....calculus... evaluate i: ∫_0 ^( (π/2)) x(√( tan(x))) dx= ??? ii:∫_0 ^( ∞) ((ln(x))/(1+x^2 +x^4 ))dx =??? m.n.july 1970


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 112797 by mnjuly1970 last updated on 09/Sep/20

$. . . . c a l c u l u s . . .$ $e v a l u a t e$ $i : \int_{0}^{\frac{π}{2}} x \sqrt{t a n (x)} d x = ? ? ?$ $i i : \int_{0}^{\infty} \frac{l n (x)}{1 + x^{2} + x^{4}} d x = ? ? ?$ $m . n . j u l y 1970$
	Answered by mathmax by abdo last updated on 09/Sep/20
	$2) I =∫_0 ^∞ ((lnx)/(x^4 +x^2 +1))dx we use the formulae ∫_0 ^∞ q(x)lnxdx =−(1/2)Re(ΣRes(q(z)ln^2 z) wehave q(x) =(1/(x^4 +x^2 +1)) we considere ϕ(z)=q(z)ln^2 z ⇒ ϕ(z) =((ln^2 z)/(z^4 +z^2 +1)) polesof ϕ? z^4 +z^2 +1 =0⇒t^2 +t +1=0 witht =z^2 Δ =1−4 =−3 ⇒t_1 =((−1+i(√3))/2)=e^((i2π)/3) and t_2 =((−1−i(√3))/2) =e^(−((i2π)/3)) ⇒ z^4 +z^2 +1 =(t−t_1 )(t−t_2 )=(z^2 −e^((i2π)/3) )(z^2 −e^(−((i2π)/3)) ) ⇒ϕ(z) =((ln^2 z)/((z−e^((iπ)/3) )(z+e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z+e^(−((iπ)/3)) ))) ⇒∫_0 ^∞ q(x)lnx dx =−(1/2)Re(Res(ϕ,e^((iπ)/3) )+Res(ϕ,−e^((iπ)/3) )+Res(ϕ,e^(−((iπ)/3)) ) +Res(ϕ,−e^(−((iπ)/3)) )} Res(ϕ,e^((iπ)/3) ) =((ln^2 (e^((iπ)/3) ))/(2e^((iπ)/3) (2i sin(((2π)/3))))) =(((((iπ)/3))^2 )/(4i(((√3)/2)))) e^(−((iπ)/3)) =−(π^2 /9).(e^(−((iπ)/3)) /(2i(√3))) Res(ϕ,−e^((iπ)/3) ) =((ln^2 (−e^((iπ)/3) ))/(−2e^((iπ)/3) (2isin(((2π)/3))))) =(((ln(−1)+((iπ)/3))^2 )/(−4i ((√3)/2))) e^(−((iπ)/3)) =(((iπ+((iπ)/3))^2 )/(−2i(√3))) e^(−((iπ)/3)) =−(((((4π)/3))^2 )/(−2i(√3))) e^(−((iπ)/3)) =((8π^2 )/(9i(√3))) e^(−((iπ)/3)) Res(ϕ,e^(−((iπ)/3)) ) =((ln^2 (e^(−((iπ)/3)) ))/((−2isin(((2π)/3)))(2e^(−((iπ)/3)) ))) =(((((−iπ)/3))^2 )/(−4i(((√3)/2)))) e^((iπ)/3) =(π^2 /9) ×(e^((iπ)/3) /(2i(√3))) Res(ϕ,−e^(−((iπ)/3)) ) =((ln^2 (−e^(−((iπ)/3)) ))/(−2e^(−((iπ)/3)) (−2isin(((2π)/3))))) =(((iπ−((iπ)/3))^2 )/(4i×((√3)/2))) e^((iπ)/3) =−(((((2π)/3))^2 )/(2i(√3))) e^((iπ)/3) =−((2π^2 )/9) .(e^((iπ)/3) /(i(√3))) ⇒ Σ Res(ϕ a_l )=−(π^2 /(18i(√3))) e^(−((iπ)/3)) +((8π^2 )/(9i(√3))) e^(−((iπ)/3)) +(π^2 /(18i(√3))) e^((iπ)/3) −((2π^2 )/(9i(√3))) e^((iπ)/3) =(π^2 /(18i(√3)))(2i sin((π/3))) −((iπ^2 )/(18(√3))){(1/2) +i((√3)/2)}+2i(π^2 /(9(√3))){(1/2)+((i(√3))/2)} =(π^2 /(9(√3)))(((√3)/2)) −((iπ^2 )/(36(√3))) +(π^2 /(36)) +((2iπ^2 )/(18(√3)))−(π^2 /9) ⇒ Re(Σ Res)=(π^2 /(18)) +(π^2 /(36))−(π^2 /9) =((2π^2 +π^2 −4π^2 )/(36)) =−(π^2 /(36)) ⇒ −(1/2)Re(Σ Res(ϕ..)) =(π^2 /(72)) ⇒∫_0 ^∞ ((lnx)/(x^4 +x^2 +1))dx =(π^2 /(72))$
