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Question Number 139290 by mathdanisur last updated on 25/Apr/21

∫_( 0) ^( ∞) ((sin(x^2 −arctan((1/x^2 ))))/( (√(1 + x^4 )))) dx

$$\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{{sin}\left({x}^{\mathrm{2}} −{arctan}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\right)}{\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{4}} }}\:{dx} \\ $$

Answered by Ar Brandon last updated on 25/Apr/21

u=(1/x) ⇒du=−(dx/x^2 )=−u^2 dx I=∫_0 ^∞ ((sin((1/u^2 )−arctan(u^2 )))/( (√(1+u^4 ))))du 2I=∫_0 ^∞ ((2sin[(x^2 /2)+(1/(2x^2 ))−(1/2)(arctan((1/x^2 ))+arctan(x^2 ))]cos[(x^2 /2)−(1/(2x^2 ))+(1/2)((arctan(x^2 )−arctan((1/x^2 )))])/( (√(1+x^4 ))))dx ...

$$\mathrm{u}=\frac{\mathrm{1}}{\mathrm{x}}\:\Rightarrow\mathrm{du}=−\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} }=−\mathrm{u}^{\mathrm{2}} \mathrm{dx} \\ $$$$\mathcal{I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }−\mathrm{arctan}\left(\mathrm{u}^{\mathrm{2}} \right)\right)}{\:\sqrt{\mathrm{1}+\mathrm{u}^{\mathrm{4}} }}\mathrm{du} \\ $$$$\mathrm{2}\mathcal{I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2sin}\left[\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)+\mathrm{arctan}\left(\mathrm{x}^{\mathrm{2}} \right)\right)\right]\mathrm{cos}\left[\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\left(\left(\mathrm{arctan}\left(\mathrm{x}^{\mathrm{2}} \right)−\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)\right)\right]\right.}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }}\mathrm{dx} \\ $$$$... \\ $$

Commented by mathdanisur last updated on 25/Apr/21

Sir, how is the continuation...

