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Question Number 84759 by mathmax by abdo last updated on 15/Mar/20
calculate ∫_(−∞) ^(+∞) ((arctan(2x^2 ))/(1+x^2 ))dx
$${calculate}\:\:\int_{−\infty} ^{+\infty} \:\:\frac{{arctan}\left(\mathrm{2}{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$
Answered by mind is power last updated on 17/Mar/20
let f(t)=∫_(−∞) ^(+∞) ((arctan(tx^2 ))/(1+x^2 ))dx f(0)=0 f′(t)=∫_(−∞) ^(+∞) ((x^2 dx)/((1+x^2 )(1+t^2 x^4 )))= f′(t)=2iπRes(i,(e^(i(π/4)) /( (√t))),(e^(i((3π)/4)) /( (√t)))) =2iπ.((−1)/(2i(1+t^2 )))+2iπ.((i/t)/((1+(i/t))(4(√t)e^((3iπ)/4) )))+2iπ.(((−i)/t)/((1−(i/t))4(√t)e^(i(π/4)) )) =−(π/(1+t^2 ))+((2π)/((t+i)(4(√t)e^(−((iπ)/4)) )))+((2π)/((t−i)(4(√t)e^(i(π/4)) ))) −(π/(1+t^2 ))+4π.Re((1/((t+i)(4(√t)e^(−((iπ)/4)) )))) −(π/(1+t^2 ))+4π.Re((((t−i)(((√2)/2)+i((√2)/2)))/((t^2 +1)(4(√t))))) −(π/(1+t^2 ))+π.((√2)/2).((√t)/(t^2 +1))+((π(√2))/(2(√t)(1+t^2 ))) f(t)=−πarctan(t)+(π/( (√2))){∫(((√t)dt)/(t^2 +1))+∫(dt/( (√t)(1+t^2 ))) ∫((√t)+(1/( (√t))))(dt/((1+t^2 ))) =∫((t+1)/( (√t)(1+t^2 )))dt u=(√t)⇒ =2∫((u^2 +1)/((1+u^4 )))du=∫((2(u^2 +1))/((u^2 −(√2)u+1)(u^2 +(√2)u+1)))du =∫(du/(u^2 −(√2)u+1))+∫(du/(u^2 +(√2)u+1)) =∫(du/((u−(1/( (√2))))^2 +(1/2)))+∫(du/((u+(1/( (√2))))^2 +(1/2))) =∫((2du)/(((√2)u−1)^2 +1))+∫((2du)/(((√2)u+1)^2 +1)) =(√2){tan^− ((√2)u−1)+tan^− ((√2)u+1)} f(t)=−πtan^− (t)+π{tan^− ((√(2t))−1)+tan^− ((√(2t))+1)}+c f(0)=0⇒ c=0 f(t)=−πtan^− (t)+π{tan^− ((√(2t))−1)+tan^− ((√(2t))+1) f(2)=∫((tan^− (2x^2 ))/(1+x^2 ))dx=−πtan^− (2)+π{tan^− (3)+tan^− (5)} =π{−tan^− (2)+tan^− (3)+tan^− (5)}
$${let}\:{f}\left({t}\right)=\int_{−\infty} ^{+\infty} \frac{{arctan}\left({tx}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}'\left({t}\right)=\int_{−\infty} ^{+\infty} \frac{{x}^{\mathrm{2}} {dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{4}} \right)}= \\ $$$${f}'\left({t}\right)=\mathrm{2}{i}\pi{Res}\left({i},\frac{{e}^{{i}\frac{\pi}{\mathrm{4}}} }{\:\sqrt{{t}}},\frac{{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} }{\:\sqrt{{t}}}\right) \\ $$$$=\mathrm{2}{i}\pi.\frac{−\mathrm{1}}{\mathrm{2}{i}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}+\mathrm{2}{i}\pi.\frac{\frac{{i}}{{t}}}{\left(\mathrm{1}+\frac{{i}}{{t}}\right)\left(\mathrm{4}\sqrt{{t}}{e}^{\frac{\mathrm{3}{i}\pi}{\mathrm{4}}} \right)}+\mathrm{2}{i}\pi.\frac{\frac{−{i}}{{t}}}{\left(\mathrm{1}−\frac{{i}}{{t}}\right)\mathrm{4}\sqrt{{t}}{e}^{{i}\frac{\pi}{\mathrm{4}}} } \\ $$$$=−\frac{\pi}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{2}\pi}{\left({t}+{i}\right)\left(\mathrm{4}\sqrt{{t}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}+\frac{\mathrm{2}\pi}{\left({t}−{i}\right)\left(\mathrm{4}\sqrt{{t}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)} \\ $$$$−\frac{\pi}{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{4}\pi.{Re}\left(\frac{\mathrm{1}}{\left({t}+{i}\right)\left(\mathrm{4}\sqrt{{t}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}\right) \\ $$$$−\frac{\pi}{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{4}\pi.