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Question Number 133433 by greg_ed last updated on 22/Feb/21
hi, everybody ! with I=∫_0 ^∞ (1/(x^4 +1)) dx, prove that : 2I=∫_0 ^∞ ((x^2 +1)/(x^4 +1)) dx.
$$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{everybody}}\:! \\ $$$$\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{I}}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{1}}\:\boldsymbol{{dx}}, \\ $$$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\::\:\mathrm{2}\boldsymbol{\mathrm{I}}=\int_{\mathrm{0}} ^{\infty} \:\frac{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{1}}\:\boldsymbol{{dx}}. \\ $$
Answered by Dwaipayan Shikari last updated on 22/Feb/21
I=∫_0 ^∞ (1/(x^4 +1))dx=(1/4)∫_0 ^∞ (u^((1/4)−1) /((u+1)^((3/4)+(1/4)) ))du x^4 =u =(1/4)Γ((3/4))Γ((1/4))=(π/(2(√2))) Φ=∫_0 ^∞ ((x^2 +1)/(x^4 +1))dx=∫_0 ^∞ (x^2 /(x^4 +1))+I =(1/4)∫_0 ^∞ (u^((3/4)−1) /((1+u)^((3/4)+(1/4)) ))du+I =(1/4)Γ((3/4))Γ((1/4))+I=(π/(2(√2)))+I=I+I=2I
$${I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\left({u}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}} }{du}\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{4}} ={u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\Phi=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\mathrm{1}}+{I} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{1}} }{\left(\mathrm{1}+{u}\right)^{\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}} }{du}+{I} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)+{I}=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}+{I}={I}+{I}=\mathrm{2}{I} \\ $$
Answered by Dwaipayan Shikari last updated on 22/Feb/21
Φ=∫_0 ^∞ (1/((x^4 +1)))dx x=(1/t) =∫_0 ^∞ (t^2 /(t^4 +1))dt=Φ 2Φ=∫_0 ^∞ (t^2 /(t^4 +1))+(1/(1+t^4 ))dt
$$\Phi=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({x}^{\mathrm{4}} +\mathrm{1}\right)}{dx}\:\:\:\:\:{x}=\frac{\mathrm{1}}{{t}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{1}}{dt}=\Phi \\ $$$$\mathrm{2}\Phi=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$

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