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Question Number 149673 by mnjuly1970 last updated on 06/Aug/21

$solve ::$ $[1] ϕ := \int_{0}^{\infty} \frac{{l n}^{2} (e x)}{e^{4} + x^{2}} d x = \frac{π k}{e^{2}}$ $k := ?$ $[2] Ω := \int_{0}^{\infty} \frac{{l n}^{3} (x)}{e^{2} + x^{2}} d x = ?$
	Answered by mathmax by abdo last updated on 06/Aug/21
	$Φ=∫_0 ^∞ ((ln^2 (ex))/(x^2 +e^4 ))dx changement x=e^2 t give Φ=∫_0 ^∞ ((ln^2 (e^3 t))/(e^4 t^2 +e^4 ))(e^2 dt) =(1/e^2 )∫_0 ^∞ (((3+lnt)^2 )/(t^2 +1))dt ⇒e^2 Φ=∫_0 ^∞ ((ln^2 t+6lnt +9)/(t^2 +1))dt =∫_0 ^∞ ((ln^2 t)/(1+t^2 ))dt+6∫_0 ^∞ ((lnt)/(t^2 +1))dt +9∫_0 ^∞ (dt/(t^2 +1)) ∫_0 ^∞ ((lnt)/(t^2 +1))dt=0(proved) ,∫_0 ^∞ (dt/(t^(2 ) +1))=(π/2) f(a)=∫_0 ^∞ (t^a /(1+t^2 ))dt =∫_0 ^∞ (e^(alnt) /(1+t^2 ))dt ⇒ f^′ (a)=∫_0 ^∞ ((lnt e^(alnt) )/(1+t^2 ))dt ⇒f^((2)) (a)=∫_0 ^∞ ((t^a ln^2 t)/(t^2 +1))dt ⇒ f^((2)) (0) =∫_0 ^∞ ((ln^2 t)/(t^2 +1))dt changement t^2 =y give t=y^(1/2) f(a)=∫_0 ^∞ (t^(a/2) /(1+y))(1/2)y^((1/2)−1) dy =(1/2)∫_0 ^∞ (t^(((a+1)/2)−1) /(1+y))dy =(1/2)×(π/(sin((((a+1)/2))π))) =(π/(2cos(((πa)/2)))) ⇒ f^′ (a)=(π/2)(−(((π/2)sin(((πa)/2)))/(cos^2 (((πa)/2)))))=−(π^2 /4)×((sin(((πa)/2)))/(cos^2 (((πa)/2)))) ⇒ f^((2)) (a)=−(π^2 /4){(((π/2)cos(((πa)/2))cos^2 (((πa)/2))+2.(π/2)sin(((πa)/2))cos(((πa)/2)))/(cos^4 (((πa)/2))))} =−(π^3 /8){((cos^2 (((πa)/2))+2sin(((πa)/2)))/(cos^3 (((πa)/2))))} ⇒f^((2)) (0)=−(π^3 /8) ⇒ e^2 Φ=−(π^3 /8)+((9π)/2) ⇒Φ=(π/e^2 )(−(π^2 /8)+(9/2))=((πk)/e^2 ) ⇒k=(9/2)−(π^2 /8)$
	$Φ = \int_{0}^{\infty} \frac{\ln^{2} (ex)}{x^{2} + e^{4}} dx changement x = e^{2} t give$ $Φ = \int_{0}^{\infty} \frac{\ln^{2} (e^{3} t)}{e^{4} t^{2} + e^{4}} (e^{2} dt) = \frac{1}{e^{2}} \int_{0}^{\infty} \frac{{(3 + lnt)}^{2}}{t^{2} + 1} dt$ $\Rightarrow e^{2} Φ = \int_{0}^{\infty} \frac{\ln^{2} t + 6 lnt + 9}{t^{2} + 1} dt$ $= \int_{0}^{\infty} \frac{\ln^{2} t}{1 + t^{2}} dt + 6 \int_{0}^{\infty} \frac{lnt}{t^{2} + 1} dt + 9 \int_{0}^{\infty} \frac{dt}{t^{2} + 1}$ $\int_{0}^{\infty} \frac{lnt}{t^{2} + 1} dt = 0 (proved), \int_{0}^{\infty} \frac{dt}{t^{2} + 1} = \frac{π}{2}$ $f (a) = \int_{0}^{\infty} \frac{t^{a}}{1 + t^{2}} dt = \int_{0}^{\infty} \frac{e^{alnt}}{1 + t^{2}} dt \Rightarrow$ $f^{^{'}} (a) = \int_{0}^{\infty} \frac{lnt e^{alnt}}{1 + t^{2}} dt \Rightarrow f^{(2)} (a) = \int_{0}^{\infty} \frac{t^{a} \ln^{2} t}{t^{2} + 1} dt \Rightarrow$ $f^{(2)} (0) = \int_{0}^{\infty} \frac{\ln^{2} t}{t^{2} + 1} dt changement t^{2} = y give t = y^{\frac{1}{2}}$ $f (a) = \int_{0}^{\infty} \frac{t^{\frac{a}{2}}}{1 + y} \frac{1}{2} y^{\frac{1}{2} - 1} dy = \frac{1}{2} \int_{0}^{\infty} \frac{t^{\frac{a + 1}{2} - 1}}{1 + y} dy$ $= \frac{1}{2} \times \frac{π}{\sin ((\frac{a + 1}{2}) π)} = \frac{π}{2 \cos (\frac{π a}{2})} \Rightarrow$ $f^{^{'}} (a) = \frac{π}{2} (- \frac{\frac{π}{2} \sin (\frac{π a}{2})}{\cos^{2} (\frac{π a}{2})}) = - \frac{π^{2}}{4} \times \frac{\sin (\frac{π a}{2})}{\cos^{2} (\frac{π a}{2})} \Rightarrow$ $f^{(2)} (a) = - \frac{π^{2}}{4} {\frac{\frac{π}{2} \cos (\frac{π a}{2}) \cos^{2} (\frac{π a}{2}) + 2 . \frac{π}{2} \sin (\frac{π a}{2}) \cos (\frac{π a}{2})}{\cos^{4} (\frac{π a}{2})}}$ $= - \frac{π^{3}}{8} {\frac{\cos^{2} (\frac{π a}{2}) + 2 \sin (\frac{π a}{2})}{\cos^{3} (\frac{π a}{2})}} \Rightarrow f^{(2)} (0) = - \frac{π^{3}}{8} \Rightarrow$ $e^{2} Φ = - \frac{π^{3}}{8} + \frac{9 π}{2} \Rightarrow Φ = \frac{π}{e^{2}} (- \frac{π^{2}}{8} + \frac{9}{2}) = \frac{π k}{e^{2}} \Rightarrow k = \frac{9}{2} - \frac{π^{2}}{8}$
	Commented by Ar Brandon last updated on 07/Aug/21
	Bravo ��
	Commented by Tawa11 last updated on 06/Aug/21

