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Question Number 142430 by Mathspace last updated on 31/May/21
calculate ∫_0 ^∞ ((log^2 x)/(1+x^2 ))dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{log}^{\mathrm{2}} {x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$
Answered by Dwaipayan Shikari last updated on 31/May/21
∫_0 ^∞ (x^a /(1+x^2 ))dx=(π/(2sin(((π(1−a))/2)))) ∫_0 ^∞ ((x^a log^2 (x))/(1+x^2 ))dx=(∂^2 /∂a^2 )((π/2)cosec((π/2)(1−a))) ∫_0 ^∞ ((log^2 (x))/(1+x^2 ))dx=(∂^2 /∂a^2 )∣_(a=0) (π/2)cosec((π/2)(1−a))
$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{a}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\pi}{\mathrm{2}{sin}\left(\frac{\pi\left(\mathrm{1}−{a}\right)}{\mathrm{2}}\right)} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{a}} {log}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\partial^{\mathrm{2}} }{\partial{a}^{\mathrm{2}} }\left(\frac{\pi}{\mathrm{2}}{cosec}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−{a}\right)\right)\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{log}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\partial^{\mathrm{2}} }{\partial{a}^{\mathrm{2}} }\mid_{{a}=\mathrm{0}} \frac{\pi}{\mathrm{2}}{cosec}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−{a}\right)\right) \\ $$
Answered by mathmax by abdo last updated on 01/Jun/21
Φ=∫_0 ^∞ ((log^2 x)/(1+x^2 ))dx changement x=t^(1/2) [give Φ=(1/4)∫_0 ^∞ ((log^2 t)/(1+t))(1/2)t^(−(1/2)) dt =(1/8)∫_0 ^∞ ((t^(−(1/2)) log^2 t)/(1+t))dt let f(a)=∫_0 ^∞ (t^(a−1) /(1+t))dt ⇒f^′ (a)=∫_0 ^∞ (∂/∂a)((e^((a−1)logt) /(1+t)))dt =∫_0 ^∞ ((t^(a−1) logt)/(1+t)) ⇒f^((2)) (a)=∫_0 ^∞ ((t^(a−1) log^2 t)/(1+t))dt ⇒ f^((2)) ((1/2))=∫_0 ^∞ ((t^(−(1/2)) log^2 t)/(1+t))dt=8Φ we have f(a)=(π/(sin(πa))) ⇒f^′ (a)=−((π^2 cos(πa))/(sin^2 (πa))) (0<a<1) and f^((2)) (a)=−π^2 .((−πsin(πa)sin^2 (πa)−cos(πa).2sin(πa)πcos(πa))/(sin^4 (πa))) 8Φ=f^((2)) ((1/2))=−π^2 ×((−π)/1)=π^3 ⇒Φ=(π^3 /8)
$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}^{\mathrm{2}} \mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\mathrm{changement}\:\mathrm{x}=\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{2}}} \left[\mathrm{give}\right. \\ $$$$\Phi=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}^{\mathrm{2}} \mathrm{t}}{\mathrm{1}+\mathrm{t}}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{log}^{\mathrm{2}} \mathrm{t}}{\mathrm{1}+\mathrm{t}}\mathrm{dt}\: \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\partial}{\partial\mathrm{a}}\left(\frac{\mathrm{e}^{\left(\mathrm{a}−\mathrm{1}\right)\mathrm{logt}} }{\mathrm{1}+\mathrm{t}}\right)\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} \mathrm{logt}}{\mathrm{1}+\mathrm{t}}\:\Rightarrow\mathrm{f}^{\left(\mathrm{2}\right)} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} \:\mathrm{log}^{\mathrm{2}} \mathrm{t}}{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{log}^{\mathrm{2}} \mathrm{t}}{\mathrm{1}+\mathrm{t}}\mathrm{dt}=\mathrm{8}\Phi \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{a}\right)}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)=−\frac{\pi^{\mathrm{2}} \:\mathrm{cos}\left(\pi\mathrm{a}\right)}{\mathrm{sin}^{\mathrm{2}} \left(\pi\mathrm{a}\right)}\:\:\:\left(\mathrm{0}<\mathrm{a}<\mathrm{1}\right) \\ $$$$\mathrm{and}\:\mathrm{f}^{\left(\mathrm{2}\right)} \left(\mathrm{a}\right)=−\pi^{\mathrm{2}} .\frac{−\pi\mathrm{sin}\left(\pi\mathrm{a}\right)\mathrm{sin}^{\mathrm{2}} \left(\pi\mathrm{a}\right)−\mathrm{cos}\left(\pi\mathrm{a}\right).\mathrm{2sin}\left(\pi\mathrm{a}\right)\pi\mathrm{cos}\left(\pi\mathrm{a}\right)}{\mathrm{sin}^{\mathrm{4}} \left(\pi\mathrm{a}\right)} \\ $$$$\mathrm{8}\Phi=\mathrm{f}^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\pi^{\mathrm{2}} \:×\frac{−\pi}{\mathrm{1}}=\pi^{\mathrm{3}} \:\Rightarrow\Phi=\frac{\pi^{\mathrm{3}} }{\mathrm{8}} \\ $$

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