Math Question and Answers


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 108506 by mnjuly1970 last updated on 17/Aug/20

	Answered by mathmax by abdo last updated on 17/Aug/20
	$A =∫_0 ^∞ ((ln(x^2 +1))/(x^2 (x^2 +3)))dx ⇒A =_(x=(√3)t) ∫_0 ^∞ ((ln(1+3t^2 ))/(3t^2 (3t^2 +3)))(√3)dt ⇒9A =(√3)∫_0 ^∞ ((ln(3t^2 +1))/(t^2 (t^2 +1)))dt =(√3){∫_0 ^∞ ((1/t^2 )−(1/(t^2 +1)))ln(3t^2 +1)dt} =(√3)∫_0 ^∞ ((ln(3t^2 +1))/t^2 )dt −(√3)∫_0 ^∞ ((ln(3t^2 +1))/(t^2 +1))dt =(√3)H−(√3)K H = ∫_0 ^∞ ((ln(3t^2 +1))/t^2 ) dt =[−(1/t)ln(1+3t^2 )]_0 ^∞ +∫_0 ^∞ ((6t)/(t(1+3t^2 )))dt =6 ∫_0 ^∞ (dt/(1+3t^2 )) =_((√3)t =u) 6 ∫_0 ^∞ (du/((√3)(1+u^2 ))) =(6/(√3))×(π/2) =((3π)/(√3)) =π(√3) let f(a) =∫_0 ^∞ ((ln(a+3t^2 ))/(t^2 +1))dt with a>0 we have f(1) =K f^′ (a) =∫_0 ^∞ (1/((a+3t^2 )(t^2 +1)))dt let decompose F(t) =(1/((3t^2 +a)(t^2 +1))) ⇒F(t) =((αt +β)/(3t^2 +a)) +((et +f)/(t^2 +1)) F(−t)=F(t) ⇒((−αt +β)/(3t^2 +a)) +((−et +f)/(t^2 +1)) =F(t) ⇒α=e =0 ⇒ F(t) =(β/(3t^2 +a)) +(f/(t^2 +1)) ⇒βt^2 +β +3ft^2 +af =1 ⇒ (β+3f)t^2 +β +af =1 ⇒ { ((β =−3f)),((−3f+af =1)) :} ⇒ { ((f =(1/(a−3)) ⇒)),((β =((−3)/(a−3)))) :} F(t) =((−3)/((a−3)(3t^2 +a))) +(1/((a−3)(t^2 +1))) ⇒ f^′ (a) =((−3)/(a−3))∫_0 ^∞ (dt/(3t^2 +a)) +(1/(a−3))∫_0 ^∞ (dt/(t^2 +1)) but ∫_0 ^∞ (dt/(3t^2 +a)) =(1/3)∫_0 ^∞ (dt/(t^2 +(a/3))) =_(t =(√(a/3))u) (1/3).(3/a)∫_0 ^∞ (1/(u^2 +1))((√a)/(√3))du =(1/(√(3a)))×(π/2) ⇒f^′ (a) =((−3)/(a−3)).(π/(2(√3)(√a))) +(π/(2(a−3))) =−((π(√3))/2).(1/((a−3)(√a))) +(π/(2(a−3))) ⇒f(a) =−((π(√3))/2) ∫ (da/((a−3)(√a))) +(π/2)ln∣a−3∣ +c ∫ (da/((√a)(a−3))) =_((√a)=z) ∫ ((2zdz)/(z(z^2 −3))) =∫ ((2dz)/((z−(√3))(z+(√3)))) =(1/(√3))∫ ((1/(z−(√3)))−(1/(z+(√3))))dz =(1/(√3))ln∣(((√a)−(√3))/((√a)+(√3)))∣ ⇒ f(a) =−(π/2)ln∣(((√a)−(√3))/((√a)+(√3)))∣ +(π/2)ln∣a−3∣ +c we have f(0) =(π/2)ln(3)+c =∫_0 ^∞ ((ln(3t^2 ))/(1+t^2 )) dt =ln(3)(π/2) +2 ∫_0 ^∞ ((lnt)/(1+t^2 )) dt(→0) =(π/2)ln3 ⇒c=0 ⇒ f(a) =(π/2)ln∣a−3∣−(π/2)ln∣((a−3)/(((√a)+(√3))^2 ))∣ =(π/2)ln∣a−3∣−(π/2)ln∣a−3∣ +(π/2)ln(((√a)+(√3))^2 ) =πln((√a)+(√3)) ⇒K =f(1) =πln(1+(√3)) 9A =(√3)H −(√3)K =3π −(√3)π ln(1+(√3)) ⇒ A =(1/9)(3π −π(√3)ln(1+(√3)))$
