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Question-130560

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Question Number 130560 by rs4089 last updated on 26/Jan/21
Answered by Dwaipayan Shikari last updated on 26/Jan/21
(π/(2(r+r^3 +r^6 +r^(10) +....)))
$$\frac{\pi}{\mathrm{2}\left({r}+{r}^{\mathrm{3}} +{r}^{\mathrm{6}} +{r}^{\mathrm{10}} +….\right)} \\ $$
Answered by mathmax by abdo last updated on 27/Jan/21
A_n =∫_0 ^∞ (dx/((x^2 +1)(r^2 x^2 +1)(r^4 x^2 +1)....(r^(2n) x^2 +1))) ⇒ 2A_n =∫_(−∞) ^(+∞) (dx/((x^2 +1)(r^2 x^2 +1).....(r^(2n) x^2 +1))) let ϕ(z)=(1/((z^2 +1)(r^2 z^2 +1)....(r^(2n) z^2 +1))) =(1/(Π_(k=1) ^n (r^(2k) z^2 +1))) r^(2k) x^2 +1 =0 ⇒x^2 =−(1/r^(2k) ) ⇒x =+^− i(1/r^k ) ⇒ ϕ(z) =(1/(Π_(k=1) ^n r^(2k) (z−(i/r^k ))(z+(i/r^k )))) ⇒∫_R ϕ(z)dz=2iπΣ_(k=1) ^n Res(ϕ,(i/r^k )) ϕ(z) =(1/(r^(2Σ_1 ^n k) Π_(k=1) ^n (z−(i/r^k ))(z+(i/r^k ))))=(1/(r^(n(n+1)) Π_(k=1) ^n (z−(i/r^k ))(z+(i/r^k )))) Res(ϕ,(i/r^k ))=lim_(z→(i/r^k )) (z−(i/r^k ))ϕ(z) =(1/(r^(n(n+1)) Π_(p=1 and p≠k) ^n (z−(i/r^p ))(z+(i/r^p )))) ⇒ A_n =((iπ)/r^(n(n+1)) )Σ_(k=1) ^n (1/(Π_(p=1,p≠k) ^n (z^2 +(1/r^(2p) ))))
$$\mathrm{A}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{r}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{r}^{\mathrm{4}} \mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)….\left(\mathrm{r}^{\mathrm{2n}} \mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\mathrm{2A}_{\mathrm{n}} =\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{r}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)…..\left(\mathrm{r}^{\mathrm{2n}} \mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\mathrm{let} \\ $$$$\varphi\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{r}^{\mathrm{2}} \mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)….\left(\mathrm{r}^{\mathrm{2n}} \:\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{\mathrm{1}}{\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\mathrm{r}^{\mathrm{2k}} \mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$$\mathrm{r}^{\mathrm{2k}} \mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow\mathrm{x}^{\mathrm{2}} \:=−\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2k}} }\:\:\Rightarrow\mathrm{x}\:=\overset{−} {+}\mathrm{i}\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{k}} }\:\Rightarrow \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{1}}{\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \mathrm{r}^{\mathrm{2k}} \left(\mathrm{z}−\frac{\mathrm{i}}{\mathrm{r}^{\mathrm{k}} }\right)\left(\mathrm{z}+\frac{\mathrm{i}}{\mathrm{r}^{\mathrm{k}} }\right)}\:\Rightarrow\int_{\mathrm{R}} \varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{Res}\left(\varphi,\frac{\mathrm{i}}{\mathrm{r}^{\mathrm{k}} }\right) \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}\sum_{\mathrm{1}} ^{\mathrm{n}} \mathrm{k}} \prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\mathrm{z}−\frac{\mathrm{i}}{\mathrm{r}^{\mathrm{k}} }\right)\left(\mathrm{z}+\frac{\mathrm{i}}{\mathrm{r}^{\mathrm{k}} }\right)}=\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)} \prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\mathrm{z}−\frac{\mathrm{i}}{\mathrm{r}^{\mathrm{k}} }\right)\left(\mathrm{z}+\frac{\mathrm{i}}{\mathrm{r}^{\mathrm{k}} }\right)} \\ $$$$\mathrm{Res}\left(\varphi,\frac{\mathrm{i}}{\mathrm{r}^{\mathrm{k}} }\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\frac{\mathrm{i}}{\mathrm{r}^{\mathrm{k}} }} \:\:\left(\mathrm{z}−\frac{\mathrm{i}}{\mathrm{r}^{\mathrm{k}} }\right)\varphi\left(\mathrm{z}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)} \prod_{\mathrm{p}=\mathrm{1}\:\mathrm{and}\:\mathrm{p}\neq\mathrm{k}} ^{\mathrm{n}} \left(\mathrm{z}−\frac{\mathrm{i}}{\mathrm{r}^{\mathrm{p}} }\right)\left(\mathrm{z}+\frac{\mathrm{i}}{\mathrm{r}^{\mathrm{p}} }\right)}\:\Rightarrow \\ $$$$\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{i}\pi}{\mathrm{r}^{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)} }\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \frac{\mathrm{1}}{\prod_{\mathrm{p}=\mathrm{1},\mathrm{p}\neq\mathrm{k}} ^{\mathrm{n}} \left(\mathrm{z}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2p}} }\right)} \\ $$

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