Question-210572

Question Number 210572 by Spillover last updated on 12/Aug/24

Answered by mathmax last updated on 13/Aug/24

I=∫_0 ^∞ ((ln^2 x)/(1+x^4 ))dx changement x=t^(1/4) give I=(1/(16))∫_0 ^∞ ((ln^2 t)/(1+t))(1/4)t^((1/4)−1) dt⇒ 64 I =∫_0 ^∞ (t^((1/4)−1) /(1+t))ln^2 (t)dt let f(λ)=∫_0 ^∞ (t^(λ−1) /(1+t))dt with 0<λ<1 f^((2)) (λ)=∫_0 ^∞ ((ln^2 t .t^(λ−1) )/(1+t))dt ⇒f^((2)) ((1/4))=64I f(λ)=(π/(sin(πλ))) ⇒f^′ (λ)=−((π^2 cos(πλ))/(sin^2 (πλ))) f^(′ ′) (λ)=−π^2 ×((−πsin(πλ)sin^2 (πλ)−πcos(πλ)2sin(πλ)cos(πλ))/(sin^4 (πλ))) =−π^3 ×((sin^2 (πλ)+2cos^2 (πλ))/(sin^3 (πλ))) =−π^3 ×((1+cos^2 (πλ))/(sin^3 (πλ))) ⇒ f^((2)) ((1/4))=−π^3 ×((1+cos^2 ((π/4)))/(sin^3 ((π/4)))) =−π^3 ×((1+(1/2))/(1/(2(√2))))=−π^3 (2(√2))(3/2)=−3(√2)π^3 ⇒64 I=−3(√2)π^3 ⇒I=−((3(√2))/(64))π^3

$${I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}^{\mathrm{2}} {x}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:\:{changement}\:{x}={t}^{\frac{\mathrm{1}}{\mathrm{4}}} {give} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}^{\mathrm{2}} {t}}{\mathrm{1}+{t}}\frac{\mathrm{1}}{\mathrm{4}}{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} {dt}\Rightarrow \\ $$$$\mathrm{64}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+{t}}{ln}^{\mathrm{2}} \left({t}\right){dt} \\ $$$${let}\:{f}\left(\lambda\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\lambda−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:\:{with}\:\mathrm{0}<\lambda<\mathrm{1} \\ $$$${f}^{\left(\mathrm{2}\right)} \left(\lambda\right)=\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{\mathrm{2}} {t}\:.{t}^{\lambda−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\Rightarrow{f}^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{64}{I} \\ $$$${f}\left(\lambda\right)=\frac{\pi}{{sin}\left(\pi\lambda\right)}\:\Rightarrow{f}^{'} \left(\lambda\right)=−\frac{\pi^{\mathrm{2}} {cos}\left(\pi\lambda\right)}{{sin}^{\mathrm{2}} \left(\pi\lambda\right)} \\ $$$${f}^{'\:'} \left(\lambda\right)=−\pi^{\mathrm{2}} ×\frac{−\pi{sin}\left(\pi\lambda\right){sin}^{\mathrm{2}} \left(\pi\lambda\right)−\pi{cos}\left(\pi\lambda\right)\mathrm{2}{sin}\left(\pi\lambda\right){cos}\left(\pi\lambda\right)}{{sin}^{\mathrm{4}} \left(\pi\lambda\right)} \\ $$$$=−\pi^{\mathrm{3}} ×\frac{{sin}^{\mathrm{2}} \left(\pi\lambda\right)+\mathrm{2}{cos}^{\mathrm{2}} \left(\pi\lambda\right)}{{sin}^{\mathrm{3}} \left(\pi\lambda\right)} \\ $$$$=−\pi^{\mathrm{3}} ×\frac{\mathrm{1}+{cos}^{\mathrm{2}} \left(\pi\lambda\right)}{{sin}^{\mathrm{3}} \left(\pi\lambda\right)}\:\Rightarrow \\ $$$${f}^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=−\pi^{\mathrm{3}} ×\frac{\mathrm{1}+{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)}{{sin}^{\mathrm{3}} \left(\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=−\pi^{\mathrm{3}} ×\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}}=−\pi^{\mathrm{3}} \left(\mathrm{2}\sqrt{\mathrm{2}}\right)\frac{\mathrm{3}}{\mathrm{2}}=−\mathrm{3}\sqrt{\mathrm{2}}\pi^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{64}\:{I}=−\mathrm{3}\sqrt{\mathrm{2}}\pi^{\mathrm{3}} \:\Rightarrow{I}=−\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{64}}\pi^{\mathrm{3}} \\ $$

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