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Question Number 44515 by maxmathsup by imad last updated on 30/Sep/18
let g(x) =∫_0 ^∞ ((t ln(t)dt)/((1+xt)^3 )) with x>0 1) give a explicit form of g(x) 2) calculate ∫_0 ^∞ ((t ln(t))/((1+t)^3 ))dt 3) calculate ∫_0 ^∞ ((tln(t))/((1+2t)^3 )) dt 4) calculate A(θ) =∫_0 ^∞ ((t ln(t))/((1+t sinθ)^3 ))dt with 0<θ<(π/2)
$${let}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}\:{ln}\left({t}\right){dt}}{\left(\mathrm{1}+{xt}\right)^{\mathrm{3}} }\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{give}\:{a}\:{explicit}\:{form}\:{of}\:{g}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}\:{ln}\left({t}\right)}{\left(\mathrm{1}+{t}\right)^{\mathrm{3}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{tln}\left({t}\right)}{\left(\mathrm{1}+\mathrm{2}{t}\right)^{\mathrm{3}} }\:{dt} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:{A}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}\:{ln}\left({t}\right)}{\left(\mathrm{1}+{t}\:{sin}\theta\right)^{\mathrm{3}} }{dt}\:\:{with}\:\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}} \\ $$
Answered by maxmathsup by imad last updated on 02/Oct/18
1) let f(x) =∫_0 ^∞ ((ln(t))/((1+tx)^2 ))dt we have proved that f(x)=−((ln(x))/x) ⇒f^′ (x) = −∫_0 ^∞ ((2t(1+tx)ln(t))/((1+tx)^4 )) =−∫_0 ^∞ ((2tln(t))/((1+tx)^3 ))dt ⇒ g(x) =−(1/2)f^′ (x) but f^′ (x) =−((1−ln(x))/x^2 ) ⇒g(x)=((1−ln(x))/(2x^2 )) 2) ∫_0 ^∞ ((tln(t))/((1+t)^3 )) =g(1)=0 3) ∫_0 ^∞ ((t ln(t))/((1+2t)^3 ))dt =g(2) =((1−ln(2))/8) 4)∫_0 ^∞ ((tln(t))/((1+tsinθ)^3 ))dt =g(sinθ) =((1−ln(sinθ))/(2sin^2 θ)) .
$$\left.\mathrm{1}\right)\:{let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({t}\right)}{\left(\mathrm{1}+{tx}\right)^{\mathrm{2}} }{dt}\:\:{we}\:{have}\:{proved}\:{that}\:{f}\left({x}\right)=−\frac{{ln}\left({x}\right)}{{x}} \\ $$$$\Rightarrow{f}^{'} \left({x}\right)\:=\:−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{t}\left(\mathrm{1}+{tx}\right){ln}\left({t}\right)}{\left(\mathrm{1}+{tx}\right)^{\mathrm{4}} }\:=−\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}{tln}\left({t}\right)}{\left(\mathrm{1}+{tx}\right)^{\mathrm{3}} }{dt}\:\Rightarrow \\ $$$${g}\left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}{f}^{'} \left({x}\right)\:{but}\:{f}^{'} \left({x}\right)\:=−\frac{\mathrm{1}−{ln}\left({x}\right)}{{x}^{\mathrm{2}} }\:\Rightarrow{g}\left({x}\right)=\frac{\mathrm{1}−{ln}\left({x}\right)}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{tln}\left({t}\right)}{\left(\mathrm{1}+{t}\right)^{\mathrm{3}} }\:={g}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}\:{ln}\left({t}\right)}{\left(\mathrm{1}+\mathrm{2}{t}\right)^{\mathrm{3}} }{dt}\:={g}\left(\mathrm{2}\right)\:=\frac{\mathrm{1}−{ln}\left(\mathrm{2}\right)}{\mathrm{8}} \\ $$$$\left.\mathrm{4}\right)\int_{\mathrm{0}} ^{\infty} \:\:\frac{{tln}\left({t}\right)}{\left(\mathrm{1}+{tsin}\theta\right)^{\mathrm{3}} }{dt}\:={g}\left({sin}\theta\right)\:=\frac{\mathrm{1}−{ln}\left({sin}\theta\right)}{\mathrm{2}{sin}^{\mathrm{2}} \theta}\:. \\ $$

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