Menu Close

calculate-0-dx-x-4-x-2-2-

Tinku Tara 0 Comments
Pin




Question Number 138973 by mathmax by abdo last updated on 20/Apr/21
calculate ∫_0 ^(∞ ) (dx/(x^4 −x^2 +2))
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty\:} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} +\mathrm{2}} \\ $$
Answered by mathmax by abdo last updated on 22/Apr/21
Φ=∫_0 ^∞ (dx/(x^4 −x^2 +2)) let f(x)=(1/(x^4 −x^2 +2)) x^4 −x^2 +2=0 →t^2 −t+2=0 →Δ=1−8=−7 ⇒t_1 =((1+i(√7))/2) ant t_2 =((1−i(√7))/2) ⇒f(x)=(1/((x^2 −t_1 )(x^2 −t_2 ))) =(1/(t_1 −t_2 ))((1/(x^2 −t_1 ))−(1/(x^2 −t_2 ))) =(1/(i(√7)))((1/(x^2 −t_1 ))−(1/(x^2 −t_2 ))) ⇒Φ=(1/(i(√7)))∫_0 ^∞ (dx/(x^2 −t_1 ))−(1/(i(√7)))∫_0 ^∞ (dx/(x^2 −t_2 )) =Φ_1 −Φ_2 Φ_1 =(1/(2i(√7)))∫_(−∞) ^(+∞) (dx/(x^2 −t_1 )) let w(z)=(1/(z^2 −((1+i(√7))/2))) we have ∣((1+i(√7))/2)∣=(1/2)(√(1+7))=((2(√2))/2)=(√2) ⇒((1+i(√7))/2)=(√2)e^(iarctan((√7))) ⇒w(z)=(1/((z−(^4 (√2))e^((i/2)arctan((√7))) )(z+(^4 (√2))e^((i/2)arctan((√7))) ))) ∫_(−∞) ^(+∞) w(z)dx =2iπ Res(w,(^4 (√2)e^((i/2)arctan((√7))) ) =2iπ.(1/(2(^4 (√2))e^((i/2)arctan((√7))) )) =((iπ)/((^4 (√2))))e^(−(i/2)arctan((√7))) ⇒ Φ_1 =(1/(2i(√7)))×((iπ)/((^4 (√2))))e^(−(i/2)arctan((√7))) =(π/(2(√7)(^4 (√2))))e^(−(i/2)arctan((√7))) Φ_2 =(1/(i(√7)))∫_0 ^∞ (dx/(x−t_2 )) =(1/(2i(√7)))×conj(∫_(−∞) ^∞ (dx/(x^2 −t_1 ))) =(1/(2i(√7)))×(−((iπ)/((^4 (√2))))e^((i/2)arctan((√7))) ) =−(π/(2(√7)(^4 (√2))))e^((i/2)arctan((√7))) ⇒ Φ=Φ_1 −Φ_2 =(π/(2(√7)(^4 (√2))))(e^((i/2)arctan((√7))) +e^(−(i/2)arctan((√7))) ) =(π/(2(√7)(^4 (√2))))(2cos(arctan((√7))) =(π/( (√7)(^4 (√2))))cos(arctan((√7))) we have cos(arctanx) =(1/( (√(1+x^2 )))) ⇒cos(arctan((√7))) =(1/( (√(1+7)))) =(1/(2(√2))) ⇒Φ =(π/( (√7)(^4 (√2))))×(1/(2(√2))) =(π/(2(√(14))(^4 (√2))))
$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} \:+\mathrm{2}} \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} +\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} +\mathrm{2}=\mathrm{0}\:\rightarrow\mathrm{t}^{\mathrm{2}} −\mathrm{t}+\mathrm{2}=\mathrm{0}\:\rightarrow\Delta=\mathrm{1}−\mathrm{8}=−\mathrm{7}\:\Rightarrow\mathrm{t}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{i}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$\mathrm{ant}\:\mathrm{t}_{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{t}_{\mathrm{1}} \right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{t}_{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\mathrm{t}_{\mathrm{1}} −\mathrm{t}_{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{t}_{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{i}\sqrt{\mathrm{7}}}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{t}_{\mathrm{2}} }\right)\:\Rightarrow\Phi=\frac{\mathrm{1}}{\mathrm{i}\sqrt{\mathrm{7}}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} −\mathrm{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{i}\sqrt{\mathrm{7}}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} −\mathrm{t}_{\mathrm{2}} } \\ $$$$=\Phi_{\mathrm{1}} −\Phi_{\mathrm{2}} \\ $$$$\Phi_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2i}\sqrt{\mathrm{7}}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} −\mathrm{t}_{\mathrm{1}} }\:\mathrm{let}\:\mathrm{w}\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{2}} −\frac{\mathrm{1}+\mathrm{i}\sqrt{\mathrm{7}}}{\mathrm{2}}} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mid\frac{\mathrm{1}+\mathrm{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\mid=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+\mathrm{7}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}=\sqrt{\mathrm{2}}\:\Rightarrow\frac{\mathrm{1}+\mathrm{i}\sqrt{\mathrm{7}}}{\mathrm{2}}=\sqrt{\mathrm{2}}\mathrm{e}^{\mathrm{iarctan}\left(\sqrt{\mathrm{7}}\right)} \\ $$$$\Rightarrow\mathrm{w}\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{z}−\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\left(\mathrm{z}+\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\mathrm{w}\left(\mathrm{z}\right)\mathrm{dx}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\mathrm{w},\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\right. \\ $$$$=\mathrm{2i}\pi.