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Question Number 34021 by prof Abdo imad last updated on 29/Apr/18

find the value of ∫_0 ^(+∞) ((cos(αx))/((x^2 +1)( x^2 +2)(x^2 +3)))dx 2) calculate ∫_0 ^∞ (dx/((x^2 +1)(x^2 +2)(x^2 +3)))

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

Commented by prof Abdo imad last updated on 30/Apr/18

let put I =∫_0 ^∞ ((cos(αx))/((x^2 +1)(x^2 +2)(x^2 +3{))dx 2I = ∫_(−∞) ^(+∞) ((cos(αx))/((x^2 +1)(x^2 +2)(x^2 +3)))dx =Re( ∫_(−∞) ^(+∞) (e^(iαx) /((x^2 +1)(x^2 +2)(x^2 +3)))dx)let consider the complex function ϕ(z) = (e^(iαz) /((z^2 +1)(z^2 +2)(z^2 +3))) the poles of ϕ are i,−i,i(2)^(1/�) ,−i(√2) ,i(√3),−i(√(3 )) and Residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ( Res(ϕ,i) +Res(ϕ,i(√2))+Res(ϕ,i(√3)) ϕ(z)= (e^(iαz) /((z−i)(z+i)(z−i(√2))(z+i(√2))(z−i(√3))(z+i(√3)))) Res(ϕ,i) = (e^(−α) /(2i(−1+2)(−1+3))) = (e^(−α) /(4i)) Res(ϕ,i(√2)) = (e^(−α(√2)) /((2i(√2))(−2+1)(−2+3))) =− (e^(−α(√2)) /(2i(√2))) Res(ϕ,i(√3)) = (e^(−α(√3)) /(2i(√3)(−3+1)(−3+2))) = (e^(−α(√3)) /(4i(√3))) ∫_(−∞) ^(+∞) ϕ(z)dz = 2iπ( (e^(−α) /(4i)) − (e^(−α(√2)) /(2i(√2))) +(e^(−α(√3)) /(4i(√3)))) =(π/2) e^(−α) −(π/(√2)) e^(−α(√2)) +(π/(2(√3))) e^(−α(√3)) ⇒ I = (π/4) e^(−α) −(π/(2(√2))) e^(−α(√2)) + (π/(4(√3))) e^(−α(√3)) .

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

Commented by abdo imad last updated on 30/Apr/18

2) let take α=0 ⇒ ∫_0 ^∞ (dx/((x^2 +1)(x^2 +2)(x^2 +3))) =(π/4) −(π/(2(√2))) +(π/(4(√3))) .

$$\left.\mathrm{2}\right)\:{let}\:{take}\:\alpha=\mathrm{0}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{3}\right)}\:=\frac{\pi}{\mathrm{4}}\:−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:+\frac{\pi}{\mathrm{4}\sqrt{\mathrm{3}}}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18

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

$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)}.\frac{\left({x}^{\mathrm{2}} +\mathrm{2}\right)−\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)}.\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}−\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)}.\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{2}\right)}{dx} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18

1/2∫_0 ^∞ {(1/((x^2 +1)))−(1/((x^2 +3)))}dx −∫_0 ^∞ ((1/(x^2 +2))−(1/(x^2 +3)))._ dx first intregl1/2{tan^(−1) x −(1/(√3))tan^(−1) ((x/(√3)))} second 1/(√2)tan^(−1) (x/(√(2))) −1/(√3) tan^(−1) (x/(√3) ) add them ...

$$\mathrm{1}/\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\left\{\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)}\right\}{dx}\:−\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{2}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{3}}\right)._{} {dx} \\ $$$${first}\:{intregl}\mathrm{1}/\mathrm{2}\left\{{tan}^{−\mathrm{1}} {x}\:−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \left(\frac{{x}}{\sqrt{\mathrm{3}}}\right)\right\} \\ $$$${second}\:\:\mathrm{1}/\sqrt{\mathrm{2}}{tan}^{−\mathrm{1}} \left({x}/\sqrt{\left.\mathrm{2}\right)}\:−\mathrm{1}/\sqrt{\mathrm{3}}\:{tan}^{−\mathrm{1}} \left({x}/\sqrt{\mathrm{3}}\:\right)\right. \\ $$$${add}\:{them}\:... \\ $$$$ \\ $$$$ \\ $$

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