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Question Number 48042 by maxmathsup by imad last updated on 18/Nov/18

find the value of ∫_(−∞) ^(+∞) ((2x+1)/((x^2 +i)(x^2 +4)))dx (i^2 =−1)

$${find}\:{the}\:{value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{2}{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} +{i}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx}\:\:\:\left({i}^{\mathrm{2}} =−\mathrm{1}\right) \\ $$

Commented by maxmathsup by imad last updated on 20/Nov/18

let A =∫_(−∞) ^(+∞) ((2x+1)/((x^2 +i)(x^2 +4)))dx and the complex function ϕ(z) =((2z+1)/((z^2 +i)(z^2 +4))) ⇒ϕ(z)=((2z+1)/((z^2 −((√(−i)))^2 )(z^2 −(2i)^2 ))) =((2z+1)/((z−(√(−i)))(z+(√(−i)))(z−2i)(z+2i))) =((2z+1)/((z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) )(z−2i)(z+2i))) tbe ples of are +^− e^(−((iπ)/4)) and +^− (2i) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,−e^(−((iπ)/4)) ) +Res(ϕ,2i)} Res(ϕ,−e^(−((iπ)/4)) ) =((−2e^(−((iπ)/4)) +1)/(−2 e^(−((iπ)/4)) (e^(−((iπ)/2)) +4))) =((2 e^(−((iπ)/4)) −1)/(2 e^(−((iπ)/4)) (4−i))) Res(ϕ,2i) =((4i +1)/((4i)((2i)^2 +i))) =((4i+1)/(4i{−4+i})) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ ((2 e^(−((iπ)/4)) −1)/(2 e^(−((iπ)/4)) (4−i))) +((4i+1)/(4i(−4+i)))} =A .

$${let}\:{A}\:=\int_{−\infty} ^{+\infty} \:\:\frac{\mathrm{2}{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} +{i}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx}\:{and}\:{the}\:{complex}\:{function}\: \\ $$$$\varphi\left({z}\right)\:=\frac{\mathrm{2}{z}+\mathrm{1}}{\left({z}^{\mathrm{2}} +{i}\right)\left({z}^{\mathrm{2}} +\mathrm{4}\right)}\:\Rightarrow\varphi\left({z}\right)=\frac{\mathrm{2}{z}+\mathrm{1}}{\left({z}^{\mathrm{2}} −\left(\sqrt{−{i}}\right)^{\mathrm{2}} \right)\left({z}^{\mathrm{2}} −\left(\mathrm{2}{i}\right)^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}{z}+\mathrm{1}}{\left({z}−\sqrt{−{i}}\right)\left({z}+\sqrt{−{i}}\right)\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)}\:=\frac{\mathrm{2}{z}+\mathrm{1}}{\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)} \\ $$$${tbe}\:{ples}\:{of}\:{are}\:\overset{−} {+}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:{and}\:\overset{−} {+}\left(\mathrm{2}{i}\right)\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:+{Res}\left(\varphi,\mathrm{2}{i}\right)\right\} \\ $$$${Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:=\frac{−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:+\mathrm{1}}{−\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \left({e}^{−\frac{{i}\pi}{\mathrm{2}}} \:+\mathrm{4}\right)}\:=\frac{\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} −\mathrm{1}}{\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \left(\mathrm{4}−{i}\right)} \\ $$$${Res}\left(\varphi,\mathrm{2}{i}\right)\:=\frac{\mathrm{4}{i}\:+\mathrm{1}}{\left(\mathrm{4}{i}\right)\left(\left(\mathrm{2}{i}\right)^{\mathrm{2}} +{i}\right)}\:=\frac{\mathrm{4}{i}+\mathrm{1}}{\mathrm{4}{i}\left\{−\mathrm{4}+{i}\right\}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\frac{\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} −\mathrm{1}}{\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \left(\mathrm{4}−{i}\right)}\:+\frac{\mathrm{4}{i}+\mathrm{1}}{\mathrm{4}{i}\left(−\mathrm{4}+{i}\right)}\right\}\:={A}\:. \\ $$

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