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Question Number 121659 by Dwaipayan Shikari last updated on 10/Nov/20
∫_0 ^∞ (dx/((1+x^2 )(1+a^2 x^2 )(1+a^4 x^2 )(1+a^6 x^2 ).....))
$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{4}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{6}} {x}^{\mathrm{2}} \right)…..} \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 10/Nov/20
∫_0 ^∞ (dx/((1+x^2 )(1+a^2 x^2 )))=(π/(2(a+1))) ∫_0 ^∞ (1/((1+x^2 )(1+a^2 x^2 )(1+a^4 x^2 )))=(π/(2(1+a+a^3 ))) ∫_0 ^∞ (1/((1+x^2 )(1+a^2 x^2 )(1+a^4 x^2 )(1+a^6 x^2 )))=(π/(2(1+a+a^3 +a^6 ))) ∫_0 ^∞ (1/((1+x^2 )(1+a^2 x^2 )(1+a^4 x^2 )....))=(π/(2(1+a+a^3 +a^6 +a^(10) +C)))
$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}=\frac{\pi}{\mathrm{2}\left({a}+\mathrm{1}\right)} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{4}} {x}^{\mathrm{2}} \right)}=\frac{\pi}{\mathrm{2}\left(\mathrm{1}+{a}+{a}^{\mathrm{3}} \right)} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{4}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{6}} {x}^{\mathrm{2}} \right)}=\frac{\pi}{\mathrm{2}\left(\mathrm{1}+{a}+{a}^{\mathrm{3}} +{a}^{\mathrm{6}} \right)} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{4}} {x}^{\mathrm{2}} \right)….}=\frac{\pi}{\mathrm{2}\left(\mathrm{1}+{a}+{a}^{\mathrm{3}} +{a}^{\mathrm{6}} +{a}^{\mathrm{10}} +{C}\right)} \\ $$
Commented by TANMAY PANACEA last updated on 10/Nov/20
excellent
$${excellent} \\ $$
Commented by Dwaipayan Shikari last updated on 10/Nov/20
Ramanujun showed this result without any proof. I have tried to prove the result
$${Ramanujun}\:{showed}\:{this}\:{result}\:{without}\:{any}\:{proof}. \\ $$$${I}\:{have}\:{tried}\:{to}\:{prove}\:{the}\:{result} \\ $$
Commented by Dwaipayan Shikari last updated on 10/Nov/20
Answered by mindispower last updated on 11/Nov/20
i found ∫_0 ^∞ (dx/(Π_(k=0) ^n (1+a^(2k) x^(2k) )))=(π/2)Σ_(k=0) ^n (1/(a^k Π_(j=0,j≠k) ^n (1−(a^(2j) /a^(2k) ))))
$${i}\:{found} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}\left(\mathrm{1}+{a}^{\mathrm{2}{k}} {x}^{\mathrm{2}{k}} \right)}=\frac{\pi}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{a}^{{k}} \underset{{j}=\mathrm{0},{j}\neq{k}} {\overset{{n}} {\prod}}\left(\mathrm{1}−\frac{{a}^{\mathrm{2}{j}} }{{a}^{\mathrm{2}{k}} }\right)} \\ $$
Commented by Dwaipayan Shikari last updated on 11/Nov/20
x is not varying
$${x}\:{is}\:{not}\:{varying} \\ $$
Answered by mindispower last updated on 11/Nov/20
let f(n)=∫_0 ^∞ (dx/((1+x^2 )......(1+x^2 a^(2n) ))) f(n)=(1/2)∫_(−∞) ^∞ (dx/((1+x^2 ).....(1+x^2 a^(2n) )))=(1/2)∫_(−∞) ^∞ g(x)dx integrat in complex uper half plan Im(z)>0 1+a^(2k) x^2 =0⇒x=+_− (i/a^k ) poles aret_k = (i/a^k ),k∈[0,...n] Res(g,(i/a^k ))=lim_(x→(i/a^k )) (x−(i/a^k ))g(x)=(1/(2ia^k )).Π_(j=0,j≠k) ^n (1/((1−(a^(2j) /a^(2k) )))) f(n)=(1/2).2iπRes(g,t_k ,k∈[0,n]) =(1/2).2iπΣ_(k≤n) (1/(2ia_k ))=(π/2)Σ_(k≤n) (1/a^k ).Π_(j=0,j≠k) ^n (1/((1−((a^j /a^k ))^2 )))
$${let}\:{f}\left({n}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)……\left(\mathrm{1}+{x}^{\mathrm{2}} {a}^{\mathrm{2}{n}} \right)} \\ $$$${f}\left({n}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)…..\left(\mathrm{1}+{x}^{\mathrm{2}} {a}^{\mathrm{2}{n}} \right)}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\infty} {g}\left({x}\right){dx} \\ $$$${integrat}\:{in}\:{complex}\:{uper}\:{half}\:{plan}\:{Im}\left({z}\right)>\mathrm{0} \\ $$$$\mathrm{1}+{a}^{\mathrm{2}{k}} {x}^{\mathrm{2}} =\mathrm{0}\Rightarrow{x}=\underset{−} {+}\frac{{i}}{{a}^{{k}} } \\ $$$${poles}\:{aret}_{{k}} =\:\frac{{i}}{{a}^{{k}} },{k}\in\left[\mathrm{0},…{n}\right] \\ $$$${Res}\left({g},\frac{{i}}{{a}^{{k}} }\right)=\underset{{x}\rightarrow\frac{{i}}{{a}^{{k}} }} {\mathrm{lim}}\left({x}−\frac{{i}}{{a}^{{k}} }\right){g}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{ia}^{{k}} }.\underset{{j}=\mathrm{0},{j}\neq{k}} {\overset{{n}} {\prod}}\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{{a}^{\mathrm{2}{j}} }{{a}^{\mathrm{2}{k}} }\right)} \\ $$$${f}\left({n}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}{i}\pi{Res}\left({g},{t}_{{k}} ,{k}\in\left[\mathrm{0},{n}\right]\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}{i}\pi\underset{{k}\leqslant{n}} {\sum}\frac{\mathrm{1}}{\mathrm{2}{ia}_{{k}} }=\frac{\pi}{\mathrm{2}}\underset{{k}\leqslant{n}} {\sum}\frac{\mathrm{1}}{{a}^{{k}} }.\underset{{j}=\mathrm{0},{j}\neq{k}} {\overset{{n}} {\prod}}\frac{\mathrm{1}}{\left(\mathrm{1}−\left(\frac{{a}^{{j}} }{{a}^{{k}} }\right)^{\mathrm{2}} \right)} \\ $$

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