	$2) I = \int_{0}^{\infty} \frac{lnx}{x^{4} + x^{2} + 1} dx we use the formulae$ $\int_{0}^{\infty} q (x) lnxdx = - \frac{1}{2} Re (Σ Res (q (z) \ln^{2} z)$ $wehave q (x) = \frac{1}{x^{4} + x^{2} + 1} we considere φ (z) = q (z) \ln^{2} z \Rightarrow$ $φ (z) = \frac{\ln^{2} z}{z^{4} + z^{2} + 1} polesof φ ?$ $z^{4} + z^{2} + 1 = 0 \Rightarrow t^{2} + t + 1 = 0 witht = z^{2}$ $Δ = 1 - 4 = - 3 \Rightarrow t_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}} and t_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}}$ $\Rightarrow z^{4} + z^{2} + 1 = (t - t_{1}) (t - t_{2}) = (z^{2} - e^{\frac{i 2 π}{3}}) (z^{2} - e^{- \frac{i 2 π}{3}})$ $\Rightarrow φ (z) = \frac{\ln^{2} z}{(z - e^{\frac{i π}{3}}) (z + e^{\frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) (z + e^{- \frac{i π}{3}})}$ $\Rightarrow \int_{0}^{\infty} q (x) lnx dx = - \frac{1}{2} Re (Res (φ, e^{\frac{i π}{3}}) + Res (φ, - e^{\frac{i π}{3}}) + Res (φ, e^{- \frac{i π}{3}})$ $+ Res (φ, - e^{- \frac{i π}{3}})}$ $Res (φ, e^{\frac{i π}{3}}) = \frac{\ln^{2} (e^{\frac{i π}{3}})}{{2 e}^{\frac{i π}{3}} (2 i \sin (\frac{2 π}{3}))} = \frac{{(\frac{i π}{3})}^{2}}{4 i (\frac{\sqrt{3}}{2})} e^{- \frac{i π}{3}} = - \frac{π^{2}}{9} . \frac{e^{- \frac{i π}{3}}}{2 i \sqrt{3}}$ $Res (φ, - e^{\frac{i π}{3}}) = \frac{\ln^{2} (- e^{\frac{i π}{3}})}{- {2 e}^{\frac{i π}{3}} (2 isin (\frac{2 π}{3}))} = \frac{{(\ln (- 1) + \frac{i π}{3})}^{2}}{- 4 i \frac{\sqrt{3}}{2}} e^{- \frac{i π}{3}}$ $= \frac{{(i π + \frac{i π}{3})}^{2}}{- 2 i \sqrt{3}} e^{- \frac{i π}{3}} = - \frac{{(\frac{4 π}{3})}^{2}}{- 2 i \sqrt{3}} e^{- \frac{i π}{3}} = \frac{8 π^{2}}{9 i \sqrt{3}} e^{- \frac{i π}{3}}$ $Res (φ, e^{- \frac{i π}{3}}) = \frac{\ln^{2} (e^{- \frac{i π}{3}})}{(- 2 isin (\frac{2 π}{3})) ({2 e}^{- \frac{i π}{3}})} = \frac{{(\frac{- i π}{3})}^{2}}{- 4 i (\frac{\sqrt{3}}{2})} e^{\frac{i π}{3}}$ $= \frac{π^{2}}{9} \times \frac{e^{\frac{i π}{3}}}{2 i \sqrt{3}}$ $Res (φ, - e^{- \frac{i π}{3}}) = \frac{\ln^{2} (- e^{- \frac{i π}{3}})}{- {2 e}^{- \frac{i π}{3}} (- 2 isin (\frac{2 π}{3}))} = \frac{{(i π - \frac{i π}{3})}^{2}}{4 i \times \frac{\sqrt{3}}{2}} e^{\frac{i π}{3}}$ $= - \frac{{(\frac{2 π}{3})}^{2}}{2 i \sqrt{3}} e^{\frac{i π}{3}} = - \frac{2 π^{2}}{9} . \frac{e^{\frac{i π}{3}}}{i \sqrt{3}} \Rightarrow$ $Σ Res (φ a_{l}) = - \frac{π^{2}}{18 i \sqrt{3}} e^{- \frac{i π}{3}} + \frac{8 π^{2}}{9 i \sqrt{3}} e^{- \frac{i π}{3}} + \frac{π^{2}}{18 i \sqrt{3}} e^{\frac{i π}{3}} - \frac{2 π^{2}}{9 i \sqrt{3}} e^{\frac{i π}{3}}$ $= \frac{π^{2}}{18 i \sqrt{3}} (2 i \sin (\frac{π}{3})) - \frac{i π^{2}}{18 \sqrt{3}} {\frac{1}{2} + i \frac{\sqrt{3}}{2}} + 2 i \frac{π^{2}}{9 \sqrt{3}} {\frac{1}{2} + \frac{i \sqrt{3}}{2}}$ $= \frac{π^{2}}{9 \sqrt{3}} (\frac{\sqrt{3}}{2}) - \frac{i π^{2}}{36 \sqrt{3}} + \frac{π^{2}}{36} + \frac{2 i π^{2}}{18 \sqrt{3}} - \frac{π^{2}}{9} \Rightarrow$ $Re (Σ Res) = \frac{π^{2}}{18} + \frac{π^{2}}{36} - \frac{π^{2}}{9} = \frac{2 π^{2} + π^{2} - 4 π^{2}}{36} = - \frac{π^{2}}{36} \Rightarrow$ $- \frac{1}{2} Re (Σ Res (φ . .)) = \frac{π^{2}}{72} \Rightarrow \int_{0}^{\infty} \frac{lnx}{x^{4} + x^{2} + 1} dx = \frac{π^{2}}{72}$