	$great$
	Commented by Ar Brandon last updated on 06/Aug/21

	$Sir, f (a) = \frac{π}{2 \cos (\frac{π a}{2})} \Rightarrow f^{'} (a) = \frac{π^{2}}{4} \cdot \frac{\sin (\frac{π}{2} a)}{\cos^{2} (\frac{π}{2} a)}$ $instead of - \frac{π^{2}}{4} \cdot \frac{\sin (\frac{π}{2} a)}{\cos^{2} (\frac{π}{2} a)} . Please check!$ $Line 10 \to Line 11$
	Commented by mathmax by abdo last updated on 07/Aug/21

	$sorry f^{^{'}} (a) = \frac{π^{2}}{4} \times \frac{\sin (\frac{π a}{2})}{\cos^{2} (\frac{π a}{2})} \Rightarrow$ $f^{(2)} (a) = \frac{π^{3}}{8} \times \frac{\cos^{2} (\frac{π a}{2}) + 2 \sin (\frac{π a}{2})}{\cos^{3} (\frac{π a}{2})} \Rightarrow f^{(2)} (0) = \frac{π^{3}}{8} \Rightarrow$ $e^{2} Φ = \frac{π^{3}}{8} + \frac{9 π}{2} \Rightarrow Φ = \frac{π}{e^{2}} (\frac{π^{2}}{8} + \frac{9}{2}) \Rightarrow k = \frac{π^{2}}{8} + \frac{9}{2}$
	Commented by mathmax by abdo last updated on 07/Aug/21

