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Question Number 202393 by sonukgindia last updated on 26/Dec/23
Answered by Mathspace last updated on 26/Dec/23
I=∫_0 ^∞ ((lnx)/(a+bx^2 ))dx (a>0,b>0) I=(1/a)∫_0 ^∞ ((lnx)/(1+(b/a)x^2 ))dx ((√(b/a))x=t) =(1/a)∫_0 ^∞ ((ln((√(a/b))t))/(1+t^2 ))×(√(a/b))dt =(1/( (√(ab))))∫_0 ^∞ ((ln((√(a/b)))+lnt)/(1+t^2 ))dt =(1/(2(√(ab))))(lna−lnb)∫_0 ^∞ (dt/(1+t^2 )) +(1/( (√(ab))))∫_0 ^∞ ((lnt)/(1+t^2 ))dt but ∫_0 ^∞ ((lnt)/(1+t^2 ))dt=0 ⇒I=((lna−lnb)/(2(√(ab))))×(π/2) =(π/(4(√(ab))))(lna−lnb)
$${I}=\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{{a}+{bx}^{\mathrm{2}} }{dx}\:\:\:\:\:\:\:\:\:\left({a}>\mathrm{0},{b}>\mathrm{0}\right) \\ $$$${I}=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{\mathrm{1}+\frac{{b}}{{a}}{x}^{\mathrm{2}} }{dx}\:\:\left(\sqrt{\frac{{b}}{{a}}}{x}={t}\right) \\ $$$$=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\sqrt{\frac{{a}}{{b}}}{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }×\sqrt{\frac{{a}}{{b}}}{dt} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{ab}}}\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left(\sqrt{\frac{{a}}{{b}}}\right)+{lnt}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{ab}}}\left({lna}−{lnb}\right)\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$+\frac{\mathrm{1}}{\:\sqrt{{ab}}}\int_{\mathrm{0}} ^{\infty} \frac{{lnt}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:{but}\:\int_{\mathrm{0}} ^{\infty} \frac{{lnt}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\mathrm{0} \\ $$$$\Rightarrow{I}=\frac{{lna}−{lnb}}{\mathrm{2}\sqrt{{ab}}}×\frac{\pi}{\mathrm{2}} \\ $$$$=\frac{\pi}{\mathrm{4}\sqrt{{ab}}}\left({lna}−{lnb}\right) \\ $$
Answered by Mathspace last updated on 27/Dec/23
J=∫_0 ^∞ ((ln^2 x)/(a+bx^2 ))dx ⇒ J=(1/a)∫_0 ^∞ ((ln^2 x)/(1+(b/a)x^2 ))dx ((√(b/a))x=t) =(1/a)∫_0 ^∞ ((ln^2 ((√(a/b))t))/(1+t^2 ))(√(a/b))dt =(1/( (√(ab))))∫_0 ^∞ (((ln((√(a/b))+lnt)^2 )/(1+t^2 ))dt (√(ab))J=∫_0 ^∞ ((ln^2 ((√(a/b)))+2ln((√(a/b)))lnt +ln^2 t)/(1+t^2 ))dt =(π/2)ln^2 ((√(a/b)))+2ln((√(a/b)))∫_0 ^∞ ((lnt)/(1+t^2 ))dt(→0) +∫_0 ^∞ ((ln^2 t)/(1+t^2 ))dt =(π/2)ln^2 ((√(a/b)))+∫_0 ^∞ ((ln^2 t)/(1+t^2 ))dt K=∫_0 ^∞ ((ln^2 t)/(1+t^2 ))dt (t=z^(1/2) ) K=(1/8)∫_0 ^∞ ((ln^2 z)/(1+z))z^((1/2)−1) dz
$${J}=\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}^{\mathrm{2}} {x}}{{a}+{bx}^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$${J}=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}^{\mathrm{2}} {x}}{\mathrm{1}+\frac{{b}}{{a}}{x}^{\mathrm{2}} }{dx}\:\:\left(\sqrt{\frac{{b}}{{a}}}{x}={t}\right) \\ $$$$=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{\mathrm{2}} \left(\sqrt{\frac{{a}}{{b}}}{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }\sqrt{\frac{{a}}{{b}}}{dt} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{ab}}}\int_{\mathrm{0}} ^{\infty} \frac{\left({ln}\left(\sqrt{\frac{{a}}{{b}}}+{lnt}\right)^{\mathrm{2}} \right.