Question Number 28035 by abdo imad last updated on 18/Jan/18
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:{x}\frac{{arctan}\left(\mathrm{2}{x}\right)}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:. \\ $$
Commented by abdo imad last updated on 23/Jan/18
![let integratr by parts I= ((−1)/(2(2+x^2 ))) arctan(2x)]^(+∞) _0 +∫_0 ^∞ ((2dx)/(2(2+x^2 )(1+4x^2 ))) = ∫_0 ^∞ (dx/((2+x^2 )(1+4x^2 ))) =(1/2)∫_(R ) (dx/((2+x^2 )(1+4x^2 ))) let introduce the complex function f(z)= (1/((z^2 +2)(4z^2 +1))) poles of f? f(z)= (/(4(z−(√2)i)(x+(√2)i)(z−(i/2))(z+(i/2)))) the poles of f are (√2)i,−(√2)i,(i/2) and ((−i)/2) ∫_R f(x)dz=2iπ(Res(f,(√2)i)+Res(f,(i/2))) Res(f,(√2)i)= (1/(4(2(√2)i))(((√2)i)^2 +(1/4)))) = (1/(8(√2)i(−2 +(1/4))))= (1/(8(√2)i .((−7)/4))) = ((−1)/(14(√2)i)) Res(f,(i/2))= (1/(4((i/2) −(√2)i)((i/2)+(√2)i)i)) = (1/(4( −(1/4)+2)i))= (1/(7i)) ∫_R ^ f(z)dz=2iπ( ((−1)/(14(√2)i)) + (1/(7i))) = ((−π)/(7(√2))) +((2π)/7)=((2π(√2)−π)/(7(√2))) .](https://www.tinkutara.com/question/Q28269.png)
$${let}\:{integratr}\:{by}\:{parts}\: \\ $$$$\left.{I}=\:\frac{−\mathrm{1}}{\mathrm{2}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)}\:{arctan}\left(\mathrm{2}{x}\right)\underset{\mathrm{0}} {\right]}^{+\infty} \:+\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{dx}}{\mathrm{2}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{{R}\:} \:\frac{{dx}}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)}\:{let}\:{introduce} \\ $$$${the}\:{complex}\:{function} \\ $$$${f}\left({z}\right)=\:\:\:\:\:\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{2}\right)\left(\mathrm{4}{z}^{\mathrm{2}} +\mathrm{1}\right)}\:\:{poles}\:{of}\:{f}? \\ $$$${f}\left({z}\right)=\:\:\:\frac{}{\mathrm{4}\left({z}−\sqrt{\mathrm{2}}{i}\right)\left({x}+\sqrt{\mathrm{2}}{i}\right)\left({z}−\frac{{i}}{\mathrm{2}}\right)\left({z}+\frac{{i}}{\mathrm{2}}\right)} \\ $$$${the}\:{poles}\:{of}\:{f}\:{are}\:\sqrt{\mathrm{2}}{i},−\sqrt{\mathrm{2}}{i},\frac{{i}}{\mathrm{2}}\:{and}\:\frac{−{i}}{\mathrm{2}} \\ $$$$\int_{{R}} {f}\left({x}\right){dz}=\mathrm{2}{i}\pi\left({Res}\left({f},\sqrt{\mathrm{2}}{i}\right)+{Res}\left({f},\frac{{i}}{\mathrm{2}}\right)\right) \\ $$$${Res}\left({f},\sqrt{\mathrm{2}}{i}\right)=\:\:\:\:\frac{\mathrm{1}}{\left.\mathrm{4}\left(\mathrm{2}\sqrt{\mathrm{2}}{i}\right)\right)\left(\left(\sqrt{\mathrm{2}}{i}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$$$=\:\:\:\:\frac{\mathrm{1}}{\mathrm{8}\sqrt{\mathrm{2}}{i}\left(−\mathrm{2}\:+\frac{\mathrm{1}}{\mathrm{4}}\right)}=\:\frac{\mathrm{1}}{\mathrm{8}\sqrt{\mathrm{2}}{i}\:.\frac{−\mathrm{7}}{\mathrm{4}}} \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{14}\sqrt{\mathrm{2}}{i}} \\ $$$${Res}\left({f},\frac{{i}}{\mathrm{2}}\right)=\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}\left(\frac{{i}}{\mathrm{2}}\:−\sqrt{\mathrm{2}}{i}\right)\left(\frac{{i}}{\mathrm{2}}+\sqrt{\mathrm{2}}{i}\right){i}} \\ $$$$=\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}\left(\:−\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}\right){i}}=\:\:\frac{\mathrm{1}}{\mathrm{7}{i}} \\ $$$$\int_{{R}} ^{} {f}\left({z}\right){dz}=\mathrm{2}{i}\pi\left(\:\:\frac{−\mathrm{1}}{\mathrm{14}\sqrt{\mathrm{2}}{i}}\:+\:\frac{\mathrm{1}}{\mathrm{7}{i}}\right) \\ $$$$=\:\frac{−\pi}{\mathrm{7}\sqrt{\mathrm{2}}}\:\:+\frac{\mathrm{2}\pi}{\mathrm{7}}=\frac{\mathrm{2}\pi\sqrt{\mathrm{2}}−\pi}{\mathrm{7}\sqrt{\mathrm{2}}}\:. \\ $$
Commented by abdo imad last updated on 23/Jan/18
$${I}=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{{R}} {f}\left({z}\right){dz}. \\ $$