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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

$$\frac{\pi }{2(r+r^3+r^6+r^{10}+....)}$$

Answered by mathmax by abdo last updated on 27/Jan/21

$${\mathrm{A}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{({\mathrm{x}}^2+1)({\mathrm{r}}^2{\mathrm{x}}^2+1)({\mathrm{r}}^4{\mathrm{x}}^2+1)....({\mathrm{r}}^{2\mathrm{n}}{\mathrm{x}}^2+1)}\Rightarrow$$$${2\mathrm{A}}_{\mathrm{n}}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{({\mathrm{x}}^2+1)({\mathrm{r}}^2{\mathrm{x}}^2+1).....({\mathrm{r}}^{2\mathrm{n}}{\mathrm{x}}^2+1)}\mathrm{let}$$$$\phi \left(\mathrm{z}\right)=\frac{1}{({\mathrm{z}}^2+1)({\mathrm{r}}^2{\mathrm{z}}^2+1)....({\mathrm{r}}^{2\mathrm{n}}{\mathrm{z}}^2+1)}=\frac{1}{\prod _{\mathrm{k}=1}^{\mathrm{n}}({\mathrm{r}}^{2\mathrm{k}}{\mathrm{z}}^2+1)}$$$${\mathrm{r}}^{2\mathrm{k}}{\mathrm{x}}^2+1=0\Rightarrow {\mathrm{x}}^2=-\frac{1}{{\mathrm{r}}^{2\mathrm{k}}}\Rightarrow \mathrm{x}=\stackrel{-}{+}\mathrm{i}\frac{1}{{\mathrm{r}}^{\mathrm{k}}}\Rightarrow$$$$\phi \left(\mathrm{z}\right)=\frac{1}{\prod _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{r}}^{2\mathrm{k}}(\mathrm{z}-\frac{\mathrm{i}}{{\mathrm{r}}^{\mathrm{k}}})(\mathrm{z}+\frac{\mathrm{i}}{{\mathrm{r}}^{\mathrm{k}}})}\Rightarrow {\int }_{\mathrm{R}}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{Res}(\phi ,\frac{\mathrm{i}}{{\mathrm{r}}^{\mathrm{k}}})$$$$\phi \left(\mathrm{z}\right)=\frac{1}{{\mathrm{r}}^{2\sum _1^{\mathrm{n}}\mathrm{k}}\prod _{\mathrm{k}=1}^{\mathrm{n}}(\mathrm{z}-\frac{\mathrm{i}}{{\mathrm{r}}^{\mathrm{k}}})(\mathrm{z}+\frac{\mathrm{i}}{{\mathrm{r}}^{\mathrm{k}}})}=\frac{1}{{\mathrm{r}}^{\mathrm{n}(\mathrm{n}+1)}\prod _{\mathrm{k}=1}^{\mathrm{n}}(\mathrm{z}-\frac{\mathrm{i}}{{\mathrm{r}}^{\mathrm{k}}})(\mathrm{z}+\frac{\mathrm{i}}{{\mathrm{r}}^{\mathrm{k}}})}$$$$\mathrm{Res}(\phi ,\frac{\mathrm{i}}{{\mathrm{r}}^{\mathrm{k}}})={\lim }_{\mathrm{z}\to \frac{\mathrm{i}}{{\mathrm{r}}^{\mathrm{k}}}}(\mathrm{z}-\frac{\mathrm{i}}{{\mathrm{r}}^{\mathrm{k}}})\phi \left(\mathrm{z}\right)$$$$=\frac{1}{{\mathrm{r}}^{\mathrm{n}(\mathrm{n}+1)}\prod _{\mathrm{p}=1\mathrm{and}\mathrm{p}\ne \mathrm{k}}^{\mathrm{n}}(\mathrm{z}-\frac{\mathrm{i}}{{\mathrm{r}}^{\mathrm{p}}})(\mathrm{z}+\frac{\mathrm{i}}{{\mathrm{r}}^{\mathrm{p}}})}\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}=\frac{\mathrm{i}\pi }{{\mathrm{r}}^{\mathrm{n}(\mathrm{n}+1)}}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\prod _{\mathrm{p}=1,\mathrm{p}\ne \mathrm{k}}^{\mathrm{n}}({\mathrm{z}}^2+\frac{1}{{\mathrm{r}}^{2\mathrm{p}}})}$$

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