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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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