Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 161899 by HongKing last updated on 23/Dec/21

Find: 𝛀 =∫_( -∞) ^( ∞) (1/((1 + x^(2n) )^2 )) dx ; n∈Z

$$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:=\underset{\:-\infty} {\overset{\:\infty} {\int}}\frac{\mathrm{1}}{\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}\boldsymbol{\mathrm{n}}} \right)^{\mathrm{2}} }\:\mathrm{dx}\:\:;\:\:\mathrm{n}\in\mathbb{Z} \\ $$

Answered by Ar Brandon last updated on 25/Dec/21

Ω=∫_(−∞) ^∞ (dx/((1+x^(2n) )^2 ))=2∫_0 ^∞ (dx/((1+x^(2n) )^2 ))=(1/n)∫_0 ^∞ (u^((1/(2n))−1) /((1+u)^2 ))du =(1/n)β((1/(2n)), 2−(1/(2n)))=(1/n)∙Γ((1/(2n)))Γ(2−(1/(2n))) =(1/n)(1−(1/(2n)))Γ((1/(2n)))Γ(1−(1/(2n))) =((2n−1)/(2n^2 ))∙(π/(sin((π/(2n))))) n≠(1/(2(k+1))), k∈Z\{−1}

$$\Omega=\int_{−\infty} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}{n}} \right)^{\mathrm{2}} }=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}{n}} \right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}{n}}−\mathrm{1}} }{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }{du} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{{n}}\beta\left(\frac{\mathrm{1}}{\mathrm{2}{n}},\:\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right)=\frac{\mathrm{1}}{{n}}\centerdot\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\Gamma\left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{{n}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }\centerdot\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)} \\ $$$$\:\:\:\:\:{n}\neq\frac{\mathrm{1}}{\mathrm{2}\left({k}+\mathrm{1}\right)},\:{k}\in\mathbb{Z}\backslash\left\{−\mathrm{1}\right\} \\ $$

Answered by mathmax by abdo last updated on 25/Dec/21

Υ=∫_(−∞) ^(+∞) (dx/((x^(2n) +1)^2 )) let f(a)=∫_(−∞) ^(+∞) (dx/(x^(2n) +a^(2n) )) (a>0) ⇒ f^′ (a)=−2n∫_(−∞) ^(+∞) (a^(2n−1) /((x^(2n) +a^(2n) )^2 ))dx ⇒f^′ (1)=−2n∫_(−∞) ^(+∞) (dx/((x^(2n) +1)^2 )) ⇒∫_(−∞) ^(+∞) (dx/((x^(2n) +1)^2 ))=−(1/(2n))f^′ (1) on f(a)=_(x=at) ∫_(−∞) ^(+∞) ((adt)/(a^(2n) t^(2n) +a^(2n) )) =(2/a^(2n−1) )∫_0 ^(+∞) (dt/(t^(2n) +1)) =_(t=z^(1/(2n)) ) (2/a^(2n−1) ) ∫_0 ^∞ (1/(z+1))(1/(2n))z^((1/(2n))−1) dz =(1/(na^(2n−1) ))∫_0 ^∞ (z^((1/(2n))−1) /(z+1))=(1/(na^(2n−1) )).(π/(sin((π/(2n)))))=(π/(nsin((π/(2n)))))a^(−2n+1) ⇒f^′ (a) =(−2n+1).(π/(nsin((π/(2n)))))a^(−2n) ⇒ f^′ (1) =((π(−2n+1))/(nsin((π/(2n))))) ⇒ ∫_(−∞) ^(+∞) (dx/((x^(2n) +1)^2 ))=−(1/(2n))×((π(−2n+1))/(nsin((π/(2n))))) =((π(2n−1))/(2n^2 sin((π/(2n)))))

$$\Upsilon=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2n}} +\mathrm{1}\right)^{\mathrm{2}} }\:\:\mathrm{let}\:\:\mathrm{f}\left(\mathrm{a}\right)=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2n}} \:+\mathrm{a}^{\mathrm{2n}} }\:\:\:\:\left(\mathrm{a}>\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)=−\mathrm{2n}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{a}^{\mathrm{2n}−\mathrm{1}} }{\left(\mathrm{x}^{\mathrm{2n}} \:+\mathrm{a}^{\mathrm{2n}} \right)^{\mathrm{2}} }\mathrm{dx}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{1}\right)=−\mathrm{2n}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2n}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2n}} +\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{2n}}\mathrm{f}^{'} \left(\mathrm{1}\right) \\ $$$$\mathrm{on}\:\mathrm{f}\left(\mathrm{a}\right)=_{\mathrm{x}=\mathrm{at}} \:\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{adt}}{\mathrm{a}^{\mathrm{2n}} \mathrm{t}^{\mathrm{2n}} \:+\mathrm{a}^{\mathrm{2n}} }\:=\frac{\mathrm{2}}{\mathrm{a}^{\mathrm{2n}−\mathrm{1}} }\int_{\mathrm{0}} ^{+\infty} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2n}} +\mathrm{1}} \\ $$$$=_{\mathrm{t}=\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{2n}}} } \:\:\:\frac{\mathrm{2}}{\mathrm{a}^{\mathrm{2n}−\mathrm{1}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{z}+\mathrm{1}}\frac{\mathrm{1}}{\mathrm{2n}}\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{2n}}−\mathrm{1}} \mathrm{dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{na}^{\mathrm{2n}−\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{2n}}−\mathrm{1}} }{\mathrm{z}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{na}^{\mathrm{2n}−\mathrm{1}} }.\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{2n}}\right)}=\frac{\pi}{\mathrm{nsin}\left(\frac{\pi}{\mathrm{2n}}\right)}\mathrm{a}^{−\mathrm{2n}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\left(−\mathrm{2n}+\mathrm{1}\right).\frac{\pi}{\mathrm{nsin}\left(\frac{\pi}{\mathrm{2n}}\right)}\mathrm{a}^{−\mathrm{2n}} \:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{1}\right)\:=\frac{\pi\left(−\mathrm{2n}+\mathrm{1}\right)}{\mathrm{nsin}\left(\frac{\pi}{\mathrm{2n}}\right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2n}} +\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{2n}}×\frac{\pi\left(−\mathrm{2n}+\mathrm{1}\right)}{\mathrm{nsin}\left(\frac{\pi}{\mathrm{2n}}\right)} \\ $$$$=\frac{\pi\left(\mathrm{2n}−\mathrm{1}\right)}{\mathrm{2n}^{\mathrm{2}} \mathrm{sin}\left(\frac{\pi}{\mathrm{2n}}\right)} \\ $$

Commented by HongKing last updated on 25/Dec/21

thank you so much cool my dear Sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com