	Commented by mnjuly1970 last updated on 10/Sep/20

	$t h a n k y o u v e r y m u c h m r$ $m a x . g r a t e f u l . .$
	Answered by mathmax by abdo last updated on 09/Sep/20
	$1) A =∫_0 ^(π/2) x(√(tanx))dx we do the changement (√(tanx))=t ⇒tanx =t^2 ⇒x =arctan(t^2 ) ⇒A =∫_0 ^∞ t ((arctan(t^2 ))/(1+t^4 )) (2t)dt =∫_0 ^∞ (t^2 /(1+t^4 )) arctan(t^2 )dt =(1/2)∫_(−∞) ^(+∞) ((t^2 arctan(t^2 ))/(t^4 +1)) dt let w(z) =((z^2 arctan(z^2 ))/(z^4 +1)) poles of w! W(z) =((z^2 arctan(z^2 ))/((z^2 −i)(z^2 +i))) =((z^2 arctan(z^2 ))/((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) residus theorem give ∫_(−∞) ^(+∞) w(z)dz =2iπ{ Res(w,e^((iπ)/4) )+Res(w,−e^(−((iπ)/4)) )} Res(w,e^((iπ)/4) ) =((i arctan(i))/(2e^((iπ)/4) (2i))) =(1/4) arctan(i)e^(−((iπ)/4)) Res(w,−e^(−((iπ)/4)) ) =((−i arctan(−i))/((−2i)(−2e^(−((iπ)/4)) ))) =(1/4) arctan(i)e^((iπ)/4) ⇒ ∫_(−∞) ^(+∞) w(z)dz =2iπ{(1/4) arctan(i)e^(−((iπ)/4)) +(1/4) arctani e^((iπ)/4) } =((iπ)/2) arctan(i)(2cos((π/4))) =iπ arctan(i)×((√2)/2) =((iπ(√2))/2) arctan(i) =2A ⇒ A =((iπ(√2))/4) arctan(i) rest to find arctan(i)!.....be continued....$
	$1) A = \int_{0}^{\frac{π}{2}} x \sqrt{tanx} dx we do the changement \sqrt{tanx} = t \Rightarrow tanx = t^{2}$ $\Rightarrow x = \arctan (t^{2}) \Rightarrow A = \int_{0}^{\infty} t \frac{\arctan (t^{2})}{1 + t^{4}} (2 t) dt$ $= \int_{0}^{\infty} \frac{t^{2}}{1 + t^{4}} \arctan (t^{2}) dt = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{t^{2} \arctan (t^{2})}{t^{4} + 1} dt let$ $w (z) = \frac{z^{2} \arctan (z^{2})}{z^{4} + 1} poles of w!$ $W (z) = \frac{z^{2} \arctan (z^{2})}{(z^{2} - i) (z^{2} + i)} = \frac{z^{2} \arctan (z^{2})}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})}$ $residus theorem give$ $\int_{- \infty}^{+ \infty} w (z) dz = 2 i π {Res (w, e^{\frac{i π}{4}}) + Res (w, - e^{- \frac{i π}{4}})}$ $Res (w, e^{\frac{i π}{4}}) = \frac{i \arctan (i)}{{2 e}^{\frac{i π}{4}} (2 i)} = \frac{1}{4} \arctan (i) e^{- \frac{i π}{4}}$ $Res (w, - e^{- \frac{i π}{4}}) = \frac{- i \arctan (- i)}{(- 2 i) (- {2 e}^{- \frac{i π}{4}})} = \frac{1}{4} \arctan (i) e^{\frac{i π}{4}} \Rightarrow$ $\int_{- \infty}^{+ \infty} w (z) dz = 2 i π {\frac{1}{4} \arctan (i) e^{- \frac{i π}{4}} + \frac{1}{4} arctani e^{\frac{i π}{4}}}$ $= \frac{i π}{2} \arctan (i) (2 \cos (\frac{π}{4})) = i π \arctan (i) \times \frac{\sqrt{2}}{2}$ $= \frac{i π \sqrt{2}}{2} \arctan (i) = 2 A \Rightarrow A = \frac{i π \sqrt{2}}{4} \arctan (i)$ $rest to find \arctan (i)! . . . . . be continued . . . .$
	Commented by mathmax by abdo last updated on 10/Sep/20

	$sorry A = \int_{- \infty}^{+ \infty} \frac{t^{2} \arctan (t^{2})}{t^{4} + 1} dt \Rightarrow A = \frac{i π \sqrt{2}}{2} \arctan (i)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum