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Question Number 112797 by mnjuly1970 last updated on 09/Sep/20
....calculus... evaluate i: ∫_0 ^( (π/2)) x(√( tan(x))) dx= ??? ii:∫_0 ^( ∞) ((ln(x))/(1+x^2 +x^4 ))dx =??? m.n.july 1970
$$\:\:\:\:\:\:\:\:\:….{calculus}… \\ $$$$\:\:\:{evaluate} \\ $$$$ \\ $$$${i}:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {x}\sqrt{\:{tan}\left({x}\right)}\:{dx}=\:???\: \\ $$$${ii}:\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx}\:=???\: \\ $$$$\:\:\:{m}.{n}.{july}\:\mathrm{1970} \\ $$
Answered by mathmax by abdo last updated on 09/Sep/20
2) I =∫_0 ^∞ ((lnx)/(x^4 +x^2 +1))dx we use the formulae ∫_0 ^∞ q(x)lnxdx =−(1/2)Re(ΣRes(q(z)ln^2 z) wehave q(x) =(1/(x^4 +x^2 +1)) we considere ϕ(z)=q(z)ln^2 z ⇒ ϕ(z) =((ln^2 z)/(z^4 +z^2 +1)) polesof ϕ? z^4 +z^2 +1 =0⇒t^2 +t +1=0 witht =z^2 Δ =1−4 =−3 ⇒t_1 =((−1+i(√3))/2)=e^((i2π)/3) and t_2 =((−1−i(√3))/2) =e^(−((i2π)/3)) ⇒ z^4 +z^2 +1 =(t−t_1 )(t−t_2 )=(z^2 −e^((i2π)/3) )(z^2 −e^(−((i2π)/3)) ) ⇒ϕ(z) =((ln^2 z)/((z−e^((iπ)/3) )(z+e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z+e^(−((iπ)/3)) ))) ⇒∫_0 ^∞ q(x)lnx dx =−(1/2)Re(Res(ϕ,e^((iπ)/3) )+Res(ϕ,−e^((iπ)/3) )+Res(ϕ,e^(−((iπ)/3)) ) +Res(ϕ,−e^(−((iπ)/3)) )} Res(ϕ,e^((iπ)/3) ) =((ln^2 (e^((iπ)/3) ))/(2e^((iπ)/3) (2i sin(((2π)/3))))) =(((((iπ)/3))^2 )/(4i(((√3)/2)))) e^(−((iπ)/3)) =−(π^2 /9).(e^(−((iπ)/3)) /(2i(√3))) Res(ϕ,−e^((iπ)/3) ) =((ln^2 (−e^((iπ)/3) ))/(−2e^((iπ)/3) (2isin(((2π)/3))))) =(((ln(−1)+((iπ)/3))^2 )/(−4i ((√3)/2))) e^(−((iπ)/3)) =(((iπ+((iπ)/3))^2 )/(−2i(√3))) e^(−((iπ)/3)) =−(((((4π)/3))^2 )/(−2i(√3))) e^(−((iπ)/3)) =((8π^2 )/(9i(√3))) e^(−((iπ)/3)) Res(ϕ,e^(−((iπ)/3)) ) =((ln^2 (e^(−((iπ)/3)) ))/((−2isin(((2π)/3)))(2e^(−((iπ)/3)) ))) =(((((−iπ)/3))^2 )/(−4i(((√3)/2)))) e^((iπ)/3) =(π^2 /9) ×(e^((iπ)/3) /(2i(√3))) Res(ϕ,−e^(−((iπ)/3)) ) =((ln^2 (−e^(−((iπ)/3)) ))/(−2e^(−((iπ)/3)) (−2isin(((2π)/3))))) =(((iπ−((iπ)/3))^2 )/(4i×((√3)/2))) e^((iπ)/3) =−(((((2π)/3))^2 )/(2i(√3))) e^((iπ)/3) =−((2π^2 )/9) .(e^((iπ)/3) /(i(√3))) ⇒ Σ Res(ϕ a_l )=−(π^2 /(18i(√3))) e^(−((iπ)/3)) +((8π^2 )/(9i(√3))) e^(−((iπ)/3)) +(π^2 /(18i(√3))) e^((iπ)/3) −((2π^2 )/(9i(√3))) e^((iπ)/3) =(π^2 /(18i(√3)))(2i sin((π/3))) −((iπ^2 )/(18(√3))){(1/2) +i((√3)/2)}+2i(π^2 /(9(√3))){(1/2)+((i(√3))/2)} =(π^2 /(9(√3)))(((√3)/2)) −((iπ^2 )/(36(√3))) +(π^2 /(36)) +((2iπ^2 )/(18(√3)))−(π^2 /9) ⇒ Re(Σ Res)=(π^2 /(18)) +(π^2 /(36))−(π^2 /9) =((2π^2 +π^2 −4π^2 )/(36)) =−(π^2 /(36)) ⇒ −(1/2)Re(Σ Res(ϕ..)) =(π^2 /(72)) ⇒∫_0 ^∞ ((lnx)/(x^4 +x^2 +1))dx =(π^2 /(72))
$$\left.\mathrm{2}\right)\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnx}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:\:\mathrm{we}\:\mathrm{use}\:\mathrm{the}\:\mathrm{formulae} \\ $$$$\int_{\mathrm{0}} ^{\infty} \mathrm{q}\left(\mathrm{x}\right)\mathrm{lnxdx}\:=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Re}\left(\Sigma\mathrm{Res}\left(\mathrm{q}\left(\mathrm{z}\right)\mathrm{ln}^{\mathrm{2}} \mathrm{z}\right)\right. \\ $$$$\mathrm{wehave}\:\mathrm{q}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\:\mathrm{we}\:\mathrm{considere}\:\varphi\left(\mathrm{z}\right)=\mathrm{q}\left(\mathrm{z}\right)\mathrm{ln}^{\mathrm{2}} \mathrm{z}\:\Rightarrow \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{z}}{\mathrm{z}^{\mathrm{4}} \:+\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\:\:\mathrm{polesof}\:\varphi? \\ $$$$\mathrm{z}^{\mathrm{4}} \:+\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\:=\mathrm{0}\Rightarrow\mathrm{t}^{\mathrm{2}} \:+\mathrm{t}\:+\mathrm{1}=\mathrm{0}\:\mathrm{witht}\:=\mathrm{z}^{\mathrm{2}} \\ $$$$\Delta\:=\mathrm{1}−\mathrm{4}\:=−\mathrm{3}\:\Rightarrow\mathrm{t}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \:\mathrm{and}\:\mathrm{t}_{\mathrm{2}} \:=\frac{−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \\ $$$$\Rightarrow\:\mathrm{z}^{\mathrm{4}} \:+\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\:=\left(\mathrm{t}−\mathrm{t}_{\mathrm{1}} \right)\left(\mathrm{t}−\mathrm{t}_{\mathrm{2}} \right)=\left(\mathrm{z}^{\mathrm{2}} −\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}^{\mathrm{2}} \:−\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right) \\ $$$$\Rightarrow\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{z}}{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \mathrm{q}\left(\mathrm{x}\right)\mathrm{lnx}\:\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Re}\left(\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)+\mathrm{Res}\left(\varphi,−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)+\mathrm{Res}\left(\varphi,\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\right. \\ $$$$\left.+\mathrm{Res}\left(\varphi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\right\} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\:=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)}{\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \left(\mathrm{2i}\:\mathrm{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)}\:=\frac{\left(\frac{\mathrm{i}\pi}{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{4i}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}\:\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:=−\frac{\pi^{\mathrm{2}} }{\mathrm{9}}.\frac{\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{2i}\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Res}\left(\varphi,−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\:=\frac{\mathrm{ln}^{\mathrm{2}} \left(−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)}{−\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \left(\mathrm{2isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)}\:=\frac{\left(\mathrm{ln}\left(−\mathrm{1}\right)+\frac{\mathrm{i}\pi}{\mathrm{3}}\right)^{\mathrm{2}} }{−\mathrm{4i}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \\ $$$$=\frac{\left(\mathrm{i}\pi+\frac{\mathrm{i}\pi}{\mathrm{3}}\right)^{\mathrm{2}} }{−\mathrm{2i}\sqrt{\mathrm{3}}}\:\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:=−\frac{\left(\frac{\mathrm{4}\pi}{\mathrm{3}}\right)^{\mathrm{2}} }{−\mathrm{2i}\sqrt{\mathrm{3}}}\:\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:=\frac{\mathrm{8}\pi^{\mathrm{2}} }{\mathrm{9i}\sqrt{\mathrm{3}}}\:\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\:=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)}{\left(−\mathrm{2isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)\left(\mathrm{2e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)}\:=\frac{\left(\frac{−\mathrm{i}\pi}{\mathrm{3}}\right)^{\mathrm{2}} }{−\mathrm{4i}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{9}}\:×\frac{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{2i}\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Res}\left(\varphi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\:=\frac{\mathrm{ln}^{\mathrm{2}} \left(−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)}{−\mathrm{2e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \left(−\mathrm{2isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)}\:=\frac{\left(\mathrm{i}\pi−\frac{\mathrm{i}\pi}{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{4i}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \\ $$$$=−\frac{\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{2i}\sqrt{\mathrm{3}}}\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \:=−\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{9}}\:.