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Question Number 26757 by abdo imad last updated on 28/Dec/17

give the decomposition of F(x) = (1/(x^(2n) +1)) inside C[x] then find the value of ∫_0 ^∞ (dx/(1+x^(2n) )) n∈N and n≠o

$${give}\:{the}\:{decomposition}\:{of}\:{F}\left({x}\right)\:=\:\:\:\frac{\mathrm{1}}{{x}^{\mathrm{2}{n}} +\mathrm{1}}\:\:{inside}\:\mathbb{C}\left[{x}\right] \\ $$$${then}\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}{n}} }\:\:\:\:\:\:{n}\in\mathbb{N}\:\:{and}\:{n}\neq{o} \\ $$

Commented by abdo imad last updated on 03/Jan/18

let find the poles of F z^(2n) +1=0 ⇔ z^(2n) = e^(i(2k+1)π) so the poles of F are z_k = e^(i(((2k+1)π)/(2n))) k∈[[0,2n−1]] F(x)= Σ_(k=0) ^(2n−1) (λ_k /(x−z_k )) and λ_k = (1/(2n z_k ^(2n−1) )) =−(1/(2n)) z_k F(x)= −(1/(2n)) Σ_(k=0) ^(2n−1) (z_k /(x−z_k )) but z_0 = e^(i(π/(2n))) z_0 ^− = e^(−i(π/(2n ))) = e^(i(((2π−(π/(2n))))/)) = z_(2n−1) , z_1 ^− =z_(2n−2) ..... ⇒ F(x) =Σ_(k=0) ^(n−1) ( (z_k /(x−z_k )) + (z_k ^− /(x−z_k ^− )) ) .

$${let}\:{find}\:{the}\:{poles}\:{of}\:\:{F} \\ $$$${z}^{\mathrm{2}{n}} +\mathrm{1}=\mathrm{0}\:\Leftrightarrow\:\:\:{z}^{\mathrm{2}{n}} \:=\:{e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:{so}\:{the}\:{poles}\:{of}\:{F}\:{are} \\ $$$${z}_{{k}} =\:\:{e}^{{i}\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}} \:\:{k}\in\left[\left[\mathrm{0},\mathrm{2}{n}−\mathrm{1}\right]\right] \\ $$$${F}\left({x}\right)=\:\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}−\mathrm{1}} \:\frac{\lambda_{{k}} }{{x}−{z}_{{k}} }\:\:\:\:\:{and}\:\:\:\lambda_{{k}} \:=\:\:\frac{\mathrm{1}}{\mathrm{2}{n}\:{z}_{{k}} ^{\mathrm{2}{n}−\mathrm{1}} }\:=−\frac{\mathrm{1}}{\mathrm{2}{n}}\:{z}_{{k}} \\ $$$${F}\left({x}\right)=\:−\frac{\mathrm{1}}{\mathrm{2}{n}}\:\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}−\mathrm{1}} \:\:\frac{{z}_{{k}} }{{x}−{z}_{{k}} }\:\:{but}\:\:{z}_{\mathrm{0}} \:=\:{e}^{{i}\frac{\pi}{\mathrm{2}{n}}} \:\:\:\:\:{z}_{\mathrm{0}} ^{−} =\:{e}^{−{i}\frac{\pi}{\mathrm{2}{n}\:}} \:=\:{e}^{{i}\frac{\left(\mathrm{2}\pi−\frac{\pi}{\mathrm{2}{n}}\right)}{}} \\ $$$$=\:{z}_{\mathrm{2}{n}−\mathrm{1}} \:\:\:\:,\:\:{z}_{\mathrm{1}} ^{−} \:\:={z}_{\mathrm{2}{n}−\mathrm{2}} ..... \\ $$$$\Rightarrow\:\:\:{F}\left({x}\right)\:\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\:\:\frac{{z}_{{k}} }{{x}−{z}_{{k}} }\:\:+\:\frac{{z}_{{k}} ^{−} }{{x}−{z}_{{k}} ^{−} }\:\right)\:\:. \\ $$

