Menu Close

let-give-n-inyehr-natural-1-find-tbe-value-of-A-n-0-dx-x-2-1-x-2-2-x-2-n-

Tinku Tara 0 Comments
Pin




Question Number 37601 by prof Abdo imad last updated on 15/Jun/18
let give n inyehr natural≥1 find tbe value of A_n = ∫_0 ^∞ (dx/((x^2 +1)(x^(2 ) +2)....(x^2 +n)))
$${let}\:{give}\:{n}\:{inyehr}\:{natural}\geqslant\mathrm{1}\:{find}\:{tbe}\:{value}\:{of} \\ $$$${A}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}\:} +\mathrm{2}\right)….\left({x}^{\mathrm{2}} \:+{n}\right)} \\ $$
Commented by math khazana by abdo last updated on 17/Jun/18
2 A_n =∫_(−∞) ^(+∞) (dx/((x^2 +1)(x^2 +2)....(x^2 +n))) let ϕ(z) = (1/((z^2 +1)(z^2 +2).....(z^2 +n))) ϕ(z)= (1/(Π_(k=1) ^n (z^2 +k)))= (1/(Π_(k=1) ^n (z−i(√k))(z+i(√k)))) = (1/(Π_(k=1) ^n (z−i(√k))Π_(k=1) ^n (z +i(√k)))) so the poles of ϕ are i(√k) and −i(√k) with k∈{1,2,...,n} ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Σ_(z_k ∈w^+ ) Res(ϕ,z_k )with w^+ ={z∈C/ Im(z)>0} ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Σ_(k=1) ^n Res(ϕ,i(√k)) Res(ϕ,i(√k)) = (1/(p^′ (i(√k)))) with p(x)=(x^2 +1)(x^2 +2)...(x^2 +n) let q(t)=(t+1)(t+2)....(t+n) ⇒ q^′ (t) = Σ_(k=1) ^n Π_(p=1_(p≠−k) ) ^n (x+p) p(x)=q(x^2 )⇒p^′ (x) =2x q^′ (x^2 ) ⇒ p^′ (x) =2x Σ_(k=1) ^n Π_(p=1_(p≠−k) ) ^n (x^2 +p) ⇒ p^′ (i(√m)) =2i(√m)Σ_(k=1) ^n Π_(p=1_(p≠−k) ) ^n (p−m) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Σ_(m=1) ^n (1/(2i(√m))){Σ_(k=1) ^n Π_(p=1_(p≠−k) ) ^n (p−m)} = (π/( (√m))) Σ_(m=1) ^n { Σ_(k=1) ^n Π_(p=1_(p≠−k) ) ^n (p−m)}.
$$\mathrm{2}\:{A}_{{n}} =\int_{−\infty} ^{+\infty} \:\:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)….\left({x}^{\mathrm{2}} \:+{n}\right)} \\ $$$${let}\:\varphi\left({z}\right)\:\:=\:\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}^{\mathrm{2}} \:+\mathrm{2}\right)…..\left({z}^{\mathrm{2}} \:+{n}\right)} \\ $$$$\varphi\left({z}\right)=\:\frac{\mathrm{1}}{\prod_{{k}=\mathrm{1}} ^{{n}} \left({z}^{\mathrm{2}} \:+{k}\right)}=\:\frac{\mathrm{1}}{\prod_{{k}=\mathrm{1}} ^{{n}} \left({z}−{i}\sqrt{{k}}\right)\left({z}+{i}\sqrt{{k}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\prod_{{k}=\mathrm{1}} ^{{n}} \left({z}−{i}\sqrt{{k}}\right)\prod_{{k}=\mathrm{1}} ^{{n}} \:\left({z}\:+{i}\sqrt{{k}}\right)}\:{so}\:{the}\:{poles}\:{of} \\ $$$$\varphi\:{are}\:{i}\sqrt{{k}}\:\:{and}\:−{i}\sqrt{{k}}\:\:\:{with}\:{k}\in\left\{\mathrm{1},\mathrm{2},…,{n}\right\} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\sum_{{z}_{{k}} \in{w}^{+} } \:{Res}\left(\varphi,{z}_{{k}} \right){with} \\ $$$${w}^{+} =\left\{{z}\in{C}/\:{Im}\left({z}\right)>\mathrm{0}\right\} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{Res}\left(\varphi,{i}\sqrt{{k}}\right) \\ $$$${Res}\left(\varphi,{i}\sqrt{{k}}\right)\:=\:\frac{\mathrm{1}}{{p}^{'} \left({i}\sqrt{{k}}\right)}\:{with}\: \\ $$$${p}\left({x}\right)=\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)…\left({x}^{\mathrm{2}} \:+{n}\right) \\ $$$${let}\:{q}\left({t}\right)=\left({t}+\mathrm{1}\right)\left({t}+\mathrm{2}\right)….\left({t}+{n}\right)\:\Rightarrow \\ $$$${q}^{'} \left({t}\right)\:=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\prod_{{p}=\mathrm{1}_{{p}\neq−{k}} } ^{{n}} \left({x}+{p}\right) \\ $$$${p}\left({x}\right)={q}\left({x}^{\mathrm{2}} \right)\Rightarrow{p}^{'} \left({x}\right)\:=\mathrm{2}{x}\:{q}^{'} \left({x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${p}^{'} \left({x}\right)\:=\mathrm{2}{x}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\prod_{{p}=\mathrm{1}_{{p}\neq−{k}} } ^{{n}} \left({x}^{\mathrm{2}} \:+{p}\right)\:\Rightarrow \\ $$$${p}^{'} \left({i}\sqrt{{m}}\right)\:=\mathrm{2}{i}\sqrt{{m}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\prod_{{p}=\mathrm{1}_{{p}\neq−{k}} } ^{{n}} \left({p}−{m}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\sum_{{m}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{{m}}}\left\{\sum_{{k}=\mathrm{1}} ^{{n}} \prod_{{p}=\mathrm{1}_{{p}\neq−{k}} } ^{{n}} \left({p}−{m}\right)\right\} \\ $$$$=\:\frac{\pi}{\:\sqrt{{m}}}\:\sum_{{m}=\mathrm{1}} ^{{n}} \left\{\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\prod_{{p}=\mathrm{1}_{{p}\neq−{k}} } ^{{n}} \left({p}−{m}\right)\right\}. \\ $$
Commented by math khazana by abdo last updated on 17/Jun/18
∫_(−∞) ^(+∞) ϕ(z)dz =(π/( (√m)))Σ_(m=1) ^n { Σ_(m=1) ^n Π_(p=1) ^n (p−m)}^(−1) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπΣ_(m=1) ^n (1/(2i(√m)Σ_(k=1) ^n Π_(p=1_(p≠−k) ) ^n (p−m))) =Σ_(m=1) ^n (π/( (√m))){ Σ_(k=1) ^n Π_(p=1_(p≠−k) ) ^n (p−m)}^(−1)
$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\frac{\pi}{\:\sqrt{{m}}}\sum_{{m}=\mathrm{1}} ^{{n}} \:\:\left\{\:\sum_{{m}=\mathrm{1}} ^{{n}} \:\prod_{{p}=\mathrm{1}} ^{{n}} \left({p}−{m}\right)\right\}^{−\mathrm{1}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\sum_{{m}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{{m}}\sum_{{k}=\mathrm{1}} ^{{n}} \prod_{{p}=\mathrm{1}_{{p}\neq−{k}} } ^{{n}} \left({p}−{m}\right)} \\ $$$$=\sum_{{m}=\mathrm{1}} ^{{n}} \:\:\frac{\pi}{\:\sqrt{{m}}}\left\{\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\prod_{{p}=\mathrm{1}_{{p}\neq−{k}} } ^{{n}} \left({p}−{m}\right)\right\}^{−\mathrm{1}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *