Question Number 79108 by mathmax by abdo last updated on 22/Jan/20
$${decompose}\:{F}\left({x}\right)=\frac{{nx}^{{n}} }{{x}^{\mathrm{2}{n}} \:+\mathrm{1}}\:\:{inside}\:{C}\left({x}\right)\:{and}\:{R}\left({x}\right)\:\:\left({n}\geqslant\mathrm{2}\right) \\ $$$${and}\:{determine}\:\int_{\mathrm{0}} ^{+\infty} {F}\left({x}\right){dx} \\ $$
Commented by mathmax by abdo last updated on 31/Jan/20
$${another}\:{way}\:{for}\:\int_{\mathrm{0}} ^{\infty} \:{F}\left({x}\right){dx}\:\:{we}\:{do}\:{the}\:{changement}\:{x}^{\mathrm{2}{n}} ={t}\:\Rightarrow \\ $$$${x}={t}^{\frac{\mathrm{1}}{\mathrm{2}{n}}} \:\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \:{n}\frac{{x}^{{n}} }{{x}^{\mathrm{2}{n}} +\mathrm{1}}{dx}\:={n}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{t}+\mathrm{1}}×\frac{\mathrm{1}}{\mathrm{2}{n}}{t}^{\frac{\mathrm{1}}{\mathrm{2}{n}}−\mathrm{1}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{{t}+\mathrm{1}}{dt}\:\:=\frac{\mathrm{1}}{\mathrm{2}}\:×\frac{\pi}{{sin}\left(\pi\left(\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)}\:=\frac{\pi}{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{2}{n}}+\frac{\pi}{\mathrm{2}}\right)} \\ $$$$\left.=\left.\frac{\pi}{\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\:\:\:\left\{\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\pi}{{sin}\left(\pi{a}\right)}=\Gamma\left({a}\right)\Gamma\left(\mathrm{1}−{a}\right)\right)\:\:{with}\:\mathrm{0}<{a}<\mathrm{1}\right)\right\} \\ $$
Answered by mind is power last updated on 22/Jan/20
![over C(x) x^(2n) +1=0⇒x^(2n) =−1=e^(iπ+2kπ) x_k =e^(((i(1+2k)π)/(2n)) ) ,k∈[0,2n−1] x_(2n−k−1) =e^(i(((1+2(2n−k−1))/(2n)))π) =e^(−i(((1+2k)/(2n)))π) ⇒x_k =x_(2n−1−k) ^− ,E ((nx^n )/(x^(2n) +1))=Σ(a_k /(x−x_k )) a_k =((nx_k )/(2nx_k ^(2n−1) ))=(1/(2x_k ^(2n−2) ))=−(x_k ^2 /2) ((nx^n )/(x^(2n) +1))=−(1/2)Σ_(k=0) ^(2n−1) (x_k ^2 /((x−x_k ))) over R(X) −(1/2)Σ_(k=0) ^(2n−1) (x_k ^2 /((x−x_k )))=−(1/2).{Σ_(k=0) ^(n−1) (x_k ^2 /(x−x_k ))+Σ_(k=n) ^(2n−1) (x_k ^2 /(x−x_k ))} =−(1/2){Σ_(k=0) ^(n−1) [(x_k ^2 /(x−x_k ))+(x_(2n−1−k) ^2 /(x−x_(2n−1−k) ))]}ByE⇒ =−(1/2){Σ_(k=0) ^(n−1) [(x_k ^2 /(x−x_k ))+(x_k ^(−2) /(x−x_k ^− ))]} =−(1/2)Σ_(k=0) ^(n−1) [((x(x_k ^2 +x_k ^(−2) )−x_k ^− x_k ^2 −x_k x_k ^(−2) )/(x^2 −2Re(x_k )x+∣x_k ∣^2 )) =−(1/2)Σ_(k=0) ^(n−1) ((2x(Re(x_k ^2 ))−2Re(x_k ))/(x^2 −2Re(x_k )+1)) =−(1/2)Σ_(k=0) ^(n−1) ((2x(cos(((1+2k)/n)π))−2cos(((1+2k)/(2n))π))/(x^2 −2cos(((1+2k)/(2n))π)+1)) over IR(x) ∫_0 ^(+∞) n(x^n /(x^(2n) +1))dx x=tg^(1/n) (t)⇒dx=(1/n)(1+tg^2 (t))tg^((1/n)−1) (t)dt =∫_0 ^(π/2) tg(t)(((1+tg^2 (t))tg^((1/n)−1) )/(1+tg^2 (t)))dt =∫_0 ^(π/2) tg^(1/n) (t)dt=∫_0 ^(π/2) sin^(1/n) (t)cos^(−(1/n)) (t)dt =(1/2)β(((1+n)/(2n)),((n−1)/(2n)))=(1/2).((Γ(((1+n)/(2n)))Γ(((n−1)/(2n))))/(Γ(((n+1)/(2n))+((n−1)/(2n)))))=((Γ(((n+1)/(2n))).Γ(1−((n+1)/(2n))))/(2Γ(1))) Γ(x)Γ(1−x)=(π/(sin(πx)))⇒Γ(((n+1)/(2n)))Γ(1−((n+1)/(2n)))=(π/(sin(((n+1)/(2n))π))),Γ(1)=1 =(π/(2sin(((n+1)/(2n))π)))=(π/(2cos((π/(2n)))))](https://www.tinkutara.com/question/Q79125.png)