\frac{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{i}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$$\Sigma\:\mathrm{Res}\left(\varphi\:\mathrm{a}_{\mathrm{l}} \right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{18i}\sqrt{\mathrm{3}}}\:\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:+\frac{\mathrm{8}\pi^{\mathrm{2}} }{\mathrm{9i}\sqrt{\mathrm{3}}}\:\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:+\frac{\pi^{\mathrm{2}} }{\mathrm{18i}\sqrt{\mathrm{3}}}\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \:−\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{9i}\sqrt{\mathrm{3}}}\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{18i}\sqrt{\mathrm{3}}}\left(\mathrm{2i}\:\mathrm{sin}\left(\frac{\pi}{\mathrm{3}}\right)\right)\:−\frac{\mathrm{i}\pi^{\mathrm{2}} }{\mathrm{18}\sqrt{\mathrm{3}}}\left\{\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right\}+\mathrm{2i}\frac{\pi^{\mathrm{2}} }{\mathrm{9}\sqrt{\mathrm{3}}}\left\{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right\} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{9}\sqrt{\mathrm{3}}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:−\frac{\mathrm{i}\pi^{\mathrm{2}} }{\mathrm{36}\sqrt{\mathrm{3}}}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{36}}\:+\frac{\mathrm{2i}\pi^{\mathrm{2}} }{\mathrm{18}\sqrt{\mathrm{3}}}−\frac{\pi^{\mathrm{2}} }{\mathrm{9}}\:\Rightarrow \\ $$$$\mathrm{Re}\left(\Sigma\:\mathrm{Res}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{18}}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{36}}−\frac{\pi^{\mathrm{2}} }{\mathrm{9}}\:=\frac{\mathrm{2}\pi^{\mathrm{2}} +\pi^{\mathrm{2}} −\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{36}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{36}}\:\Rightarrow \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Re}\left(\Sigma\:\mathrm{Res}\left(\varphi..\right)\right)\:=\frac{\pi^{\mathrm{2}} }{\mathrm{72}}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{lnx}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{72}} \\ $$
Commented by mnjuly1970 last updated on 10/Sep/20
thank you very much mr max.grateful..
$${thank}\:{you}\:{very}\:{much}\:{mr} \\ $$$${max}.{grateful}.. \\ $$
Answered by mathmax by abdo last updated on 09/Sep/20
1) A =∫_0 ^(π/2) x(√(tanx))dx we do the changement (√(tanx))=t ⇒tanx =t^2 ⇒x =arctan(t^2 ) ⇒A =∫_0 ^∞ t ((arctan(t^2 ))/(1+t^4 )) (2t)dt =∫_0 ^∞ (t^2 /(1+t^4 )) arctan(t^2 )dt =(1/2)∫_(−∞) ^(+∞) ((t^2 arctan(t^2 ))/(t^4 +1)) dt let w(z) =((z^2 arctan(z^2 ))/(z^4 +1)) poles of w! W(z) =((z^2 arctan(z^2 ))/((z^2 −i)(z^2 +i))) =((z^2 arctan(z^2 ))/((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) residus theorem give ∫_(−∞) ^(+∞) w(z)dz =2iπ{ Res(w,e^((iπ)/4) )+Res(w,−e^(−((iπ)/4)) )} Res(w,e^((iπ)/4) ) =((i arctan(i))/(2e^((iπ)/4) (2i))) =(1/4) arctan(i)e^(−((iπ)/4)) Res(w,−e^(−((iπ)/4)) ) =((−i arctan(−i))/((−2i)(−2e^(−((iπ)/4)) ))) =(1/4) arctan(i)e^((iπ)/4) ⇒ ∫_(−∞) ^(+∞) w(z)dz =2iπ{(1/4) arctan(i)e^(−((iπ)/4)) +(1/4) arctani e^((iπ)/4) } =((iπ)/2) arctan(i)(2cos((π/4))) =iπ arctan(i)×((√2)/2) =((iπ(√2))/2) arctan(i) =2A ⇒ A =((iπ(√2))/4) arctan(i) rest to find arctan(i)!.....be continued....
