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](https://www.tinkutara.com/question/Q26757.png)
$${give}\:{the}\:{decomposition}\:{of}\:{F}\left({x}\right)\:=\:\:\:\frac{\mathrm{1}}{{x}^{\mathrm{2}{n}} +\mathrm{1}}\:\:{inside}\:\mathbb{C}\left[{x}\right] \\ $$$${then}\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}{n}} }\:\:\:\:\:\:{n}\in\mathbb{N}\:\:{and}\:{n}\neq{o} \\ $$
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 ^− )) ) .](https://www.tinkutara.com/question/Q27218.png)
$${let}\:{find}\:{the}\:{poles}\:{of}\:\:{F} \\ $$$${z}^{\mathrm{2}{n}} +\mathrm{1}=\mathrm{0}\:\Leftrightarrow\:\:\:{z}^{\mathrm{2}{n}} \:=\:{e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:{so}\:{the}\:{poles}\:{of}\:{F}\:{are} \\ $$$${z}_{{k}} =\:\:{e}^{{i}\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}} \:\:{k}\in\left[\left[\mathrm{0},\mathrm{2}{n}−\mathrm{1}\right]\right] \\ $$$${F}\left({x}\right)=\:\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}−\mathrm{1}} \:\frac{\lambda_{{k}} }{{x}−{z}_{{k}} }\:\:\:\:\:{and}\:\:\:\lambda_{{k}} \:=\:\:\frac{\mathrm{1}}{\mathrm{2}{n}\:{z}_{{k}} ^{\mathrm{2}{n}−\mathrm{1}} }\:=−\frac{\mathrm{1}}{\mathrm{2}{n}}\:{z}_{{k}} \\ $$$${F}\left({x}\right)=\:−\frac{\mathrm{1}}{\mathrm{2}{n}}\:\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}−\mathrm{1}} \:\:\frac{{z}_{{k}} }{{x}−{z}_{{k}} }\:\:{but}\:\:{z}_{\mathrm{0}} \:=\:{e}^{{i}\frac{\pi}{\mathrm{2}{n}}} \:\:\:\:\:{z}_{\mathrm{0}} ^{−} =\:{e}^{−{i}\frac{\pi}{\mathrm{2}{n}\:}} \:=\:{e}^{{i}\frac{\left(\mathrm{2}\pi−\frac{\pi}{\mathrm{2}{n}}\right)}{}} \\ $$$$=\:{z}_{\mathrm{2}{n}−\mathrm{1}} \:\:\:\:,\:\:{z}_{\mathrm{1}} ^{−} \:\:={z}_{\mathrm{2}{n}−\mathrm{2}} ….. \\ $$$$\Rightarrow\:\:\:{F}\left({x}\right)\:\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\:\:\frac{{z}_{{k}} }{{x}−{z}_{{k}} }\:\:+\:\frac{{z}_{{k}} ^{−} }{{x}−{z}_{{k}} ^{−} }\:\right)\:\:. \\ $$
Commented by abdo imad last updated on 03/Jan/18
$${F}\left({x}\right)=\:−\frac{\mathrm{1}}{\mathrm{2}{n}}\:\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\:\:\:\frac{{z}_{{k}} }{{x}−{z}_{{k}} }\:+\:\:\frac{{z}_{{k}} ^{−} }{{x}−{z}_{{k}} ^{−} }\:\:\right) \\ $$
Commented by abdo imad last updated on 23/Jan/18
$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}{n}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}{n}} } \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{4}{n}}\:\:\left(\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{z}_{{k}} \:\int_{{R}} \:\:\frac{{dx}}{{x}−{z}_{{k}} }\:\:+\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{z}_{{k}} ^{−} \:\:\int_{{R}} \:\frac{{dx}}{{x}−{z}_{{k}} ^{−} }\:\right) \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{4}{n}}\left(\:{i}\pi\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{z}_{{k}} −\:\:{i}\pi\:\sum_{{k}=} ^{{n}−\mathrm{1}} \:{z}_{{k}} ^{−} \:\:\right) \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{4}{n}}\left(−\mathrm{2}\pi\right)\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right) \\ $$$$=\:\frac{\pi}{\mathrm{2}{n}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)\:{let}\:{find} \\ $$$${A}=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{\left.\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)={Im}\left(\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{{i}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)} \right){but} \\ $$$$=\:{e}^{{i}\frac{\pi}{\mathrm{2}{n}}} \:\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\left({e}^{{i}\frac{\pi}{{n}}\:} \right)^{{k}} =\:\:{e}^{{i}\frac{\pi}{\mathrm{2}{n}}} \:\frac{\mathrm{2}}{\mathrm{1}−{e}^{{i}\frac{\pi}{{n}}} } \\ $$$$=\frac{\mathrm{2}\:{e}^{{i}\frac{\pi}{\mathrm{2}{n}}} }{\mathrm{1}−{cos}\left(\frac{\pi}{{n}}\right)−{isin}\left(\frac{\pi}{{n}}\right)}\:=\:\:\:\frac{{e}^{{i}\frac{\pi}{\mathrm{2}{n}}} }{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}{n}}\right)\:−\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{2}{n}}\right){cos}\left(\frac{\pi}{\mathrm{2}{n}}\right)} \\ $$$$=\:\:\:\frac{−\mathrm{1}}{{isin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}=\:\:\frac{{i}}{{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\:\Rightarrow\:{A}=\:\frac{\mathrm{1}}{{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\:\:{so} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}{n}} }=\:\:\frac{\pi}{\mathrm{2}{n}\:{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\:\:=\:\:\:\frac{\frac{\pi}{\mathrm{2}{n}}}{{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\:\:. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$