Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 136401 by mathmax by abdo last updated on 21/Mar/21

find ∫_0 ^∞ ((arctan(x^2 ))/(x^4 +1))dx

$$\mathrm{find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{arctan}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}}\mathrm{dx} \\ $$

Answered by Dwaipayan Shikari last updated on 21/Mar/21

I(α)=∫_0 ^∞ ((tan^(−1) (αx^2 ))/(x^4 +1))dx I′(α)=∫_0 ^∞ (x^2 /((1+α^2 x^4 )(x^4 +1)))dx =(1/(α^2 −1))∫_0 ^∞ (1/(x^2 (1+x^4 )))−(1/(x^2 (1+α^2 x^4 )))dx =(1/(α^2 −1))∫_0 ^∞ (1/x^2 )−(x^2 /(1+x^4 ))−(1/x^2 )+((α^2 x^2 )/(1+α^2 x^4 ))dx α^2 x^4 =t⇒4α^2 x^3 =(dt/dx) =−(1/(4(α^2 −1)))∫_0 ^∞ (u^(−(1/4)) /((1+u)))du+(1/(4(α^2 −1)))∫_0 ^∞ (((√α)t^(−(1/4)) )/((1+t)))dt =−(((√2)π)/(4(α^2 −1)))+((√(2α))/(4(α^2 −1)))π I(α)=(π/(2(√2)))∫(((√α)−1)/(α^2 −1))dα α=u^2 =(π/( (√2)))∫((u^2 −u)/(u^4 −1))du=(π/( (√2)))∫(u/((u+1)(u^2 +1)))du=(π/( 2(√2)))∫(u/(1+u))−((u^2 −u)/(1+u^2 ))du I(α)=(π/(2(√2)))tan^(−1) (u)+(π/(4(√2)))log(((1+u^2 )/((1+u)^2 )))+C α=0 ⇒I(0)=0+C=0⇒C=0 I(α)=(π/(2(√2)))tan^(−1) (u)+(π/(4(√2)))log(((1+u^2 )/((1+u)^2 ))) I(1)=(π^2 /(8(√2)))−(π/(4(√2)))log(2)

$${I}\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−\mathrm{1}} \left(\alpha{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{4}} +\mathrm{1}}{dx} \\ $$$${I}'\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{4}} \right)\left({x}^{\mathrm{4}} +\mathrm{1}\right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\alpha^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{4}} \right)}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{4}} \right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\alpha^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\alpha^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{4}} }{dx}\:\:\:\:\:\:\:\:\:\:\:\alpha^{\mathrm{2}} {x}^{\mathrm{4}} ={t}\Rightarrow\mathrm{4}\alpha^{\mathrm{2}} {x}^{\mathrm{3}} =\frac{{dt}}{{dx}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{−\frac{\mathrm{1}}{\mathrm{4}}} }{\left(\mathrm{1}+{u}\right)}{du}+\frac{\mathrm{1}}{\mathrm{4}\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\int_{\mathrm{0}} ^{\infty} \frac{\sqrt{\alpha}{t}^{−\frac{\mathrm{1}}{\mathrm{4}}} }{\left(\mathrm{1}+{t}\right)}{dt} \\ $$$$=−\frac{\sqrt{\mathrm{2}}\pi}{\mathrm{4}\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\sqrt{\mathrm{2}\alpha}}{\mathrm{4}\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\pi \\ $$$${I}\left(\alpha\right)=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\int\frac{\sqrt{\alpha}−\mathrm{1}}{\alpha^{\mathrm{2}} −\mathrm{1}}{d}\alpha\:\:\:\:\:\:\:\alpha={u}^{\mathrm{2}} \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{2}}}\int\frac{{u}^{\mathrm{2}} −{u}}{{u}^{\mathrm{4}} −\mathrm{1}}{du}=\frac{\pi}{\:\sqrt{\mathrm{2}}}\int\frac{{u}}{\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} +\mathrm{1}\right)}{du}=\frac{\pi}{\:\mathrm{2}\sqrt{\mathrm{2}}}\int\frac{{u}}{\mathrm{1}+{u}}−\frac{{u}^{\mathrm{2}} −{u}}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$${I}\left(\alpha\right)=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left({u}\right)+\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}{log}\left(\frac{\mathrm{1}+{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }\right)+{C} \\ $$$$\alpha=\mathrm{0}\:\Rightarrow{I}\left(\mathrm{0}\right)=\mathrm{0}+{C}=\mathrm{0}\Rightarrow{C}=\mathrm{0} \\ $$$${I}\left(\alpha\right)=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left({u}\right)+\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}{log}\left(\frac{\mathrm{1}+{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }\right) \\ $$$${I}\left(\mathrm{1}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{8}\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}{log}\left(\mathrm{2}\right) \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 22/Mar/21

thanks sir

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com