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Question Number 127958 by mnjuly1970 last updated on 03/Jan/21

...nice calculus... calculate Ω=∫_1 ^( ∞) ((ln(x^4 −2x^2 +2))/(x(√(x^2 −1)) )) dx=?

$$\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:{calculate} \\ $$$$\:\:\:\:\Omega=\int_{\mathrm{1}} ^{\:\infty} \frac{{ln}\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}\right)}{{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:}\:{dx}=? \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 03/Jan/21

∫_1 ^∞ ((log(x^4 −2x^2 +2))/(x(√(x^2 −1))))dx =∫_0 ^(π/2) ((log(sec^4 θ−2sec^2 θ+2))/(secθtanθ))secθtanθ =∫_0 ^(π/2) log(1−2cos^2 θ+2cos^4 θ)−4∫_0 ^(π/2) log(cosθ)dθ =∫_0 ^(π/2) log(1−(1/2)sin^2 2θ)+2πlog(2) =∫_0 ^(π/2) log(cos^2 2θ+(1/2)sin^2 2θ)+2πlog(2) =πlog(((1+(1/( (√2))))/2))+πlog(4)=πlog(2+(√2)) Lemma ∫_0 ^(π/2) log(a^2 cos^2 θ+b^2 sin^2 θ)=πlog(((a+b)/2))

$$\int_{\mathrm{1}} ^{\infty} \frac{{log}\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}\right)}{{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{log}\left({sec}^{\mathrm{4}} \theta−\mathrm{2}{sec}^{\mathrm{2}} \theta+\mathrm{2}\right)}{{sec}\theta{tan}\theta}{sec}\theta{tan}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left(\mathrm{1}−\mathrm{2}{cos}^{\mathrm{2}} \theta+\mathrm{2}{cos}^{\mathrm{4}} \theta\right)−\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({cos}\theta\right){d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \mathrm{2}\theta\right)+\mathrm{2}\pi{log}\left(\mathrm{2}\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({cos}^{\mathrm{2}} \mathrm{2}\theta+\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \mathrm{2}\theta\right)+\mathrm{2}\pi{log}\left(\mathrm{2}\right) \\ $$$$=\pi{log}\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{\mathrm{2}}\right)+\pi{log}\left(\mathrm{4}\right)=\pi{log}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right) \\ $$$${Lemma} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({a}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)=\pi{log}\left(\frac{{a}+{b}}{\mathrm{2}}\right) \\ $$

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