{Re}\left(\frac{\left({t}−{i}\right)\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{4}\sqrt{{t}}\right)}\right) \\ $$$$−\frac{\pi}{\mathrm{1}+{t}^{\mathrm{2}} }+\pi.\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}.\frac{\sqrt{{t}}}{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{{t}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$${f}\left({t}\right)=−\pi{arctan}\left({t}\right)+\frac{\pi}{\:\sqrt{\mathrm{2}}}\left\{\int\frac{\sqrt{{t}}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}+\int\frac{{dt}}{\:\sqrt{{t}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\right. \\ $$$$\int\left(\sqrt{{t}}+\frac{\mathrm{1}}{\:\sqrt{{t}}}\right)\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$=\int\frac{{t}+\mathrm{1}}{\:\sqrt{{t}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}\:\:{u}=\sqrt{{t}}\Rightarrow \\ $$$$=\mathrm{2}\int\frac{{u}^{\mathrm{2}} +\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{4}} \right)}{du}=\int\frac{\mathrm{2}\left({u}^{\mathrm{2}} +\mathrm{1}\right)}{\left({u}^{\mathrm{2}} −\sqrt{\mathrm{2}}{u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} +\sqrt{\mathrm{2}}{u}+\mathrm{1}\right)}{du} \\ $$$$=\int\frac{{du}}{{u}^{\mathrm{2}} −\sqrt{\mathrm{2}}{u}+\mathrm{1}}+\int\frac{{du}}{{u}^{\mathrm{2}} +\sqrt{\mathrm{2}}{u}+\mathrm{1}} \\ $$$$=\int\frac{{du}}{\left({u}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}}+\int\frac{{du}}{\left({u}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\int\frac{\mathrm{2}{du}}{\left(\sqrt{\mathrm{2}}{u}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}+\int\frac{\mathrm{2}{du}}{\left(\sqrt{\mathrm{2}}{u}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\sqrt{\mathrm{2}}\left\{{tan}^{−} \left(\sqrt{\mathrm{2}}{u}−\mathrm{1}\right)+{tan}^{−} \left(\sqrt{\mathrm{2}}{u}+\mathrm{1}\right)\right\} \\ $$$${f}\left({t}\right)=−\pi{tan}^{−} \left({t}\right)+\pi\left\{{tan}^{−} \left(\sqrt{\mathrm{2}{t}}−\mathrm{1}\right)+{tan}^{−} \left(\sqrt{\mathrm{2}{t}}+\mathrm{1}\right)\right\}+{c} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow \\ $$$${c}=\mathrm{0} \\ $$$${f}\left({t}\right)=−\pi{tan}^{−} \left({t}\right)+\pi\left\{{tan}^{−} \left(\sqrt{\mathrm{2}{t}}−\mathrm{1}\right)+{tan}^{−} \left(\sqrt{\mathrm{2}{t}}+\mathrm{1}\right)\right. \\ $$$${f}\left(\mathrm{2}\right)=\int\frac{{tan}^{−} \left(\mathrm{2}{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=−\pi{tan}^{−} \left(\mathrm{2}\right)+\pi\left\{{tan}^{−} \left(\mathrm{3}\right)+{tan}^{−} \left(\mathrm{5}\right)\right\} \\ $$$$=\pi\left\{−{tan}^{−} \left(\mathrm{2}\right)+{tan}^{−} \left(\mathrm{3}\right)+{tan}^{−} \left(\mathrm{5}\right)\right\} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by abdomathmax last updated on 18/Mar/20
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$

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