Commented by abdo imad last updated on 03/Jan/18

F(x)= −(1/(2n)) Σ_(k=0) ^(n−1) ( (z_k /(x−z_k )) + (z_k ^− /(x−z_k ^− )) )

$${F}\left({x}\right)=\:−\frac{\mathrm{1}}{\mathrm{2}{n}}\:\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\:\:\:\frac{{z}_{{k}} }{{x}−{z}_{{k}} }\:+\:\:\frac{{z}_{{k}} ^{−} }{{x}−{z}_{{k}} ^{−} }\:\:\right) \\ $$

Commented by abdo imad last updated on 23/Jan/18

∫_0 ^∞ (dx/(1+x^(2n) )) = (1/2)∫_(−∞) ^(+∞) (dx/(1+x^(2n) )) = ((−1)/(4n)) (Σ_(k=0) ^(n−1) z_k ∫_R (dx/(x−z_k )) +Σ_(k=0) ^(n−1) z_k ^− ∫_R (dx/(x−z_k ^− )) ) = ((−1)/(4n))( iπ Σ_(k=0) ^(n−1) z_k − iπ Σ_(k=) ^(n−1) z_k ^− ) = ((−1)/(4n))(−2π) Σ_(k=0) ^(n−1) sin((((2k+1)π)/(2n))) = (π/(2n)) Σ_(k=0) ^(n−1) sin((((2k+1)π)/(2n))) let find A= Σ_(k=0) ^(n−1) sin(((2k+1)π)/(2n)))=Im( Σ_(k=0) ^(n−1) e^(i((((2k+1)π)/(2n)))) )but = e^(i(π/(2n))) Σ_(k=0) ^(n−1) (e^(i(π/n) ) )^k = e^(i(π/(2n))) (2/(1−e^(i(π/n)) )) =((2 e^(i(π/(2n))) )/(1−cos((π/n))−isin((π/n)))) = (e^(i(π/(2n))) /(sin^2 ((π/(2n))) −2isin((π/(2n)))cos((π/(2n))))) = ((−1)/(isin((π/(2n)))))= (i/(sin((π/(2n))))) ⇒ A= (1/(sin((π/(2n))))) so ∫_0 ^∞ (dx/(1+x^(2n) ))= (π/(2n sin((π/(2n))))) = ((π/(2n))/(sin((π/(2n))))) .

$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}{n}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}{n}} } \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{4}{n}}\:\:\left(\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{z}_{{k}} \:\int_{{R}} \:\:\frac{{dx}}{{x}−{z}_{{k}} }\:\:+\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{z}_{{k}} ^{−} \:\:\int_{{R}} \:\frac{{dx}}{{x}−{z}_{{k}} ^{−} }\:\right) \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{4}{n}}\left(\:{i}\pi\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{z}_{{k}} −\:\:{i}\pi\:\sum_{{k}=} ^{{n}−\mathrm{1}} \:{z}_{{k}} ^{−} \:\:\right) \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{4}{n}}\left(−\mathrm{2}\pi\right)\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right) \\ $$$$=\:\frac{\pi}{\mathrm{2}{n}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)\:{let}\:{find} \\ $$$${A}=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{\left.\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)={Im}\left(\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{{i}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)} \right){but} \\ $$$$=\:{e}^{{i}\frac{\pi}{\mathrm{2}{n}}} \:\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\left({e}^{{i}\frac{\pi}{{n}}\:} \right)^{{k}} =\:\:{e}^{{i}\frac{\pi}{\mathrm{2}{n}}} \:\frac{\mathrm{2}}{\mathrm{1}−{e}^{{i}\frac{\pi}{{n}}} } \\ $$$$=\frac{\mathrm{2}\:{e}^{{i}\frac{\pi}{\mathrm{2}{n}}} }{\mathrm{1}−{cos}\left(\frac{\pi}{{n}}\right)−{isin}\left(\frac{\pi}{{n}}\right)}\:=\:\:\:\frac{{e}^{{i}\frac{\pi}{\mathrm{2}{n}}} }{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}{n}}\right)\:−\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{2}{n}}\right){cos}\left(\frac{\pi}{\mathrm{2}{n}}\right)} \\ $$$$=\:\:\:\frac{−\mathrm{1}}{{isin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}=\:\:\frac{{i}}{{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\:\Rightarrow\:{A}=\:\frac{\mathrm{1}}{{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\:\:{so} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}{n}} }=\:\:\frac{\pi}{\mathrm{2}{n}\:{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\:\:=\:\:\:\frac{\frac{\pi}{\mathrm{2}{n}}}{{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\:\:. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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