Question Number 210820 by Ghisom last updated on 19/Aug/24
$$\mathrm{prove} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{arctan}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}\:{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$
Answered by BHOOPENDRA last updated on 20/Aug/24
![I=∫_0 ^∞ ((arctan(√(x^2 +2)))/((x^2 +1)((√(x^2 +2))))) dx Using Feyman integration I(α)=∫_0 ^∞ ((arctanα((√(x^2 +2))))/((x^2 +1)((√(x^2 +2)))) dx I′(α)=∫_0 ^∞ (1/((x^2 +1)((√(x^2 +2))))) ∂_𝛂 arctan(α(√(x^2 +2))) =∫_0 ^∞ (1/((x^2 +1)(α^2 x^2 +2α^2 +1))) dx = ∫_0 ^1 [∫_0 ^∞ ((1/(1+α^2 ))/((1+x^2 ))) dx− ∫_0 ^∞ ((α^2 /(1+α^2 ))/((α^2 x^2 +x^2 +(2α^2 +1)))dx]dα =∫_0 ^1 (1/((1+α^2 )))[∫_0 ^∞ (1/(1+x^2 ))dx−(α^2 /α^2 )∫_0 ^∞ (dx/(x^2 +((√((2α^2 +1)/α^2 )))^2 ))]dα =∫_0 ^1 (1/(1+α^2 ))[(π/2)−((π/2)×(α/( (√(2α^2 +1)))))] =(π/2)[∫_0 ^1 (1/(1+α^2 ))dα−∫_0 ^1 (α/((1+α^2 )((√(2α^2 +1)))))dα] =(π/2)[(π/4) −(π/(12))] =((π×2π)/(2×12)) =(π^2 /(12))](https://www.tinkutara.com/question/Q210853.png)
$${I}=\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\sqrt{\left.{x}^{\mathrm{2}} +\mathrm{2}\right)}\right.}\:{dx} \\ $$$${Using}\:{Feyman}\:{integration}\: \\ $$$${I}\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\alpha\left(\sqrt{\left.{x}^{\mathrm{2}} +\mathrm{2}\right)}\right.}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\right.}\:{dx} \\ $$$${I}'\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\right)}\:\:\partial_{\boldsymbol{\alpha}} \:{arctan}\left(\alpha\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\right) \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\alpha^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}\alpha^{\mathrm{2}} +\mathrm{1}\right)}\:{dx} \\ $$$$\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left[\int_{\mathrm{0}} ^{\infty} \frac{\frac{\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:{dx}−\:\int_{\mathrm{0}} ^{\infty} \frac{\alpha^{\mathrm{2}} /\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}{\left(\alpha^{\mathrm{2}} {x}^{\mathrm{2}} +{x}^{\mathrm{2}} +\left(\mathrm{2}\alpha^{\mathrm{2}} +\mathrm{1}\right)\right.}{dx}\right]{d}\alpha \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\left[\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}−\frac{\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}^{\mathrm{2}} +\left(\sqrt{\frac{\mathrm{2}\alpha^{\mathrm{2}} +\mathrm{1}}{\alpha^{\mathrm{2}} }}\right)^{\mathrm{2}} }\right]{d}\alpha \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }\left[\frac{\pi}{\mathrm{2}}−\left(\frac{\pi}{\mathrm{2}}×\frac{\alpha}{\:\sqrt{\mathrm{2}\alpha^{\mathrm{2}} +\mathrm{1}}}\right)\right] \\ $$$$=\frac{\pi}{\mathrm{2}}\left[\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }{d}\alpha−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\alpha}{\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\left(\sqrt{\mathrm{2}\alpha^{\mathrm{2}} +\mathrm{1}}\right)}{d}\alpha\right] \\ $$$$=\frac{\pi}{\mathrm{2}}\left[\frac{\pi}{\mathrm{4}}\:−\frac{\pi}{\mathrm{12}}\right] \\ $$$$=\frac{\pi×\mathrm{2}\pi}{\mathrm{2}×\mathrm{12}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$
Commented by Ghisom last updated on 20/Aug/24
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by BHOOPENDRA last updated on 20/Aug/24
$${This}\:{can}\:{also}\:{be}\:{solved}\:{by}\:{complex}\:{residues} \\ $$