$${Sir},\:{how}\:{is}\:{the}\:{continuation}... \\ $$

Answered by mathmax by abdo last updated on 26/Apr/21

Φ =∫_0 ^∞ ((sin(x^2 −arctan((1/x^2 )))/( (√(1+x^4 ))))dx ⇒Φ =_((1/x)=t) −∫_0 ^∞ ((sin((1/t^2 )−arctan(t^2 )))/( (√(1+(1/t^4 )))))(−(dt/t^2 )) =∫_0 ^∞ ((sin((1/t^2 )−arctan(t^2 )))/( (√(1+t^4 ))))dt =∫_0 ^∞ ((sin((1/t^2 ))cos(arctan(t^2 ))−cos((1/t^2 ))sinarctan(t^2 ))/( (√(1+t^4 ))))dt =∫_0 ^∞ (((1/( (√(1+t^4 ))))sin((1/t^2 )))/( (√(1+t^4 ))))dt−∫_0 ^∞ (((t^2 /( (√(1+t^4 ))))cos((1/t^2 )))/( (√(1+t^4 ))))dt =∫_0 ^∞ ((sin((1/t^2 )))/((1+t^4 )))dt−∫_0 ^∞ ((t^2 cos((1/t^2 )))/(1+t^4 ))dt =Φ_1 −Φ_2 Φ_1 =(1/2) Im(∫_(−∞) ^(+∞) (e^(i/t^2 ) /(1+t^4 ))dt) let w(z) =(e^(i/z^2 ) /(z^4 +1)) ⇒ w(z)=(e^(i/z^2 ) /((z^2 −i)(z^2 +i))) =(e^(i/z^2 ) /((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) ∫_R w(z)dz =2iπ{ Res(w,e^((iπ)/4) ) +Res(w,−e^(−((iπ)/4)) )} Res(w,e^((iπ)/4) ) =(e^(i(e^((iπ)/4) )^(−2) ) /(2e^((iπ)/4) (2i))) =(1/(4i))e^(−((iπ)/4)) .e^(i(−i)) =(e/(4i))e^(−((iπ)/4)) Res(w,−e^(−((iπ)/4)) ) =(e^(i(−e^(−((iπ)/4)) )^(−2) ) /(−2e^(−((iπ)/4)) (−2i))) =(1/(4i))e^((iπ)/4) e^(−1) ⇒ ∫_R w(z)dz =2iπ{(e/(4i))e^(−((iπ)/4)) +(e^(−1) /(4i))e^((iπ)/4) } =(π/2){ e((1/( (√2)))−(i/( (√2))))+e^(−1) ((1/( (√2)))+(i/( (√2))))} =(π/2){(e/( (√2)))+(e^(−1) /( (√2)))+i((e^(−1) /( (√2)))−(e/( (√2))))} ⇒Φ_1 =(π/4)((e^(−1) /( (√2)))−(e/( (√2)))) Φ_2 =(1/2)Re(∫_(−∞) ^(+∞) ((x^2 e^(i/x^2 ) )/(x^4 +1))dx) let Ψ(z) =((z^2 e^(i/z^2 ) )/(z^4 +1)) ⇒Ψ(z)=((z^2 e^(i/z^2 ) )/((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) residus theorem give ∫_R Ψ(z)dz =2iπ{ Res(Ψ,e^((iπ)/4) ) +Res(Ψ,−e^(−((iπ)/4)) )} Res(Ψ,e^((iπ)/4) ) =((i e^(i(e^((iπ)/4) )^(−2) ) )/(2e^((iπ)/4) (2i))) =(1/4)e^(−((iπ)/4)) .e Res(Ψ,−e^(−((iπ)/4)) ) =((−i e^(i(−e^(−((iπ)/4)) )^(−2) ) )/(−2e^(−((iπ)/4)) (−2i)))=−(1/4)e^((iπ)/4) e^(−1) ⇒∫_R Ψ(z)dz =2iπ{(e/4)e^(−((iπ)/4)) −(e^(−1) /4)e^((iπ)/4) } =((iπ)/2){ e((1/( (√2)))−(i/( (√2))))−e^(−1) ((1/( (√2)))+(i/( (√2))))} =((iπ)/2){ (e/( (√2)))−((ie)/( (√2)))−(e^(−1) /( (√2)))−((ie^(−1) )/( (√2)))} =(π/(2(√2))){ie +e−ie^(−1) +e^(−1) } ⇒Φ_2 =(π/(4(√2)))(e+e^(−1) ) Φ =Φ_1 −Φ_2 =(π/(4(√2)))( e^(−1) −e) +(π/(4(√2)))(e+e^(−1) ) =(π/(4(√2)))(2e^(−1) ) =(π/(2e(√2)))

