt{\mathrm{k}}}\right)+\mathrm{c} \\ $$$$\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{k}}=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{k}}} \\ $$$$\mathrm{A}_{\mathrm{n}} =\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\underset{\mathrm{l}=\mathrm{1},\mathrm{l}\neq\mathrm{k}} {\overset{\mathrm{n}} {\prod}}\frac{\pi}{\mathrm{2}\sqrt{\mathrm{k}}.\left(\mathrm{l}−\mathrm{k}\right)} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 18/Oct/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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