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Question Number 33735 by prof Abdo imad last updated on 22/Apr/18

calculate ∫_0 ^∞ ((cos(2x)dx)/((x^2 +1)( 2x^2 +3))) .

$${calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{cos}\left(\mathrm{2}{x}\right){dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\:\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{3}\right)}\:. \\ $$

Commented by prof Abdo imad last updated on 25/Apr/18

let put I = ∫_0 ^∞ ((cos(2x))/((x^2 +1)(2x^2 +3)))dx 2I = ∫_(−∞) ^(+∞) ((cos(2x))/((x^2 +1)(2x^2 +3)))dx =Re( ∫_(−∞) ^(+∞) (e^(i2x) /((x^2 +1)(2x^2 +3)))dx) let consider the complex function ϕ(z)= (e^(i2z) /((z^2 +1)(2z^2 +3))) ϕ(z) = (e^(i2z) /(2(z^2 +1)(z^2 +(3/2)))) = (e^(i2z) /(2(z−i)(z+i)(z−i(√((3/2))(z +i((√3)/2)))))) so the poles of ϕ are i,−i,i(√(3/2)) ,−i(√(3/2)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (Res(ϕ,i) +Res(ϕ,i(√(3/2)) )) Res(ϕ,i) = (e^(−2) /(2(2i)(−1+(3/2)))) = (e^(−2) /(4i.(1/2))) = (e^(−2) /(2i)) Res(ϕ,i(√(3/2))) = (e^(−2(√(3/2))) /(2(−(3/2)+1)(2i(√(3/2))))) = (e^(−(√6)) /(−2i((√3)/(√2)))) =−(e^(−(√6)) /(i(√6))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ( (e^(−2) /(2i)) −(e^(−(√6)) /(i(√6)))) = π e^(−2) −((2π)/(√6)) e^(−(√6)) ⇒ I = (π/2) e^(−2) −(π/(√6)) e^(−(√6)) .

$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)}{dx} \\ $$$$\mathrm{2}{I}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)}{dx} \\ $$$$={Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{i}\mathrm{2}{x}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)}{dx}\right)\:{let}\:{consider}\:{the} \\ $$$${complex}\:{function}\:\varphi\left({z}\right)=\:\frac{{e}^{{i}\mathrm{2}{z}} }{\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}{z}^{\mathrm{2}} +\mathrm{3}\right)} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\mathrm{2}{z}} }{\mathrm{2}\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left({z}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}\right)}\:=\:\:\frac{{e}^{{i}\mathrm{2}{z}} }{\mathrm{2}\left({z}−{i}\right)\left({z}+{i}\right)\left({z}−{i}\sqrt{\left.\frac{\mathrm{3}}{\mathrm{2}}\right)\left({z}\:+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}\right.} \\ $$$${so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{i},−{i},{i}\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\:,−{i}\sqrt{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left({Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,{i}\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\:\right)\right) \\ $$$${Res}\left(\varphi,{i}\right)\:=\:\:\:\frac{{e}^{−\mathrm{2}} }{\mathrm{2}\left(\mathrm{2}{i}\right)\left(−\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\:=\:\frac{{e}^{−\mathrm{2}} }{\mathrm{4}{i}.\frac{\mathrm{1}}{\mathrm{2}}}\:=\:\frac{{e}^{−\mathrm{2}} }{\mathrm{2}{i}} \\ $$$${Res}\left(\varphi,{i}\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\right)\:=\:\:\frac{{e}^{−\mathrm{2}\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}} }{\mathrm{2}\left(−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}\right)\left(\mathrm{2}{i}\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\right)} \\ $$$$=\:\:\:\:\frac{{e}^{−\sqrt{\mathrm{6}}} }{−\mathrm{2}{i}\frac{\sqrt{\mathrm{3}}}{\sqrt{\mathrm{2}}}}\:=−\frac{{e}^{−\sqrt{\mathrm{6}}} }{{i}\sqrt{\mathrm{6}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:\:\frac{{e}^{−\mathrm{2}} }{\mathrm{2}{i}}\:−\frac{{e}^{−\sqrt{\mathrm{6}}} }{{i}\sqrt{\mathrm{6}}}\right) \\ $$$$=\:\pi\:{e}^{−\mathrm{2}} \:−\frac{\mathrm{2}\pi}{\sqrt{\mathrm{6}}}\:{e}^{−\sqrt{\mathrm{6}}} \:\:\:\:\:\Rightarrow\:\: \\ $$$${I}\:=\:\frac{\pi}{\mathrm{2}}\:{e}^{−\mathrm{2}} \:\:−\frac{\pi}{\sqrt{\mathrm{6}}}\:{e}^{−\sqrt{\mathrm{6}}} \:. \\ $$

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