}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\sqrt{{ab}}{J}=\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{\mathrm{2}} \left(\sqrt{\frac{{a}}{{b}}}\right)+\mathrm{2}{ln}\left(\sqrt{\frac{{a}}{{b}}}\right){lnt}\:+{ln}^{\mathrm{2}} {t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\pi}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\sqrt{\frac{{a}}{{b}}}\right)+\mathrm{2}{ln}\left(\sqrt{\frac{{a}}{{b}}}\right)\int_{\mathrm{0}} ^{\infty} \frac{{lnt}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\left(\rightarrow\mathrm{0}\right) \\ $$$$+\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{\mathrm{2}} {t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\pi}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\sqrt{\frac{{a}}{{b}}}\right)+\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{\mathrm{2}} {t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$${K}=\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{\mathrm{2}} {t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:\left({t}={z}^{\frac{\mathrm{1}}{\mathrm{2}}} \right) \\ $$$${K}=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{\mathrm{2}} {z}}{\mathrm{1}+{z}}{z}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dz} \\ $$$$ \\ $$$$ \\ $$
Commented by Mathspace last updated on 27/Dec/23
f(a)=∫_0 ^∞ (z^(a−1) /(1+z))dt o<a<1 f^((2)) (a)=∫_0 ^∞ ((t^(a−1) ln^2 z)/(1+z))dt f^((2)) ((1/2))=∫_0 ^∞ ((z^((1/2)−1) ln^2 z)/(1+z))dz=8K f(a)=(π/(sin(πa))) ⇒ f^′ (a)=−π((πcos(πa))/(sin^2 (πa))) =−π^2 ((cos(πa))/(sin^2 (πa))) f^((2)) (a)=−π^2 ((−πsin(πa)sin^2 (πa)−cos(πa)2πsin(πa)cos(πa))/(sin^4 (πa))) =π^3 ((sin^2 (πa)+2cos^2 (πa))/(sin^3 (πa))) =π^3 ((1+cos^2 (πa))/(sin^3 (πa))) f^((2)) ((1/2))=π^3 ×((1+0)/1^3 )=π^3 ⇒ 8K=π^3 ⇒K=(π^3 /8) ⇒ (√(ab))J=(π/2)ln^2 ((√(a/b)))+(π^3 /8) ⇒ J=(π/(2(√(ab))))ln^2 ((√(a/b)))+(π^3 /(8(√(ab)))) by (mathsup by abdo)
$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{z}^{{a}−\mathrm{1}} }{\mathrm{1}+{z}}{dt}\:\:\:\:\:\:\:\:\:\:\:{o}<{a}<\mathrm{1} \\ $$$${f}^{\left(\mathrm{2}\right)} \left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}−\mathrm{1}} {ln}^{\mathrm{2}} {z}}{\mathrm{1}+{z}}{dt} \\ $$$${f}^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{z}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {ln}^{\mathrm{2}} {z}}{\mathrm{1}+{z}}{dz}=\mathrm{8}{K} \\ $$$${f}\left({a}\right)=\frac{\pi}{{sin}\left(\pi{a}\right)}\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)=−\pi\frac{\pi{cos}\left(\pi{a}\right)}{{sin}^{\mathrm{2}} \left(\pi{a}\right)} \\ $$$$=−\pi^{\mathrm{2}} \frac{{cos}\left(\pi{a}\right)}{{sin}^{\mathrm{2}} \left(\pi{a}\right)} \\ $$$${f}^{\left(\mathrm{2}\right)} \left({a}\right)=−\pi^{\mathrm{2}} \frac{−\pi{sin}\left(\pi{a}\right){sin}^{\mathrm{2}} \left(\pi{a}\right)−{cos}\left(\pi{a}\right)\mathrm{2}\pi{sin}\left(\pi{a}\right){cos}\left(\pi{a}\right)}{{sin}^{\mathrm{4}} \left(\pi{a}\right)} \\ $$$$=\pi^{\mathrm{3}} \frac{{sin}^{\mathrm{2}} \left(\pi{a}\right)+\mathrm{2}{cos}^{\mathrm{2}} \left(\pi{a}\right)}{{sin}^{\mathrm{3}} \left(\pi{a}\right)} \\ $$$$=\pi^{\mathrm{3}} \frac{\mathrm{1}+{cos}^{\mathrm{2}} \left(\pi{a}\right)}{{sin}^{\mathrm{3}} \left(\pi{a}\right)} \\ $$$${f}^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\pi^{\mathrm{3}} ×\frac{\mathrm{1}+\mathrm{0}}{\mathrm{1}^{\mathrm{3}} }=\pi^{\mathrm{3}} \:\Rightarrow \\ $$$$\mathrm{8}{K}=\pi^{\mathrm{3}} \Rightarrow{K}=\frac{\pi^{\mathrm{3}} }{\mathrm{8}}\:\Rightarrow \\ $$$$\sqrt{{ab}}{J}=\frac{\pi}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\sqrt{\frac{{a}}{{b}}}\right)+\frac{\pi^{\mathrm{3}} }{\mathrm{8}}\:\Rightarrow \\ $$$${J}=\frac{\pi}{\mathrm{2}\sqrt{{ab}}}{ln}^{\mathrm{2}} \left(\sqrt{\frac{{a}}{{b}}}\right)+\frac{\pi^{\mathrm{3}} }{\mathrm{8}\sqrt{{ab}}} \\ $$$${by}\:\left({mathsup}\:{by}\:{abdo}\right) \\ $$

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