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Question Number 220544 by Nicholas666 last updated on 15/May/25
Prove that inequality; ∫_( 0) ^( 1) ((ln(1 + x^2 ))/(1 + x^2 )) dx < ∫_( 0) ^( 1) ((x ln(1 + x^2 ))/(1 + x^2 )) dx + (1/3)
$$ \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{inequality}}; \\ $$$$\:\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:{dx}\:<\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{x}\:\mathrm{ln}\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)}{\mathrm{1}\:+\:{x}^{\mathrm{2}} \:}\:{dx}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by SdC355 last updated on 15/May/25
first. ∫_0 ^∞ ((z∙ln(z^2 +1))/(z^2 +1)) dz= z^2 =u ∫_0 ^1 ((zln(z^2 +1))/(z^2 +1)) dz =(1/2)∫_0 ^( 1) ((ln(u+1))/(u+1)) du u+1=w (1/2)∫_0 ^( 1) ((ln(u+1))/(u+1)) du=(1/2) ∫_0 ^( ln(2)) w dw=(1/4)(ln(2))^2 ∫_0 ^( 1) f(z)dz+(1/3)=(1/4)(ln(2))^2 +(1/3) ∫_0 ^( 1) ((ln(z^2 +1))/(z^2 +1)) dz=(π/2)ln(2)−C (C is Catalan′s Const) (π/2)ln(2)−C≈^(app) 0.1728274509745...... (1/4)(ln(2))^2 +(1/3)≈^(app) 0.453447.......... ∴ ∫_0 ^( 1) f(z)dz<∫_0 ^( 1) zf(z)dz+(1/3).
$$\mathrm{first}. \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{z}\centerdot\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z}= \\ $$$${z}^{\mathrm{2}} ={u}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{z}\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{\mathrm{ln}\left({u}+\mathrm{1}\right)}{{u}+\mathrm{1}}\:\mathrm{d}{u} \\ $$$${u}+\mathrm{1}={w} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{\mathrm{ln}\left({u}+\mathrm{1}\right)}{{u}+\mathrm{1}}\:\mathrm{d}{u}=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\mathrm{ln}\left(\mathrm{2}\right)} \:{w}\:\mathrm{d}{w}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{ln}\left(\mathrm{2}\right)\right)^{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{f}\left({z}\right)\mathrm{d}{z}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{ln}\left(\mathrm{2}\right)\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}\: \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z}=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)−\mathcal{C}\:\left(\mathcal{C}\:\mathrm{is}\:\mathrm{Catalan}'\mathrm{s}\:\mathrm{Const}\right) \\ $$$$\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)−\mathcal{C}\overset{\mathrm{app}} {\approx}\mathrm{0}.\mathrm{1728274509745}…… \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{ln}\left(\mathrm{2}\right)\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}\overset{\mathrm{app}} {\approx}\:\mathrm{0}.\mathrm{453447}………. \\ $$$$\therefore\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{f}\left({z}\right)\mathrm{d}{z}<\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{zf}\left({z}\right)\mathrm{d}{z}+\frac{\mathrm{1}}{\mathrm{3}}. \\ $$

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