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Question Number 59278 by Mr X pcx last updated on 07/May/19
let f(x) =∫_0 ^∞ ((ln(1+xt^2 ))/(2+t^2 )) dt determine a explicit form of f(x) 2)calculate ∫_0 ^∞ ((ln(1+3x^2 ))/(2+t^2 ))dt
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left(\mathrm{1}+{xt}^{\mathrm{2}} \right)}{\mathrm{2}+{t}^{\mathrm{2}} }\:{dt} \\ $$$${determine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left(\mathrm{1}+\mathrm{3}{x}^{\mathrm{2}} \right)}{\mathrm{2}+{t}^{\mathrm{2}} }{dt} \\ $$
Commented by maxmathsup by imad last updated on 13/May/19
1) we have f^′ (x) =∫_0 ^∞ (t^2 /((2+t^2 )(1+xt^2 ))) dt let decompose F(t) =(t^2 /((t^2 +2)(xt^2 +1))) case 1 x>0 F(t) =((at +b)/(t^2 +2)) +((ct +d)/(xt^2 +1)) we have F(−t)=F(t) ⇒ ((−at +b)/(t^2 +2)) +((−ct +d)/(xt^2 +1)) = F(t) ⇒a=c=0 ⇒F(t) =(b/(t^2 +2)) +(d/(xt^2 +1)) lim_(t→+∞) t^2 F(t) =(1/x) =b +(d/x) ⇒1 =bx +d ⇒d =1−bx F(0) =0 =(b/2) +d ⇒b+2d =0 ⇒b =−2d ⇒d =1−(−2d)x =1+2dx ⇒ (1−2x)d =1 ⇒d =(1/(1−2x)) (we suppose x≠(1/2)) ⇒b =((−2)/(1−2x)) =(2/(2x−1)) ⇒ F(t) =(2/((2x−1)(t^2 +2))) −(1/((2x−1)(xt^2 +1))) ⇒f^′ (x) =∫_0 ^∞ F(t)dt =(2/(2x−1)) ∫_0 ^∞ (dt/(t^2 +2)) −(1/(2x−1))∫_0 ^∞ (dt/(xt^2 +1)) but ∫_0 ^∞ (dt/(t^2 +2)) =_(t =(√2)u) ∫_0 ^∞ (((√2)du)/(2(1+u^2 ))) =(1/( (√2))) ∫_0 ^∞ (du/(1+u^2 )) =(π/(2(√2))) ∫_0 ^∞ (dt/(xt^2 +1)) =_((√x)t =u) ∫_0 ^∞ (du/( (√x)(1+u^2 ))) =(π/(2(√x))) ⇒ f^′ (x) = (2/(2x−1)) (π/(2(√2))) −(1/(2x−1)) (π/(2(√x))) ⇒f(x) =(π/( (√2))) ∫ (dx/(2x−1)) −(π/2) ∫ (dx/( (√x)(2x−1))) +c =(π/(2(√2)))ln∣2x−1∣ −(π/2) ∫ (dx/( (√x)(2x−1))) +c ∫ (dx/( (√x)(2x−1))) =_((√x)=u) ∫ ((2udu)/(u(2u^2 −1))) =2 ∫ (du/(2u^2 −1)) =2 ∫ (du/(((√2)u−1)((√2)u +1))) = ∫ ((1/( (√2)u−1)) −(1/( (√2)u +1)))du =(1/( (√2)))ln∣(((√2)u−1)/( (√2)u +1))∣ =(1/( (√2)))ln∣(((√(2x))−1)/( (√(2x)) +1))∣ ⇒ f(x) =(π/(2(√2)))ln∣2x−1∣ −(π/(2(√2)))ln∣(((√(2x))−1)/( (√(2x)) +1))∣ +c f(0) =0 =c ⇒ f(x) =(π/(2(√2))){ln∣(√(2x))−1∣ +ln∣(√(2x))+1∣−ln∣(√(2x))−1∣ +ln∣(√(2x))+1∣} =(π/(2(√2))) (2ln∣(√(2x)) +1∣ ⇒ f(x) =(π/( (√2)))ln∣(√(2x))+1∣ with x >0 rest to stydy the case x<0
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{2}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{xt}^{\mathrm{2}} \right)}\:{dt} \\ $$$${let}\:{decompose}\:{F}\left({t}\right)\:=\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} \:+\mathrm{2}\right)\left({xt}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\:\:{case}\:\mathrm{1}\:\:\:\:\:{x}>\mathrm{0} \\ $$$${F}\left({t}\right)\:=\frac{{at}\:+{b}}{{t}^{\mathrm{2}} \:+\mathrm{2}}\:+\frac{{ct}\:+{d}}{{xt}^{\mathrm{2}} \:+\mathrm{1}}\:\:{we}\:{have}\:{F}\left(−{t}\right)={F}\left({t}\right)\:\Rightarrow \\ $$$$\frac{−{at}\:+{b}}{{t}^{\mathrm{2}} \:+\mathrm{2}}\:+\frac{−{ct}\:+{d}}{{xt}^{\mathrm{2}} \:+\mathrm{1}}\:=\:{F}\left({t}\right)\:\Rightarrow{a}={c}=\mathrm{0}\:\Rightarrow{F}\left({t}\right)\:=\frac{{b}}{{t}^{\mathrm{2}} \:+\mathrm{2}}\:+\frac{{d}}{{xt}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${lim}_{{t}\rightarrow+\infty} \:{t}^{\mathrm{2}} \:{F}\left({t}\right)\:=\frac{\mathrm{1}}{{x}}\:={b}\:+\frac{{d}}{{x}}\:\Rightarrow\mathrm{1}\:={bx}\:+{d}\:\Rightarrow{d}\:=\mathrm{1}−{bx} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{0}\:=\frac{{b}}{\mathrm{2}}\:+{d}\:\Rightarrow{b}+\mathrm{2}{d}\:=\mathrm{0}\:\Rightarrow{b}\:=−\mathrm{2}{d}\:\Rightarrow{d}\:=\mathrm{1}−\left(−\mathrm{2}{d}\right){x}\:=\mathrm{1}+\mathrm{2}{dx}\:\Rightarrow \\ $$$$\left(\mathrm{1}−\mathrm{2}{x}\right){d}\:=\mathrm{1}\:\Rightarrow{d}\:=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}{x}}\:\:\:\:\left({we}\:{suppose}\:{x}\neq\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow{b}\:=\frac{−\mathrm{2}}{\mathrm{1}−\mathrm{2}{x}}\:=\frac{\mathrm{2}}{\mathrm{2}{x}−\mathrm{1}}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{2}}{\left(\mathrm{2}{x}−\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{2}\right)}\:−\frac{\mathrm{1}}{\left(\mathrm{2}{x}−\mathrm{1}\right)\left({xt}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow{f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} {F}\left({t}\right){dt} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}{x}−\mathrm{1}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \:\frac{{dt}}{{xt}^{\mathrm{2}} \:+\mathrm{1}}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}}\:=_{{t}\:=\sqrt{\mathrm{2}}{u}} \:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\sqrt{\mathrm{2}}{du}}{\mathrm{2}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{xt}^{\mathrm{2}} \:+\mathrm{1}}\:=_{\sqrt{{x}}{t}\:={u}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{du}}{\:\sqrt{{x}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\frac{\pi}{\mathrm{2}\sqrt{{x}}}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\:\frac{\mathrm{2}}{\mathrm{2}{x}−\mathrm{1}}\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:−\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\:\frac{\pi}{\mathrm{2}\sqrt{{x}}}\:\Rightarrow{f}\left({x}\right)\:=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:\int\:\:\frac{{dx}}{\mathrm{2}{x}−\mathrm{1}}\:−\frac{\pi}{\mathrm{2}}\:\int\:\:\frac{{dx}}{\:\sqrt{{x}}\left(\mathrm{2}{x}−\mathrm{1}\right)}\:+{c} \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mid\mathrm{2}{x}−\mathrm{1}\mid\:−\frac{\pi}{\mathrm{2}}\:\int\:\:\:\:\frac{{dx}}{\:\sqrt{{x}}\left(\mathrm{2}{x}−\mathrm{1}\right)}\:+{c}\: \\ $$$$\int\:\:\:\frac{{dx}}{\:\sqrt{{x}}\left(\mathrm{2}{x}−\mathrm{1}\right)}\:=_{\sqrt{{x}}={u}} \:\:\:\:\int\:\:\frac{\mathrm{2}{udu}}{{u}\left(\mathrm{2}{u}^{\mathrm{2}} −\mathrm{1}\right)}\:=\mathrm{2}\:\int\:\:\:\:\frac{{du}}{\mathrm{2}{u}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\mathrm{2}\:\int\:\:\frac{{du}}{\left(\sqrt{\mathrm{2}}{u}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}{u}\:+\mathrm{1}\right)}\:=\:\int\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}{u}−\mathrm{1}}\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}{u}\:+\mathrm{1}}\right){du} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{ln}\mid\frac{\sqrt{\mathrm{2}}{u}−\mathrm{1}}{\:\sqrt{\mathrm{2}}{u}\:+\mathrm{1}}\mid\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{ln}\mid\frac{\sqrt{\mathrm{2}{x}}−\mathrm{1}}{\:\sqrt{\mathrm{2}{x}}\:+\mathrm{1}}\mid\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mid\mathrm{2}{x}−\mathrm{1}\mid\:−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mid\frac{\sqrt{\mathrm{2}{x}}−\mathrm{1}}{\:\sqrt{\mathrm{2}{x}}\:+\mathrm{1}}\mid\:+{c} \\ $$$${f}\left(\mathrm{0}\right)\:=\mathrm{0}\:={c}\:\Rightarrow\:{f}\left({x}\right)\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{{ln}\mid\sqrt{\mathrm{2}{x}}−\mathrm{1}\mid\:+{ln}\mid\sqrt{\mathrm{2}{x}}+\mathrm{1}\mid−{ln}\mid\sqrt{\mathrm{2}{x}}−\mathrm{1}\mid\:+{ln}\mid\sqrt{\mathrm{2}{x}}+\mathrm{1}\mid\right\} \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\left(\mathrm{2}{ln}\mid\sqrt{\mathrm{2}{x}}\:+\mathrm{1}\mid\:\Rightarrow\:{f}\left({x}\right)\:=\frac{\pi}{\:\sqrt{\mathrm{2}}}{ln}\mid\sqrt{\mathrm{2}{x}}+\mathrm{1}\mid\:\:\:{with}\:{x}\:>\mathrm{0}\:\:\:{rest}\:{to}\:{stydy}\:{the}\:{case}\:{x}<\mathrm{0}\right. \\ $$
Commented by maxmathsup by imad last updated on 13/May/19
duo to x>0 we have f(x) =(π/( (√2)))ln((√(2x)) +1)
$${duo}\:{to}\:{x}>\mathrm{0}\:\:{we}\:{have}\:{f}\left({x}\right)\:=\frac{\pi}{\:\sqrt{\mathrm{2}}}{ln}\left(\sqrt{\mathrm{2}{x}}\:+\mathrm{1}\right) \\ $$
Commented by maxmathsup by imad last updated on 13/May/19
2) ∫_0 ^∞ ((ln(1+3t^2 ))/(2+t^2 )) dt =f(3) =(π/( (√2)))ln(1+(√6)) .
$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left(\mathrm{1}+\mathrm{3}{t}^{\mathrm{2}} \right)}{\mathrm{2}+{t}^{\mathrm{2}} }\:{dt}\:={f}\left(\mathrm{3}\right)\:=\frac{\pi}{\:\sqrt{\mathrm{2}}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{6}}\right)\:. \\ $$

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