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Question Number 109214 by mathmax by abdo last updated on 22/Aug/20

$${\mathrm{calculateA}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{({\mathrm{x}}^2+\mathrm{n})({\mathrm{x}}^2+2\mathrm{n})}\mathrm{with}\mathrm{n}\mathrm{integr}\mathrm{natural}⩾1$$

Answered by mathmax by abdo last updated on 22/Aug/20

$${2\mathrm{A}}_{\mathrm{n}}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{({\mathrm{x}}^2+\mathrm{n})({\mathrm{x}}^2+2\mathrm{n})}\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{1}{({\mathrm{z}}^2+\mathrm{n})({\mathrm{z}}^2+2\mathrm{n})}\Rightarrow$$$$\phi \left(\mathrm{z}\right)=\frac{1}{(\mathrm{z}-\mathrm{i}\sqrt{\mathrm{n}})(\mathrm{z}+\mathrm{i}\sqrt{\mathrm{n}})(\mathrm{z}-\mathrm{i}\sqrt{2\mathrm{n}})(\mathrm{z}+\mathrm{i}\sqrt{2\mathrm{n}})}\mathrm{residus}\mathrm{theorem}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\phi ,\mathrm{i}\sqrt{\mathrm{n}})+\mathrm{Res}(\phi ,-\mathrm{i}\sqrt{\mathrm{n}})\}$$$$\mathrm{Res}(\phi ,\mathrm{i}\sqrt{\mathrm{n}})=\frac{1}{2\mathrm{i}\sqrt{\mathrm{n}}(-\mathrm{n}+2\mathrm{n})}=\frac{1}{2\mathrm{in}\sqrt{\mathrm{n}}}$$$$\mathrm{Res}(\phi ,\mathrm{i}\sqrt{2\mathrm{n}})=\frac{1}{2\mathrm{i}\sqrt{2\mathrm{n}}(-2\mathrm{n}+\mathrm{n})}=-\frac{1}{2\mathrm{in}\sqrt{2\mathrm{n}}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\frac{1}{2\mathrm{in}\sqrt{\mathrm{n}}}-\frac{1}{2\mathrm{in}\sqrt{2\mathrm{n}}}\}=\frac{\pi }{\mathrm{n}\sqrt{\mathrm{n}}}-\frac{\pi }{\mathrm{n}\sqrt{2}\sqrt{\mathrm{n}}}$$$$=\frac{\pi }{\mathrm{n}\sqrt{\mathrm{n}}}\{1-\frac{1}{\sqrt{2}}\}=\frac{\pi (\sqrt{2}-1)}{\mathrm{n}\sqrt{2\mathrm{n}}}\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\frac{\pi (\sqrt{2}-1)}{2\mathrm{n}\sqrt{2\mathrm{n}}}$$

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