Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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))) .

$${we}\:{have}\:\mathrm{2}{I}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{i}}\:\:{let}\:\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} −{i}}\:\:{we}\:{have} \\ $$$$\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\left({z}−\sqrt{{i}}\right)\left({z}+\sqrt{{i}}\right)}\:=\:\frac{\mathrm{1}}{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}\:+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:=\:\mathrm{2}{i}\pi\:.\frac{\mathrm{1}}{\mathrm{2}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:={i}\pi\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$${I}\:=\frac{{i}\pi}{\mathrm{2}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:\:\:\:\:\:{also}\:{we}\:{have}\:\mathrm{2}{J}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+{i}}\:{let}\:\psi\left({z}\right)\:=\frac{\mathrm{1}}{{z}^{\mathrm{2}} \:+{i}} \\ $$$$\psi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−\sqrt{−{i}}\right)\left({z}+\sqrt{−{i}}\right)}\:=\frac{\mathrm{1}}{\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}\:\:{and}\: \\ $$$$\int_{−\infty} ^{+\infty} \:\psi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\psi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:=\mathrm{2}{i}\pi\:.\frac{\mathrm{1}}{−\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:=−{i}\pi\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow{J}=−\frac{{i}\pi}{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \\ $$$${another}\:{way}\:{we}\:{see}\:{that}\:{J}\:=\overset{−} {{I}}\:\Rightarrow\:{J}\:=\:−\frac{{i}\pi}{\mathrm{2}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:. \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{I}\:−{J}\:=\:\int_{\mathrm{0}} ^{\infty} \:\left\{\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{i}}\:−\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+{i}}\right\}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{i}}{{x}^{\mathrm{4}} \:+\mathrm{1}}{dx}\:\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:{I}\:−{J}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\frac{{i}\pi}{\mathrm{2}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:+\frac{{i}\pi}{\mathrm{2}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right\}\:=\:\frac{\pi}{\mathrm{4}}\left\{\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right\}\:=\frac{\pi}{\mathrm{4}}\:\left(\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{4}}\right)\right\}=\frac{\pi}{\mathrm{2}}\:.\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com