$$\Phi\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} \:−\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)\right.}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }}\mathrm{dx}\:\Rightarrow\Phi\:=_{\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{t}} \:\:−\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }−\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\right)}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{4}} }}}\left(−\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} }\right) \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }−\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\right)}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }}\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)\mathrm{cos}\left(\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\right)−\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)\mathrm{sinarctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }}\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }}\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }}\mathrm{dt}−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\frac{\mathrm{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }}\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }}\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{4}} \right)}\mathrm{dt}−\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{2}} \:\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt} \\ $$$$=\Phi_{\mathrm{1}} −\Phi_{\mathrm{2}} \\ $$$$\Phi_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{Im}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{t}^{\mathrm{2}} }} }{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\right)\:\:\mathrm{let}\:\mathrm{w}\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{z}^{\mathrm{2}} }} }{\mathrm{z}^{\mathrm{4}} +\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{w}\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{z}^{\mathrm{2}} }} }{\left(\mathrm{z}^{\mathrm{2}} −\mathrm{i}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{i}\right)}\:=\frac{\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{z}^{\mathrm{2}} }} }{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)} \\ $$$$\int_{\mathrm{R}} \mathrm{w}\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\:\mathrm{Res}\left(\mathrm{w},\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:+\mathrm{Res}\left(\mathrm{w},−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$$\mathrm{Res}\left(\mathrm{w},\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:=\frac{\mathrm{e}^{\mathrm{i}\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{−\mathrm{2}} } }{\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(\mathrm{2i}\right)}\:=\frac{\mathrm{1}}{\mathrm{4i}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \:.\mathrm{e}^{\mathrm{i}\left(−\mathrm{i}\right)} \:=\frac{\mathrm{e}}{\mathrm{4i}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \\ $$$$\mathrm{Res}\left(\mathrm{w},−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:=\frac{\mathrm{e}^{\mathrm{i}\left(−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{−\mathrm{2}} } }{−\mathrm{2e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(−\mathrm{2i}\right)}\:=\frac{\mathrm{1}}{\mathrm{4i}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\mathrm{e}^{−\mathrm{1}} \:\Rightarrow \\ $$$$\int_{\mathrm{R}} \mathrm{w}\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\frac{\mathrm{e}}{\mathrm{4i}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} +\frac{\mathrm{e}^{−\mathrm{1}} }{\mathrm{4i}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\left\{\:\mathrm{e}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{e}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)\right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\left\{\frac{\mathrm{e}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{e}^{−\mathrm{1}} }{\:\sqrt{\mathrm{2}}}+\mathrm{i}\left(\frac{\mathrm{e}^{−\mathrm{1}} }{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{e}}{\:\sqrt{\mathrm{2}}}\right)\right\}\:\Rightarrow\Phi_{\mathrm{1}} =\frac{\pi}{\mathrm{4}}\left(\frac{\mathrm{e}^{−\mathrm{1}} }{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{e}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$\Phi_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{x}^{\mathrm{2}} \:\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{x}^{\mathrm{2}} }} }{\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}}\mathrm{dx}\right)\:\mathrm{let}\:\Psi\left(\mathrm{z}\right)\:=\frac{\mathrm{z}^{\mathrm{2}} \:\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{z}^{\mathrm{2}} }} }{\mathrm{z}^{\mathrm{4}} \:+\mathrm{1}} \\ $$$$\Rightarrow\Psi\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{2}} \:\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{z}^{\mathrm{2}} }} }{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)} \\ $$$$\mathrm{residus}\:\mathrm{theorem}\:\mathrm{give} \\ $$$$\int_{\mathrm{R}} \Psi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\:\mathrm{Res}\left(\Psi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:+\mathrm{Res}\left(\Psi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$$\mathrm{Res}\left(\Psi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:=\frac{\mathrm{i}\:\mathrm{e}^{\mathrm{i}\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{−\mathrm{2}} } }{\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(\mathrm{2i}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} .\mathrm{e} \\ $$$$\mathrm{Res}\left(\Psi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:=\frac{−\mathrm{i}\:\mathrm{e}^{\mathrm{i}\left(−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{−\mathrm{2}} } }{−\mathrm{2e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(−\mathrm{2i}\right)}=−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \mathrm{e}^{−\mathrm{1}} \\ $$$$\Rightarrow\int_{\mathrm{R}} \Psi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\frac{\mathrm{e}}{\mathrm{4}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} −\frac{\mathrm{e}^{−\mathrm{1}} }{\mathrm{4}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right\} \\ $$$$=\frac{\mathrm{i}\pi}{\mathrm{2}}\left\{\:\mathrm{e}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)−\mathrm{e}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)\right\} \\ $$$$=\frac{\mathrm{i}\pi}{\mathrm{2}}\left\{\:\frac{\mathrm{e}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{ie}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{e}^{−\mathrm{1}} }{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{ie}^{−\mathrm{1}} }{\:\sqrt{\mathrm{2}}}\right\} \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\mathrm{ie}\:+\mathrm{e}−\mathrm{ie}^{−\mathrm{1}} \:+\mathrm{e}^{−\mathrm{1}} \right\}\:\Rightarrow\Phi_{\mathrm{2}} =\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\left(\mathrm{e}+\mathrm{e}^{−\mathrm{1}} \right) \\ $$$$\Phi\:=\Phi_{\mathrm{1}} −\Phi_{\mathrm{2}} =\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\left(\:\mathrm{e}^{−\mathrm{1}} −\mathrm{e}\right)\:+\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\left(\mathrm{e}+\mathrm{e}^{−\mathrm{1}} \right) \\ $$$$=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\left(\mathrm{2e}^{−\mathrm{1}} \right)\:=\frac{\pi}{\mathrm{2e}\sqrt{\mathrm{2}}}\: \\ $$

Commented by mathdanisur last updated on 26/Apr/21

Thankyou Sir, but the answer is not

$${Thankyou}\:{Sir},\:{but}\:{the}\:{answer}\:{is}\:{not}\: \\ $$

Commented by mathmax by abdo last updated on 26/Apr/21

show your work sir..

$$\mathrm{show}\:\mathrm{your}\:\mathrm{work}\:\mathrm{sir}.. \\ $$

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