	$yes error of sing . . .!$
	Answered by Ar Brandon last updated on 06/Aug/21

	$ϕ = \int_{0}^{\infty} \frac{\ln^{2} (e x)}{e^{4} + x^{2}} d x = \int_{0}^{\infty} \frac{1 + 2 \ln x + \ln^{2} x}{e^{4} + x^{2}} d x$ $= \frac{1}{e^{2}} \int_{0}^{\infty} \frac{1 + 2 \ln (e^{2} u) + \ln^{2} (e^{2} u)}{1 + u^{2}} d u$ $= \frac{π}{2 e^{2}} + \frac{1}{e^{2}} \int_{0}^{\infty} \frac{4 + 2 \ln u + 4 + 4 \ln u + \ln^{2} u}{1 + u^{2}} d u$ $= \frac{π}{2 e^{2}} + \frac{4 π}{e^{2}} + \frac{1}{e^{2}} \int_{0}^{\infty} \frac{6 \ln u + \ln^{2} u}{1 + u^{2}} d u$ $= \frac{9 π}{2 e^{2}} + \frac{6}{e^{2}} \cdot \frac{\partial}{\partial α} ∣_{α = 1} \int_{0}^{\infty} \frac{u^{α - 1}}{1 + u^{2}} d u + \frac{1}{e^{2}} \cdot \frac{\partial^{2}}{\partial α^{2}} ∣_{α = 1} \int_{0}^{\infty} \frac{u^{α - 1}}{1 + u^{2}} d u$ $= \frac{9 π}{2 e^{2}} + \frac{3}{e^{2}} \cdot \frac{\partial}{\partial α} ∣_{α = 1} \frac{π}{\sin (\frac{π α}{2})} + \frac{1}{2 e^{2}} \cdot \frac{\partial^{2}}{\partial α^{2}} ∣_{α = 1} \frac{π}{\sin (\frac{π α}{2})}$ $= \frac{9 π}{2 e^{2}} - \frac{3 π^{2}}{2 e^{2}} ∣_{α = 1} cosec (\frac{π α}{2}) \cot (\frac{π α}{2})$ $+ \frac{π^{3}}{8 e^{2}} ∣_{α = 1} {cosec}^{3} (\frac{π α}{2}) + cosec (\frac{π α}{2}) \cot^{2} (\frac{π α}{2})$ $= \frac{9 π}{2 e^{2}} + \frac{π^{3}}{8 e^{2}} = \frac{π}{e^{2}} (\frac{9}{2} + \frac{π^{2}}{8}), k = \frac{9}{2} + \frac{π^{2}}{8}$
	Commented by Ar Brandon last updated on 06/Aug/21
	It was a pleasure, Sir ��
	Commented by mnjuly1970 last updated on 06/Aug/21

	$v e r y n i c e m r b r a n d o n . .$
	Answered by Ar Brandon last updated on 06/Aug/21

	$Ω = \int_{0}^{\infty} \frac{\ln^{3} x}{e^{2} + x^{2}} d x = \frac{1}{e} \int_{0}^{\infty} \frac{\ln^{3} (e u)}{1 + u^{2}} d u$ $= \frac{1}{e} \int_{0}^{\infty} \frac{1 + 3 \ln u + {3 \ln}^{2} u + \ln^{3} u}{1 + u^{2}} d u$ $= \frac{π}{2 e} + \frac{3 π^{3}}{8 e} + \frac{1}{e} \int_{0}^{\infty} \frac{\ln^{3} u}{1 + u^{2}} d u$ $= \frac{(4 + 3 π^{2}) π}{8 e} + \frac{1}{e} \cdot \frac{\partial^{3}}{\partial α^{3}} ∣_{α = 1} \int_{0}^{\infty} \frac{u^{α - 1}}{1 + u^{2}} d u$ $= \frac{(4 + 3 π^{2}) π}{8 e} + \frac{1}{e} \cdot \frac{\partial^{3}}{\partial α^{3}} ∣_{α = 1} \frac{π}{\sin (\frac{π α}{2})}$ $= \frac{(4 + 3 π^{2}) π}{8 e} - \frac{π}{8 e} ∣_{α = 1} (\frac{3 π}{2} {cosec}^{2} (\frac{π α}{2}) cosec (\frac{π α}{2}) \cot (\frac{π α}{2})$ $+ π ({cosec}^{3} (\frac{π α}{2}) \cot (\frac{π α}{2}) + \frac{1}{2} cosec (\frac{π α}{2}) \cot^{3} (\frac{π α}{2}))$ $= \frac{(4 + 3 π^{2}) π}{8 e}$
	Commented by mnjuly1970 last updated on 06/Aug/21

	$b r a v o m r b r a n d o n . . .$
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