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Question Number 147006 by mathmax by abdo last updated on 17/Jul/21
$$\mathrm{find}\:\:\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{2}\right)......\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{n}\right)} \\ $$
Answered by mindispower last updated on 17/Jul/21
$${I}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\infty} \frac{{dx}}{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left({x}^{\mathrm{2}} +\sqrt{{k}^{\mathrm{2}} }\right)} \\ $$$${by}\:{Residue}\:{th}\:{in}\:{uper}\:{half}\:?{complex}\:{plan} \\ $$$${Im}\left({z}\right)\geqslant\mathrm{0} \\ $$$$=\mathrm{2}{i}\pi\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{Res}\left({f},{i}\sqrt{{k}}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{2}{i}\pi,\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{{k}}}.\underset{{j}=\mathrm{1},{j}\neq{k}} {\overset{{n}} {\prod}}\frac{\mathrm{1}}{\left({k}−{j}\right)} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\pi}{\:\sqrt{{k}}},\underset{{j}=\mathrm{1},{j}\neq{k}} {\overset{{n}} {\prod}}\frac{\mathrm{1}}{\left({j}−{k}\right)} \\ $$