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Question Number 35215 by abdo mathsup 649 cc last updated on 16/May/18

find the value of ∫_(−∞) ^(+∞) ((cosx +cos(2x))/(x^2 +9))dx

$${find}\:{the}\:{value}\:{of}\:\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cosx}\:+{cos}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{2}} \:+\mathrm{9}}{dx} \\ $$

Commented by abdo mathsup 649 cc last updated on 23/May/18

let put I = ∫_(−∞) ^(+∞) ((cosx +cos(2x))/(x^2 +9))dx = Re( ∫_(−∞) ^(+∞) ((e^(ix) + e^(i2x) )/(x^2 +9))dx) let consider the complex function ϕ(z) = ((e^(iz) +e^(i2z) )/(z^2 +9)) ϕ(z) = ((e^(iz) +e^(i2z) )/((z−3i)(z+3i))) so the poles of ϕ are 3i and −3i(simples) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,3i) Res(ϕ,3i) = ((e^(i(3i)) +e^(2i(3i)) )/(6i)) = ((e^(−3) + e^(−6) )/(6i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((e^(−3) +e^(−6) )/(6i)) =(π/3){ e^(−3) +e^(−6) } ⇒ ★ I = (π/3){ e^(−3) +e^(−6) }★

$${let}\:{put}\:{I}\:\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cosx}\:+{cos}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{2}} \:+\mathrm{9}}{dx} \\ $$$$=\:{Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{ix}} \:\:+\:{e}^{{i}\mathrm{2}{x}} }{{x}^{\mathrm{2}} \:+\mathrm{9}}{dx}\right)\:{let}\:{consider}\:{the}\:{complex} \\ $$$${function}\:\varphi\left({z}\right)\:=\:\frac{{e}^{{iz}} \:\:+{e}^{{i}\mathrm{2}{z}} }{{z}^{\mathrm{2}} \:+\mathrm{9}} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{{iz}} \:+{e}^{{i}\mathrm{2}{z}} }{\left({z}−\mathrm{3}{i}\right)\left({z}+\mathrm{3}{i}\right)}\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$$\mathrm{3}{i}\:{and}\:−\mathrm{3}{i}\left({simples}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\mathrm{3}{i}\right) \\ $$$${Res}\left(\varphi,\mathrm{3}{i}\right)\:=\:\frac{{e}^{{i}\left(\mathrm{3}{i}\right)} \:+{e}^{\mathrm{2}{i}\left(\mathrm{3}{i}\right)} }{\mathrm{6}{i}}\:=\:\frac{{e}^{−\mathrm{3}} \:+\:{e}^{−\mathrm{6}} }{\mathrm{6}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{−\mathrm{3}} \:+{e}^{−\mathrm{6}} }{\mathrm{6}{i}}\:=\frac{\pi}{\mathrm{3}}\left\{\:{e}^{−\mathrm{3}} \:+{e}^{−\mathrm{6}} \right\}\:\Rightarrow \\ $$$$\bigstar\:\:{I}\:\:=\:\frac{\pi}{\mathrm{3}}\left\{\:{e}^{−\mathrm{3}} \:+{e}^{−\mathrm{6}} \right\}\bigstar \\ $$

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