$$\left.\mathrm{1}\right)\:\mathrm{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{x}\sqrt{\mathrm{tanx}}\mathrm{dx}\:\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\sqrt{\mathrm{tanx}}=\mathrm{t}\:\Rightarrow\mathrm{tanx}\:=\mathrm{t}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}\:=\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\:\Rightarrow\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \mathrm{t}\:\:\frac{\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\:\left(\mathrm{2t}\right)\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\:\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{t}^{\mathrm{2}} \:\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{t}^{\mathrm{4}} \:+\mathrm{1}}\:\mathrm{dt}\:\mathrm{let} \\ $$$$\mathrm{w}\left(\mathrm{z}\right)\:=\frac{\mathrm{z}^{\mathrm{2}} \:\mathrm{arctan}\left(\mathrm{z}^{\mathrm{2}} \right)}{\mathrm{z}^{\mathrm{4}} \:+\mathrm{1}}\:\:\mathrm{poles}\:\mathrm{of}\:\mathrm{w}! \\ $$$$\mathrm{W}\left(\mathrm{z}\right)\:=\frac{\mathrm{z}^{\mathrm{2}} \:\mathrm{arctan}\left(\mathrm{z}^{\mathrm{2}} \right)}{\left(\mathrm{z}^{\mathrm{2}} −\mathrm{i}\right)\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{i}\right)}\:=\frac{\mathrm{z}^{\mathrm{2}} \:\mathrm{arctan}\left(\mathrm{z}^{\mathrm{2}} \right)}{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)} \\ $$$$\mathrm{residus}\:\mathrm{theorem}\:\mathrm{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\mathrm{w}\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\:\mathrm{Res}\left(\mathrm{w},\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)+\mathrm{Res}\left(\mathrm{w},−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$$\mathrm{Res}\left(\mathrm{w},\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:=\frac{\mathrm{i}\:\mathrm{arctan}\left(\mathrm{i}\right)}{\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(\mathrm{2i}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{arctan}\left(\mathrm{i}\right)\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \\ $$$$\mathrm{Res}\left(\mathrm{w},−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:=\frac{−\mathrm{i}\:\mathrm{arctan}\left(−\mathrm{i}\right)}{\left(−\mathrm{2i}\right)\left(−\mathrm{2e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{arctan}\left(\mathrm{i}\right)\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\mathrm{w}\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{arctan}\left(\mathrm{i}\right)\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \:+\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{arctani}\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right\} \\ $$$$=\frac{\mathrm{i}\pi}{\mathrm{2}}\:\mathrm{arctan}\left(\mathrm{i}\right)\left(\mathrm{2cos}\left(\frac{\pi}{\mathrm{4}}\right)\right)\:=\mathrm{i}\pi\:\mathrm{arctan}\left(\mathrm{i}\right)×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{i}\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{arctan}\left(\mathrm{i}\right)\:=\mathrm{2A}\:\Rightarrow\:\mathrm{A}\:=\frac{\mathrm{i}\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:\mathrm{arctan}\left(\mathrm{i}\right) \\ $$$$\mathrm{rest}\:\mathrm{to}\:\mathrm{find}\:\mathrm{arctan}\left(\mathrm{i}\right)!…..\mathrm{be}\:\mathrm{continued}…. \\ $$
Commented by mathmax by abdo last updated on 10/Sep/20
sorry A =∫_(−∞) ^(+∞) ((t^2 arctan(t^2 ))/(t^4 +1)) dt ⇒ A =((iπ(√2))/2) arctan(i)
$$\mathrm{sorry}\:\mathrm{A}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{t}^{\mathrm{2}} \:\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{t}^{\mathrm{4}} \:+\mathrm{1}}\:\mathrm{dt}\:\Rightarrow\:\mathrm{A}\:=\frac{\mathrm{i}\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{arctan}\left(\mathrm{i}\right) \\ $$

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