solve-1-0-ln-2-e-x-e-4-x-2-dx-pi-k-e-2-k-2-0-ln-3-x-e-2

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

solve ::

[1] \boldsymbol 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 = \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 \dots!

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 \dots

b r a v o m r b r a n d o n \dots