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Question Number 121659 by Dwaipayan Shikari last updated on 10/Nov/20

$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{4}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{6}} {x}^{\mathrm{2}} \right).....} \\$$ $$\\$$

Commented byDwaipayan Shikari last updated on 10/Nov/20

$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}=\frac{\pi}{\mathrm{2}\left({a}+\mathrm{1}\right)} \\$$ $$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{4}} {x}^{\mathrm{2}} \right)}=\frac{\pi}{\mathrm{2}\left(\mathrm{1}+{a}+{a}^{\mathrm{3}} \right)} \\$$ $$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{4}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{6}} {x}^{\mathrm{2}} \right)}=\frac{\pi}{\mathrm{2}\left(\mathrm{1}+{a}+{a}^{\mathrm{3}} +{a}^{\mathrm{6}} \right)} \\$$ $$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{4}} {x}^{\mathrm{2}} \right)....}=\frac{\pi}{\mathrm{2}\left(\mathrm{1}+{a}+{a}^{\mathrm{3}} +{a}^{\mathrm{6}} +{a}^{\mathrm{10}} +{C}\right)} \\$$

Commented byTANMAY PANACEA last updated on 10/Nov/20

$${excellent} \\$$

Commented byDwaipayan Shikari last updated on 10/Nov/20

$${Ramanujun}\:{showed}\:{this}\:{result}\:{without}\:{any}\:{proof}. \\$$ $${I}\:{have}\:{tried}\:{to}\:{prove}\:{the}\:{result} \\$$

Commented byDwaipayan Shikari last updated on 10/Nov/20

Answered by mindispower last updated on 11/Nov/20

$${i}\:{found} \\$$ $$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}\left(\mathrm{1}+{a}^{\mathrm{2}{k}} {x}^{\mathrm{2}{k}} \right)}=\frac{\pi}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{a}^{{k}} \underset{{j}=\mathrm{0},{j}\neq{k}} {\overset{{n}} {\prod}}\left(\mathrm{1}−\frac{{a}^{\mathrm{2}{j}} }{{a}^{\mathrm{2}{k}} }\right)} \\$$

Commented byDwaipayan Shikari last updated on 11/Nov/20

$${x}\:{is}\:{not}\:{varying} \\$$

Answered by mindispower last updated on 11/Nov/20

$${let}\:{f}\left({n}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)......\left(\mathrm{1}+{x}^{\mathrm{2}} {a}^{\mathrm{2}{n}} \right)} \\$$ $${f}\left({n}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right).....\left(\mathrm{1}+{x}^{\mathrm{2}} {a}^{\mathrm{2}{n}} \right)}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\infty} {g}\left({x}\right){dx} \\$$ $${integrat}\:{in}\:{complex}\:{uper}\:{half}\:{plan}\:{Im}\left({z}\right)>\mathrm{0} \\$$ $$\mathrm{1}+{a}^{\mathrm{2}{k}} {x}^{\mathrm{2}} =\mathrm{0}\Rightarrow{x}=\underset{−} {+}\frac{{i}}{{a}^{{k}} } \\$$ $${poles}\:{aret}_{{k}} =\:\frac{{i}}{{a}^{{k}} },{k}\in\left[\mathrm{0},...{n}\right] \\$$ $${Res}\left({g},\frac{{i}}{{a}^{{k}} }\right)=\underset{{x}\rightarrow\frac{{i}}{{a}^{{k}} }} {\mathrm{lim}}\left({x}−\frac{{i}}{{a}^{{k}} }\right){g}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{ia}^{{k}} }.\underset{{j}=\mathrm{0},{j}\neq{k}} {\overset{{n}} {\prod}}\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{{a}^{\mathrm{2}{j}} }{{a}^{\mathrm{2}{k}} }\right)} \\$$ $${f}\left({n}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}{i}\pi{Res}\left({g},{t}_{{k}} ,{k}\in\left[\mathrm{0},{n}\right]\right) \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}{i}\pi\underset{{k}\leqslant{n}} {\sum}\frac{\mathrm{1}}{\mathrm{2}{ia}_{{k}} }=\frac{\pi}{\mathrm{2}}\underset{{k}\leqslant{n}} {\sum}\frac{\mathrm{1}}{{a}^{{k}} }.\underset{{j}=\mathrm{0},{j}\neq{k}} {\overset{{n}} {\prod}}\frac{\mathrm{1}}{\left(\mathrm{1}−\left(\frac{{a}^{{j}} }{{a}^{{k}} }\right)^{\mathrm{2}} \right)} \\$$