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Question Number 35960 by Joel579 last updated on 26/May/18
Solve using Residue Theorem I = ∫_(−∞) ^(+∞) (x^2 /(x^4 + 16)) dx
$$\mathrm{Solve}\:\mathrm{using}\:\mathrm{Residue}\:\mathrm{Theorem} \\ $$$$\mathrm{I}\:=\:\int_{−\infty} ^{+\infty} \:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} \:+\:\mathrm{16}}\:{dx} \\ $$
Commented by abdo mathsup 649 cc last updated on 26/May/18
let consider the complex functionϕ(z)=(z^2 /(z^4 +16)) ϕ(z)= (z^2 /((z^2 −4i)(z^2 +4i))) = (z^2 /((z −2(√i))(z +2(√i))( z −2(√(−i)))(z+2(√(−i))))) ϕ(z) = (z^2 /((z −2e^(i(π/4)) )(z+2e^(i(π/4)) )(z −2 e^(−i(π/4)) )( z+2 e^(−i(π/4)) ))) the poles of ϕ are 2 e^(i(π/4)) , −2 e^(i(π/4)) , 2 e^(−i(π/4)) , −2 e^(−i(π/4)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res( ϕ, 2 e^(i(π/4)) ) +Res( ϕ,−2e^(−i(π/4)) )} Res(ϕ, 2 e^(i(π/4)) ) = lim_(z→2e^(i(π/4)) ) (z −2e^(i(π/4)) )ϕ(z) = ((4 e^(i(π/2)) )/(4 e^(i(π/4)) ( 4i +4i))) = ((4i)/(4 e^(i(π/4)) (8i))) = (1/8)e^(−i(π/4)) Res(ϕ ,−2 e^(−i(π/4)) ) = lim_(z→ −2e^(−i(π/4)) ) (z+e^(−i(π/4)) )ϕ(z) = ((4 e^(−i(π/2)) )/(−4 e^(−i(π/4)) ( 4 e^(−i(π/4)) −4i))) = ((−4i)/(−4 e^(−i(π/4)) (−8i))) = −(1/8) e^(i(π/4)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (1/8)( e^(−i(π/4)) −e^(i(π/4)) )} =((−iπ)/4){ e^(i(π/4)) −e^(−i(π/4)) } = ((−iπ)/4){2isin((π/4))} = (π/2) ((√2)/2) = ((π(√2))/4) so I = ((π(√2))/4) .
$${let}\:{consider}\:{the}\:{complex}\:{function}\varphi\left({z}\right)=\frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} \:+\mathrm{16}} \\ $$$$\varphi\left({z}\right)=\:\frac{{z}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} −\mathrm{4}{i}\right)\left({z}^{\mathrm{2}} \:+\mathrm{4}{i}\right)}\: \\ $$$$=\:\frac{{z}^{\mathrm{2}} }{\left({z}\:−\mathrm{2}\sqrt{{i}}\right)\left({z}\:+\mathrm{2}\sqrt{{i}}\right)\left(\:{z}\:−\mathrm{2}\sqrt{−{i}}\right)\left({z}+\mathrm{2}\sqrt{−{i}}\right)} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} }{\left({z}\:−\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\left({z}+\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\left({z}\:−\mathrm{2}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)\left(\:{z}+\mathrm{2}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)} \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:\mathrm{2}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:,\:−\mathrm{2}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\:,\:\mathrm{2}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} ,\:−\mathrm{2}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\:\varphi,\:\mathrm{2}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\:+{Res}\left(\:\varphi,−\mathrm{2}{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left(\varphi,\:\mathrm{2}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\:=\:{lim}_{{z}\rightarrow\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{4}}} } \:\:\:\left({z}\:−\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\varphi\left({z}\right) \\ $$$$=\:\:\:\frac{\mathrm{4}\:{e}^{{i}\frac{\pi}{\mathrm{2}}} }{\mathrm{4}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\left(\:\mathrm{4}{i}\:+\mathrm{4}{i}\right)}\:=\:\frac{\mathrm{4}{i}}{\mathrm{4}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\left(\mathrm{8}{i}\right)}\:=\:\:\frac{\mathrm{1}}{\mathrm{8}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} \\ $$$${Res}\left(\varphi\:,−\mathrm{2}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)\:=\:{lim}_{{z}\rightarrow\:−\mathrm{2}{e}^{−{i}\frac{\pi}{\mathrm{4}}} } \:\:\left({z}+{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)\varphi\left({z}\right) \\ $$$$=\:\:\frac{\mathrm{4}\:{e}^{−{i}\frac{\pi}{\mathrm{2}}} }{−\mathrm{4}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \left(\:\mathrm{4}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} −\mathrm{4}{i}\right)} \\ $$$$=\:\frac{−\mathrm{4}{i}}{−\mathrm{4}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \left(−\mathrm{8}{i}\right)}\:=\:−\frac{\mathrm{1}}{\mathrm{8}}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\frac{\mathrm{1}}{\mathrm{8}}\left(\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \:−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\right\} \\ $$$$=\frac{−{i}\pi}{\mathrm{4}}\left\{\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:−{e}^{−{i}\frac{\pi}{\mathrm{4}}} \:\right\}\:=\:\frac{−{i}\pi}{\mathrm{4}}\left\{\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right\} \\ $$$$=\:\frac{\pi}{\mathrm{2}}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:\:{so}\:\:{I}\:\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:\:. \\ $$
Commented by Joel579 last updated on 26/May/18
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$
Commented by abdo mathsup 649 cc last updated on 26/May/18
nevermind sir joel.
$${nevermind}\:{sir}\:{joel}. \\ $$

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