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Question Number 192407 by mnjuly1970 last updated on 17/May/23

find the value of the following integral χ = ∫_0 ^( ∞) (( ln^( 2) (x ))/(1+ x^( 2) )) dx = ? −−−−−−−−−

$$ \\ $$$$\:\:\:\:\:{find}\:\:{the}\:{value}\:{of}\:{the}\:{following}\:{integral} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\chi\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:\mathrm{ln}^{\:\mathrm{2}} \left({x}\:\right)}{\mathrm{1}+\:{x}^{\:\mathrm{2}} }\:{dx}\:=\:?\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:−−−−−−−−− \\ $$

Answered by ARUNG_Brandon_MBU last updated on 17/May/23

χ=∫_0 ^∞ ((ln^2 x)/(1+x^2 ))dx=(1/8)∫_0 ^∞ ((t^(−(1/2)) ln^2 t)/(1+t))dt =(∂^2 /∂α^2 )∣_(α=(1/2)) (1/8)∫_0 ^∞ (t^(α−1) /(1+t))dt=(∂^2 /∂α^2 )∣_(α=(1/2)) (1/8)β(α, 1−α) =(∂^2 /∂α^2 )∣_(α=(1/2)) (1/8)∙(π/(sin(πα)))=(∂/∂α)∣_(α=(1/2)) −(1/8)∙((π^2 cos(πα))/(sin^2 (πα))) =−(π^3 /8)∣_(α=(1/2)) (−(1/(sin(πα)))−((2cos^2 (πα))/(sin^3 (πα))))=(π^3 /8)

$$\chi=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}^{\mathrm{2}} {x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{ln}^{\mathrm{2}} {t}}{\mathrm{1}+{t}}{dt} \\ $$$$\:\:\:=\frac{\partial^{\mathrm{2}} }{\partial\alpha^{\mathrm{2}} }\mid_{\alpha=\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\alpha−\mathrm{1}} }{\mathrm{1}+{t}}{dt}=\frac{\partial^{\mathrm{2}} }{\partial\alpha^{\mathrm{2}} }\mid_{\alpha=\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{8}}\beta\left(\alpha,\:\mathrm{1}−\alpha\right) \\ $$$$\:\:\:=\frac{\partial^{\mathrm{2}} }{\partial\alpha^{\mathrm{2}} }\mid_{\alpha=\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{\pi}{\mathrm{sin}\left(\pi\alpha\right)}=\frac{\partial}{\partial\alpha}\mid_{\alpha=\frac{\mathrm{1}}{\mathrm{2}}} −\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{\pi^{\mathrm{2}} \mathrm{cos}\left(\pi\alpha\right)}{\mathrm{sin}^{\mathrm{2}} \left(\pi\alpha\right)} \\ $$$$\:\:\:=−\frac{\pi^{\mathrm{3}} }{\mathrm{8}}\mid_{\alpha=\frac{\mathrm{1}}{\mathrm{2}}} \left(−\frac{\mathrm{1}}{\mathrm{sin}\left(\pi\alpha\right)}−\frac{\mathrm{2cos}^{\mathrm{2}} \left(\pi\alpha\right)}{\mathrm{sin}^{\mathrm{3}} \left(\pi\alpha\right)}\right)=\frac{\pi^{\mathrm{3}} }{\mathrm{8}} \\ $$

Commented by mnjuly1970 last updated on 17/May/23

so nice solution Sir thank you very much.

$$\:\:{so}\:{nice}\:{solution}\:{Sir} \\ $$$$\:\:{thank}\:{you}\:{very}\:{much}. \\ $$

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