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Question Number 37310 by math khazana by abdo last updated on 11/Jun/18
calculate ∫_(−∞) ^(+∞) (dx/((x^2 +1)(x^2 +4)(x^2 +9))) .
$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)\left({x}^{\mathrm{2}} \:+\mathrm{9}\right)}\:. \\ $$
Commented by math khazana by abdo last updated on 12/Jun/18
let consider ϕ(z) = (1/((z^2 +1)(z^2 +4)(z^(2 ) +9))) we have ϕ(z) = (1/((z−i)(z+i)(z−2i)(z+2i)(z−3i)(z+3i))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,i) +Res(ϕ,2i) +Res(ϕ,3i)} Res(ϕ,i) = (1/(2i(−1+4)(−1+9))) = (1/(48i)) Res(ϕ,2i) = (1/(4i(−4+1)(−4+9))) = (1/(60i)) Res(ϕ, 3i) = (1/(6i(−9+1)(−9+4))) = (1/(240i)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (1/(48i)) +(1/(60i)) +(1/(240i))} = (π/(24)) + (π/(30)) +(π/(120)) = ((5π)/(120)) + ((5π)/(120)) =((10π)/(120)) = (π/(12)) .
$${let}\:{consider}\:\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}^{\mathrm{2}} \:+\mathrm{4}\right)\left({z}^{\mathrm{2}\:\:} +\mathrm{9}\right)} \\ $$$${we}\:{have}\:\varphi\left({z}\right)\:\:=\:\frac{\mathrm{1}}{\left({z}−{i}\right)\left({z}+{i}\right)\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)\left({z}−\mathrm{3}{i}\right)\left({z}+\mathrm{3}{i}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,\mathrm{2}{i}\right)\:+{Res}\left(\varphi,\mathrm{3}{i}\right)\right\} \\ $$$${Res}\left(\varphi,{i}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}{i}\left(−\mathrm{1}+\mathrm{4}\right)\left(−\mathrm{1}+\mathrm{9}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{48}{i}} \\ $$$${Res}\left(\varphi,\mathrm{2}{i}\right)\:=\:\:\frac{\mathrm{1}}{\mathrm{4}{i}\left(−\mathrm{4}+\mathrm{1}\right)\left(−\mathrm{4}+\mathrm{9}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{60}{i}} \\ $$$${Res}\left(\varphi,\:\mathrm{3}{i}\right)\:=\:\frac{\mathrm{1}}{\mathrm{6}{i}\left(−\mathrm{9}+\mathrm{1}\right)\left(−\mathrm{9}+\mathrm{4}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{240}{i}} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\:\frac{\mathrm{1}}{\mathrm{48}{i}}\:+\frac{\mathrm{1}}{\mathrm{60}{i}}\:+\frac{\mathrm{1}}{\mathrm{240}{i}}\right\} \\ $$$$=\:\frac{\pi}{\mathrm{24}}\:\:+\:\frac{\pi}{\mathrm{30}}\:\:+\frac{\pi}{\mathrm{120}}\:\:\:\:\:\: \\ $$$$=\:\frac{\mathrm{5}\pi}{\mathrm{120}}\:+\:\frac{\mathrm{5}\pi}{\mathrm{120}}\:=\frac{\mathrm{10}\pi}{\mathrm{120}}\:=\:\frac{\pi}{\mathrm{12}}\:\:. \\ $$
Answered by ajfour last updated on 11/Jun/18
=(1/(24))∫(dx/(x^2 +1))+(1/(15))∫(dx/(x^2 +4))+(1/(40))∫(dx/(x^2 +9)) =(1/(24))tan^(−1) x+(1/(30))tan^(−1) (x/2)+(1/(120))tan^(−1) (x/3)+c . With limits −∞ →+∞ I= (π/(12)) .
$$=\frac{\mathrm{1}}{\mathrm{24}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{15}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{40}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{9}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{24}}\mathrm{tan}^{−\mathrm{1}} {x}+\frac{\mathrm{1}}{\mathrm{30}}\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{120}}\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{\mathrm{3}}+{c}\:. \\ $$$${With}\:{limits}\:−\infty\:\rightarrow+\infty \\ $$$${I}=\:\frac{\pi}{\mathrm{12}}\:. \\ $$
Commented by math khazana by abdo last updated on 12/Jun/18
correct answer thanks sir Ajfour
$${correct}\:{answer}\:{thanks}\:{sir}\:{Ajfour} \\ $$

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