$${over}\:{C}\left({x}\right) \\ $$$${x}^{\mathrm{2}{n}} +\mathrm{1}=\mathrm{0}\Rightarrow{x}^{\mathrm{2}{n}} =−\mathrm{1}={e}^{{i}\pi+\mathrm{2}{k}\pi} \\ $$$${x}_{{k}} ={e}^{\frac{{i}\left(\mathrm{1}+\mathrm{2}{k}\right)\pi}{\mathrm{2}{n}}\:\:} ,{k}\in\left[\mathrm{0},\mathrm{2}{n}−\mathrm{1}\right] \\ $$$${x}_{\mathrm{2}{n}−{k}−\mathrm{1}} ={e}^{{i}\left(\frac{\mathrm{1}+\mathrm{2}\left(\mathrm{2}{n}−{k}−\mathrm{1}\right)}{\mathrm{2}{n}}\right)\pi} ={e}^{−{i}\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{2}{n}}\right)\pi} \\ $$$$\Rightarrow{x}_{{k}} =\overset{−} {{x}}_{\mathrm{2}{n}−\mathrm{1}−{k}} ,{E} \\ $$$$\frac{{nx}^{{n}} }{{x}^{\mathrm{2}{n}} +\mathrm{1}}=\Sigma\frac{{a}_{{k}} }{{x}−{x}_{{k}} } \\ $$$${a}_{{k}} =\frac{{nx}_{{k}} }{\mathrm{2}{nx}_{{k}} ^{\mathrm{2}{n}−\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{2}{x}_{{k}} ^{\mathrm{2}{n}−\mathrm{2}} }=−\frac{{x}_{{k}} ^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{{nx}^{{n}} }{{x}^{\mathrm{2}{n}} +\mathrm{1}}=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\frac{{x}_{{k}} ^{\mathrm{2}} }{\left({x}−{x}_{{k}} \right)} \\ $$$${over}\:{R}\left({X}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\frac{{x}_{{k}} ^{\mathrm{2}} }{\left({x}−{x}_{{k}} \right)}=−\frac{\mathrm{1}}{\mathrm{2}}.\left\{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{x}_{{k}} ^{\mathrm{2}} }{{x}−{x}_{{k}} }+\underset{{k}={n}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\:\frac{{x}_{{k}} ^{\mathrm{2}} }{{x}−{x}_{{k}} }\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left\{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left[\frac{{x}_{{k}} ^{\mathrm{2}} }{{x}−{x}_{{k}} }+\frac{{x}_{\mathrm{2}{n}−\mathrm{1}−{k}} ^{\mathrm{2}} }{{x}−{x}_{\mathrm{2}{n}−\mathrm{1}−{k}} }\right]\right\}{ByE}\Rightarrow \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left\{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left[\frac{{x}_{{k}} ^{\mathrm{2}} }{{x}−{x}_{{k}} }+\frac{\overset{−\mathrm{2}} {{x}}_{{k}} }{{x}−\overset{−} {{x}}_{{k}} }\right]\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left[\frac{{x}\left({x}_{{k}} ^{\mathrm{2}} +\overset{−\mathrm{2}} {{x}}_{{k}} \right)−\overset{−} {{x}}_{{k}} {x}_{{k}} ^{\mathrm{2}} −{x}_{{k}} \overset{−\mathrm{2}} {{x}}_{{k}} }{{x}^{\mathrm{2}} −\mathrm{2}{Re}\left({x}_{{k}} \right){x}+\mid{x}_{{k}} \mid^{\mathrm{2}} }\right. \