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1-calculate-I-0-dx-x-2-i-and-J-0-dx-x-2-i-2-find-the-value-of-0-dx-x-4-1-

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Question Number 43676 by maxmathsup by imad last updated on 13/Sep/18
1)calculate I = ∫_0 ^∞ (dx/(x^2 −i)) and J = ∫_0 ^∞ (dx/(x^2 +i)) 2) find the value of ∫_0 ^∞ (dx/(x^4 +1))
$$\left.\mathrm{1}\right){calculate}\:\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{i}}\:\:{and}\:\:{J}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+{i}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{1}} \\ $$
Commented by maxmathsup by imad last updated on 13/Sep/18
i^2 =−1
$${i}^{\mathrm{2}} =−\mathrm{1} \\ $$
Commented by maxmathsup by imad last updated on 15/Sep/18
we have 2I = ∫_(−∞) ^(+∞) (dx/(x^2 −i)) let ϕ(z) = (1/(z^2 −i)) we have ϕ(z) = (1/((z−(√i))(z+(√i)))) = (1/((z−e^((iπ)/4) )(z +e^((iπ)/4) ))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,e^((iπ)/4) ) = 2iπ .(1/(2 e^((iπ)/4) )) =iπ e^(−((iπ)/4)) ⇒ I =((iπ)/2) e^(−((iπ)/4)) also we have 2J =∫_(−∞) ^(+∞) (dx/(x^2 +i)) let ψ(z) =(1/(z^2 +i)) ψ(z) =(1/((z−(√(−i)))(z+(√(−i))))) =(1/((z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) and ∫_(−∞) ^(+∞) ψ(z)dz =2iπ Res(ψ,−e^(−((iπ)/4)) ) =2iπ .(1/(−2 e^(−((iπ)/4)) )) =−iπ e^((iπ)/4) ⇒J=−((iπ)/2)e^((iπ)/4) another way we see that J =I^− ⇒ J = −((iπ)/2) e^((iπ)/4) . 2) we have I −J = ∫_0 ^∞ {(1/(x^2 −i)) −(1/(x^2 +i))}dx =∫_0 ^∞ ((2i)/(x^4 +1))dx ⇒ ∫_0 ^∞ (dx/(x^4 +1)) =(1/(2i)){ I −J} =(1/(2i)){((iπ)/2) e^(−((iπ)/4)) +((iπ)/2) e^((iπ)/4) } = (π/4){ e^((iπ)/4) +e^(−((iπ)/4)) } =(π/4) (2cos((π/4))}=(π/2) .(1/( (√2))) =(π/(2(√2))) .
$${we}\:{have}\:\mathrm{2}{I}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{i}}\:\:{let}\:\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} −{i}}\:\:{we}\:{have} \\ $$$$\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\left({z}−\sqrt{{i}}\right)\left({z}+\sqrt{{i}}\right)}\:=\:\frac{\mathrm{1}}{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}\:+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:=\:\mathrm{2}{i}\pi\:.\frac{\mathrm{1}}{\mathrm{2}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:={i}\pi\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$${I}\:=\frac{{i}\pi}{\mathrm{2}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:\:\:\:\:\:{also}\:{we}\:{have}\:\mathrm{2}{J}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+{i}}\:{let}\:\psi\left({z}\right)\:=\frac{\mathrm{1}}{{z}^{\mathrm{2}} \:+{i}} \\ $$$$\psi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−\sqrt{−{i}}\right)\left({z}+\sqrt{−{i}}\right)}\:=\frac{\mathrm{1}}{\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}\:\:{and}\: \\ $$$$\int_{−\infty} ^{+\infty} \:\psi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\psi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:=\mathrm{2}{i}\pi\:.\frac{\mathrm{1}}{−\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:=−{i}\pi\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow{J}=−\frac{{i}\pi}{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \\ $$$${another}\:{way}\:{we}\:{see}\:{that}\:{J}\:=\overset{−} {{I}}\:\Rightarrow\:{J}\:=\:−\frac{{i}\pi}{\mathrm{2}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:. \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{I}\:−{J}\:=\:\int_{\mathrm{0}} ^{\infty} \:\left\{\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{i}}\:−\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+{i}}\right\}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{i}}{{x}^{\mathrm{4}} \:+\mathrm{1}}{dx}\:\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:{I}\:−{J}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\frac{{i}\pi}{\mathrm{2}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:+\frac{{i}\pi}{\mathrm{2}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right\}\:=\:\frac{\pi}{\mathrm{4}}\left\{\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right\}\:=\frac{\pi}{\mathrm{4}}\:\left(\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{4}}\right)\right\}=\frac{\pi}{\mathrm{2}}\:.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:. \\ $$

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