\frac{\mathrm{1}}{\mathrm{2}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)} }\:=\frac{\mathrm{i}\pi}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\mathrm{e}^{−\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)} \:\Rightarrow \\ $$$$\Phi_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2i}\sqrt{\mathrm{7}}}×\frac{\mathrm{i}\pi}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\mathrm{e}^{−\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)} \:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{7}}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\mathrm{e}^{−\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)} \\ $$$$\Phi_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{i}\sqrt{\mathrm{7}}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\mathrm{x}−\mathrm{t}_{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2i}\sqrt{\mathrm{7}}}×\mathrm{conj}\left(\int_{−\infty} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} −\mathrm{t}_{\mathrm{1}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2i}\sqrt{\mathrm{7}}}×\left(−\frac{\mathrm{i}\pi}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\:=−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{7}}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)} \:\Rightarrow \\ $$$$\Phi=\Phi_{\mathrm{1}} −\Phi_{\mathrm{2}} =\frac{\pi}{\mathrm{2}\sqrt{\mathrm{7}}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\left(\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)} +\mathrm{e}^{−\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)} \right) \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{7}}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\left(\mathrm{2cos}\left(\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)\right)\:=\frac{\pi}{\:\sqrt{\mathrm{7}}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\mathrm{cos}\left(\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)\right)\right. \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{cos}\left(\mathrm{arctanx}\right)\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\:\Rightarrow\mathrm{cos}\left(\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{7}}}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow\Phi\:=\frac{\pi}{\:\sqrt{\mathrm{7}}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{14}}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)} \\ $$
Commented by mathmax by abdo last updated on 23/Apr/21
sorry Φ=Φ_1 −Φ_2 =(π/(2(√7)(^4 (√2))))(2cos((1/2)arctan((√7))) =(π/( (√7)(^4 (√2)))).cos((1/2)arctan((√7)))
$$\mathrm{sorry}\:\:\Phi=\Phi_{\mathrm{1}} −\Phi_{\mathrm{2}} =\frac{\pi}{\mathrm{2}\sqrt{\mathrm{7}}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\left(\mathrm{2cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)\right)\right. \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{7}}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}.\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{7}}\right)\right) \\ $$

Pin

Leave a Reply

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