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{2}{x}\left({Re}\left({x}_{{k}} ^{\mathrm{2}} \right)\right)−\mathrm{2}{Re}\left({x}_{{k}} \right)}{{x}^{\mathrm{2}} −\mathrm{2}{Re}\left({x}_{{k}} \right)+\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{2}{x}\left({cos}\left(\frac{\mathrm{1}+\mathrm{2}{k}}{{n}}\pi\right)\right)−\mathrm{2}{cos}\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{2}{n}}\pi\right)}{{x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{2}{n}}\pi\right)+\mathrm{1}}\:\:{over}\:{IR}\left({x}\right) \\ $$$$\int_{\mathrm{0}} ^{+\infty} {n}\frac{{x}^{{n}} }{{x}^{\mathrm{2}{n}} +\mathrm{1}}{dx} \\ $$$${x}={tg}^{\frac{\mathrm{1}}{{n}}} \left({t}\right)\Rightarrow{dx}=\frac{\mathrm{1}}{{n}}\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({t}\right)\right){tg}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \left({t}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {tg}\left({t}\right)\frac{\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({t}\right)\right){tg}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \:}{\mathrm{1}+{tg}^{\mathrm{2}} \left({t}\right)}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {tg}^{\frac{\mathrm{1}}{{n}}} \left({t}\right){dt}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\frac{\mathrm{1}}{{n}}} \left({t}\right){cos}^{−\frac{\mathrm{1}}{{n}}} \left({t}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\mathrm{1}+{n}}{\mathrm{2}{n}},\frac{{n}−\mathrm{1}}{\mathrm{2}{n}}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\Gamma\left(\frac{\mathrm{1}+{n}}{\mathrm{2}{n}}\right)\Gamma\left(\frac{{n}−\mathrm{1}}{\mathrm{2}{n}}\right)}{\Gamma\left(\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}+\frac{{n}−\mathrm{1}}{\mathrm{2}{n}}\right)}=\frac{\Gamma\left(\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\right).\Gamma\left(\mathrm{1}−\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\right)}{\mathrm{2}\Gamma\left(\mathrm{1}\right)} \\ $$$$\Gamma\left({x}\right)\Gamma\left(\mathrm{1}−{x}\right)=\frac{\pi}{{sin}\left(\pi{x}\right)}\Rightarrow\Gamma\left(\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\right)\Gamma\left(\mathrm{1}−\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\right)=\frac{\pi}{{sin}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\pi\right)},\Gamma\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$ \\ $$$$=\frac{\pi}{\mathrm{2}{sin}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\pi\right)}=\frac{\pi}{\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}{n}}\right)} \\ $$
Commented by mathmax by abdo last updated on 23/Jan/20
$${thank}\:{you}\:{sir}. \\ $$
Commented by mind is power last updated on 23/Jan/20
$${y}'{re}\:{Welcom} \\ $$