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Question Number 28035 by abdo imad last updated on 18/Jan/18

$$findthevalueof{\int }_0^{\mathrm{\infty }}x\frac{arctan\left(2x\right)}{{(2+x^2)}^2}dx.$$

Commented by abdo imad last updated on 23/Jan/18

$$letintegratrbyparts$$$$\text{Missing left or extra right}$$$$={\int }_0^{\mathrm{\infty }}\frac{dx}{(2+x^2)(1+4x^2)}$$$$=\frac{1}{2}{\int }_R\frac{dx}{(2+x^2)(1+4x^2)}letintroduce$$$$thecomplexfunction$$$$f\left(z\right)=\frac{1}{(z^2+2)(4z^2+1)}polesoff?$$$$f\left(z\right)=\frac{}{4(z-\sqrt{2}i)(x+\sqrt{2}i)(z-\frac{i}{2})(z+\frac{i}{2})}$$$$thepolesoffare\sqrt{2}i,-\sqrt{2}i,\frac{i}{2}and\frac{-i}{2}$$$${\int }_{R}f\left(x\right)dz=2i\pi (Res(f,\sqrt{2}i)+Res(f,\frac{i}{2}))$$$$Res(f,\sqrt{2}i)=\frac{1}{4\left(2\sqrt{2}i\right))({\left(\sqrt{2}i\right)}^2+\frac{1}{4})}$$$$=\frac{1}{8\sqrt{2}i(-2+\frac{1}{4})}=\frac{1}{8\sqrt{2}i.\frac{-7}{4}}$$$$=\frac{-1}{14\sqrt{2}i}$$$$Res(f,\frac{i}{2})=\frac{1}{4(\frac{i}{2}-\sqrt{2}i)(\frac{i}{2}+\sqrt{2}i)i}$$$$=\frac{1}{4(-\frac{1}{4}+2)i}=\frac{1}{7i}$$$${\int }_R^{}f\left(z\right)dz=2i\pi (\frac{-1}{14\sqrt{2}i}+\frac{1}{7i})$$$$=\frac{-\pi }{7\sqrt{2}}+\frac{2\pi }{7}=\frac{2\pi \sqrt{2}-\pi }{7\sqrt{2}}.$$

Commented by abdo imad last updated on 23/Jan/18

$$I=\frac{1}{2}{\int }_{R}f\left(z\right)dz.$$

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