	$A = \int_{0}^{\infty} \frac{\ln (x^{2} + 1)}{x^{2} (x^{2} + 3)} dx \Rightarrow A =_{x = \sqrt{3} t} \int_{0}^{\infty} \frac{\ln (1 + {3 t}^{2})}{{3 t}^{2} ({3 t}^{2} + 3)} \sqrt{3} dt$ $\Rightarrow 9 A = \sqrt{3} \int_{0}^{\infty} \frac{\ln ({3 t}^{2} + 1)}{t^{2} (t^{2} + 1)} dt = \sqrt{3} {\int_{0}^{\infty} (\frac{1}{t^{2}} - \frac{1}{t^{2} + 1}) \ln ({3 t}^{2} + 1) dt}$ $= \sqrt{3} \int_{0}^{\infty} \frac{\ln ({3 t}^{2} + 1)}{t^{2}} dt - \sqrt{3} \int_{0}^{\infty} \frac{\ln ({3 t}^{2} + 1)}{t^{2} + 1} dt = \sqrt{3} H - \sqrt{3} K$ $H = \int_{0}^{\infty} \frac{\ln ({3 t}^{2} + 1)}{t^{2}} dt = {[- \frac{1}{t} \ln (1 + {3 t}^{2})]}_{0}^{\infty} + \int_{0}^{\infty} \frac{6 t}{t (1 + {3 t}^{2})} dt$ $= 6 \int_{0}^{\infty} \frac{dt}{1 + {3 t}^{2}} =_{\sqrt{3} t = u} 6 \int_{0}^{\infty} \frac{du}{\sqrt{3} (1 + u^{2})} = \frac{6}{\sqrt{3}} \times \frac{π}{2} = \frac{3 π}{\sqrt{3}} = π \sqrt{3}$ $let f (a) = \int_{0}^{\infty} \frac{\ln (a + {3 t}^{2})}{t^{2} + 1} dt with a > 0 we have f (1) = K$ $f^{^{'}} (a) = \int_{0}^{\infty} \frac{1}{(a + {3 t}^{2}) (t^{2} + 1)} dt let decompose$ $F (t) = \frac{1}{({3 t}^{2} + a) (t^{2} + 1)} \Rightarrow F (t) = \frac{α t + β}{{3 t}^{2} + a} + \frac{et + f}{t^{2} + 1}$ $F (- t) = F (t) \Rightarrow \frac{- α t + β}{{3 t}^{2} + a} + \frac{- et + f}{t^{2} + 1} = F (t) \Rightarrow α = e = 0 \Rightarrow$ $F (t) = \frac{β}{{3 t}^{2} + a} + \frac{f}{t^{2} + 1} \Rightarrow β t^{2} + β + {3 ft}^{2} + af = 1 \Rightarrow$ $(β + 3 f) t^{2} + β + af = 1 \Rightarrow {\begin{cases} β = - 3 f \\ - 3 f + af = 1 \end{cases} \Rightarrow {\begin{cases} f = \frac{1}{a - 3} \Rightarrow \\ β = \frac{- 3}{a - 3} \end{cases}$ $F (t) = \frac{- 3}{(a - 3) ({3 t}^{2} + a)} + \frac{1}{(a - 3) (t^{2} + 1)} \Rightarrow$ $f^{^{'}} (a) = \frac{- 3}{a - 3} \int_{0}^{\infty} \frac{dt}{{3 t}^{2} + a} + \frac{1}{a - 3} \int_{0}^{\infty} \frac{dt}{t^{2} + 1} but$ $\int_{0}^{\infty} \frac{dt}{{3 t}^{2} + a} = \frac{1}{3} \int_{0}^{\infty} \frac{dt}{t^{2} + \frac{a}{3}} =_{t = \sqrt{\frac{a}{3}} u} \frac{1}{3} . \frac{3}{a} \int_{0}^{\infty} \frac{1}{u^{2} + 1} \frac{\sqrt{a}}{\sqrt{3}} du$ $= \frac{1}{\sqrt{3 a}} \times \frac{π}{2} \Rightarrow f^{^{'}} (a) = \frac{- 3}{a - 3} . \frac{π}{2 \sqrt{3} \sqrt{a}} + \frac{π}{2 (a - 3)}$ $= - \frac{π \sqrt{3}}{2} . \frac{1}{(a - 3) \sqrt{a}} + \frac{π}{2 (a - 3)} \Rightarrow f (a) = - \frac{π \sqrt{3}}{2} \int \frac{da}{(a - 3) \sqrt{a}} + \frac{π}{2} \ln ∣ a - 3 ∣ + c$ $\int \frac{da}{\sqrt{a} (a - 3)} =_{\sqrt{a} = z} \int \frac{2 zdz}{z (z^{2} - 3)} = \int \frac{2 dz}{(z - \sqrt{3}) (z + \sqrt{3})}$ $= \frac{1}{\sqrt{3}} \int (\frac{1}{z - \sqrt{3}} - \frac{1}{z + \sqrt{3}}) dz = \frac{1}{\sqrt{3}} \ln ∣ \frac{\sqrt{a} - \sqrt{3}}{\sqrt{a} + \sqrt{3}} ∣ \Rightarrow$ $f (a) = - \frac{π}{2} \ln ∣ \frac{\sqrt{a} - \sqrt{3}}{\sqrt{a} + \sqrt{3}} ∣ + \frac{π}{2} \ln ∣ a - 3 ∣ + c$ $we have f (0) = \frac{π}{2} \ln (3) + c = \int_{0}^{\infty} \frac{\ln ({3 t}^{2})}{1 + t^{2}} dt$ $= \ln (3) \frac{π}{2} + 2 \int_{0}^{\infty} \frac{lnt}{1 + t^{2}} dt (\to 0) = \frac{π}{2} \ln 3 \Rightarrow c = 0 \Rightarrow$ $f (a) = \frac{π}{2} \ln ∣ a - 3 ∣ - \frac{π}{2} \ln ∣ \frac{a - 3}{{(\sqrt{a} + \sqrt{3})}^{2}} ∣ = \frac{π}{2} \ln ∣ a - 3 ∣ - \frac{π}{2} \ln ∣ a - 3 ∣$ $+ \frac{π}{2} \ln ({(\sqrt{a} + \sqrt{3})}^{2}) = π \ln (\sqrt{a} + \sqrt{3}) \Rightarrow K = f (1) = π \ln (1 + \sqrt{3})$ $9 A = \sqrt{3} H - \sqrt{3} K = 3 π - \sqrt{3} π \ln (1 + \sqrt{3}) \Rightarrow$ $A = \frac{1}{9} (3 π - π \sqrt{3} \ln (1 + \sqrt{3}))$
	Commented by mnjuly1970 last updated on 18/Aug/20

	$. . . . . ♣ M a t h e m a t i c a l A n l y s i s ♣ . . . . .$ $. . . ★ PROVE THAT ★ . . .$ $1) \frac{π}{3 \sqrt{3}} = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{5} + \frac{1}{7} - \frac{1}{8} + \frac{1}{10} - . . . . \infty$ $2) \frac{π}{2 \sqrt{2}} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - . . .$ $thank you master very thank you$ $peace be upon you and mercey . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum