Menu Close

calculate-0-cos-x-2-x-4-x-2-1-dx-

Tinku Tara 0 Comments
Pin




Question Number 76194 by abdomathmax last updated on 25/Dec/19
calculate ∫_0 ^∞ ((cos(x^2 ))/(x^4 −x^2 +1))dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$
Commented by abdomathmax last updated on 25/Dec/19
let A =∫_0 ^∞ ((cos(x^2 ))/(x^4 −x^2 +1))dx ⇒A =(1/2) ∫_(−∞) ^(+∞) ((cos(x^2 ))/(x^4 −x^2 +1))dx ⇒2A =Re (∫_(−∞) ^(+∞) (e^(ix^2 ) /(x^4 −x^2 +1))dx) let W(z) =(e^(iz^2 ) /(z^4 −z^2 +1)) poles of W? z^4 −z^2 +1 =0 ⇒t^2 −t +1 =0 with t=z^2 Δ=1−4=−3 ⇒t_1 =((1+(√3))/2) and t_2 =((1−(√3))/2) t_1 =e^((iπ)/3) and t_2 =e^((iπ)/3) ⇒W(z) =(e^(iz^2 ) /((z^2 −e^((iπ)/3) )(z^2 −e^(−((iπ)/3)) ))) =(e^(iz^2 ) /((z−e^((iπ)/6) )(z+e^((iπ)/6) )(z−e^(−((iπ)/6)) )(z+e^(−((iπ)/6)) ))) ∫_(−∞) ^(+∞) W(z)dz =2iπ { Res(W,e^((iπ)/6) ) +Res(W,−e^(−((iπ)/6)) )} we have Res(W, e^((iπ)/6) ) =(e^(i(((iπ)/3))) /(2e^((iπ)/6) (e^((iπ)/3) −e^(−((iπ)/3)) ))) =((e^(−(π/3)) e^(−((iπ)/6)) )/(4 sin((π/3)))) =((e^(−(π/3)) e^(−((iπ)/6)) )/(4((√3)/2))) =(e^(−(π/3)) /(2(√3))) e^(−((iπ)/6)) Res(W,−e^(−((iπ)/6)) ) =(e^(i(−((iπ)/3))) /((−2 e^(−((iπ)/6)) )( e^(−((iπ)/3)) −e^((iπ)/3) ))) =(e^(π/3) /(4 sin((π/3))))e^((iπ)/6) =((e^(π/3) e^((iπ)/3) )/(4 ((√3)/2))) =((e^(π/3) e^((iπ)/6) )/(2(√3))) ⇒ ∫_(−∞) ^(+∞) W(z)dz =((2iπ)/(2(√3))){ e^(−(π/3)) e^(−((iπ)/6)) +e^(π/3) e^((iπ)/6) } =((iπ)/( (√3))){ e^(−(π/3)) ( ((√3)/2)−(i/2))+e^(π/3) (((√3)/2)+(i/2))} =(π/( (√3))){i((√3)/2) e^(−(π/3)) +(1/2)e^(−(π/3)) +i((√3)/2)e^(π/3) −(1/2)e^(π/3) } ⇒ 2A =−(π/(2(√3)))( e^(π/3) −e^(−(π/3)) ) ⇒ A =−(π/(4(√3)))( e^(π/3) −e^(−(π/3)) )=(π/(4(√3)))( e^(−(π/3)) −e^(π/3) ) × ×
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:\Rightarrow{A}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$$\Rightarrow\mathrm{2}{A}\:={Re}\:\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{ix}^{\mathrm{2}} } }{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}{dx}\right)\:{let} \\ $$$${W}\left({z}\right)\:=\frac{{e}^{{iz}^{\mathrm{2}} } }{{z}^{\mathrm{4}} −{z}^{\mathrm{2}} \:+\mathrm{1}}\:\:{poles}\:{of}\:{W}? \\ $$$${z}^{\mathrm{4}} −{z}^{\mathrm{2}} \:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} −{t}\:+\mathrm{1}\:=\mathrm{0}\:\:{with}\:{t}={z}^{\mathrm{2}} \\ $$$$\Delta=\mathrm{1}−\mathrm{4}=−\mathrm{3}\:\Rightarrow{t}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\:{and}\:{t}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${t}_{\mathrm{1}} ={e}^{\frac{{i}\pi}{\mathrm{3}}} \:\:{and}\:{t}_{\mathrm{2}} ={e}^{\frac{{i}\pi}{\mathrm{3}}} \:\Rightarrow{W}\left({z}\right)\:=\frac{{e}^{{iz}^{\mathrm{2}} } }{\left({z}^{\mathrm{2}} −{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}^{\mathrm{2}} −{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)} \\ $$$$=\frac{{e}^{{iz}^{\mathrm{2}} } }{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left({W},{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)\:+{Res}\left({W},−{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)\right\} \\ $$$${we}\:{have}\:{Res}\left({W},\:{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)\:=\frac{{e}^{{i}\left(\frac{{i}\pi}{\mathrm{3}}\right)} }{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{6}}} \left({e}^{\frac{{i}\pi}{\mathrm{3}}} −{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)} \\ $$$$=\frac{{e}^{−\frac{\pi}{\mathrm{3}}} \:{e}^{−\frac{{i}\pi}{\mathrm{6}}} }{\mathrm{4}\:{sin}\left(\frac{\pi}{\mathrm{3}}\right)}\:=\frac{{e}^{−\frac{\pi}{\mathrm{3}}} \:{e}^{−\frac{{i}\pi}{\mathrm{6}}} }{\mathrm{4}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\frac{{e}^{−\frac{\pi}{\mathrm{3}}} }{\mathrm{2}\sqrt{\mathrm{3}}}\:{e}^{−\frac{{i}\pi}{\mathrm{6}}} \\ $$$${Res}\left({W},−{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)\:=\frac{{e}^{{i}\left(−\frac{{i}\pi}{\mathrm{3}}\right)} }{\left(−\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)\left(\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} −{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)} \\ $$$$=\frac{{e}^{\frac{\pi}{\mathrm{3}}} }{\mathrm{4}\:{sin}\left(\frac{\pi}{\mathrm{3}}\right)}{e}^{\frac{{i}\pi}{\mathrm{6}}} \:=\frac{{e}^{\frac{\pi}{\mathrm{3}}} \:{e}^{\frac{{i}\pi}{\mathrm{3}}} }{\mathrm{4}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\frac{{e}^{\frac{\pi}{\mathrm{3}}} \:{e}^{\frac{{i}\pi}{\mathrm{6}}} }{\mathrm{2}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\frac{\mathrm{2}{i}\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\left\{\:{e}^{−\frac{\pi}{\mathrm{3}}} \:{e}^{−\frac{{i}\pi}{\mathrm{6}}} \:+{e}^{\frac{\pi}{\mathrm{3}}} \:{e}^{\frac{{i}\pi}{\mathrm{6}}} \right\} \\ $$$$=\frac{{i}\pi}{\:\sqrt{\mathrm{3}}}\left\{\:{e}^{−\frac{\pi}{\mathrm{3}}} \left(\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{{i}}{\mathrm{2}}\right)+{e}^{\frac{\pi}{\mathrm{3}}} \left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{3}}}\left\{{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{e}^{−\frac{\pi}{\mathrm{3}}} \:+\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\pi}{\mathrm{3}}} \:+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{e}^{\frac{\pi}{\mathrm{3}}} −\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{\pi}{\mathrm{3}}} \right\}\:\Rightarrow \\ $$$$\mathrm{2}{A}\:=−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\left(\:{e}^{\frac{\pi}{\mathrm{3}}} \:−{e}^{−\frac{\pi}{\mathrm{3}}} \right)\:\Rightarrow \\ $$$${A}\:=−\frac{\pi}{\mathrm{4}\sqrt{\mathrm{3}}}\left(\:{e}^{\frac{\pi}{\mathrm{3}}} \:−{e}^{−\frac{\pi}{\mathrm{3}}} \right)=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{3}}}\left(\:{e}^{−\frac{\pi}{\mathrm{3}}} −{e}^{\frac{\pi}{\mathrm{3}}} \right) \\ $$$$× \\ $$$$× \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *