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Question Number 76780 by mathmax by abdo last updated on 30/Dec/19

$$findA={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}x{e}^{-x^2}arctan(x-\frac{1}{x})dx$$

Commented by mathmax by abdo last updated on 06/Jan/20

$$byparts{u}^{{}^{\prime }}={xe}^{-x^2}andv=arctan(x-\frac{1}{x})\Rightarrow$$$$A={[-\frac{1}{2}{e}^{-x^2}arctan(x-\frac{1}{x})]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}+\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-x^2}\times \frac{1+\frac{1}{x^2}}{1+{(x-\frac{1}{x})}^2}dx$$$$=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{(x^2+1)e^{-x^2}}{x^2\{1+\frac{{(x^2-1)}^2}{x^2}\}}dx\Rightarrow 2A={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{(x^2+1)e^{-x^2}}{x^2\left\{\frac{x^2+{(x^2-1)}^2}{x^2}\right\}}dx$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{(x^2+1)e^{-x^2}}{x^2+x^4-2x^2+1}dx={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{(x^2+1)e^{-x^2}}{x^4-x^2+1}dx$$$$letW\left(z\right)=\frac{(z^2+1)e^{-x^2}}{z^4-z^2+1}[polesofW?$$$$z^4-z^2+1=0\to t^2-t+1=0(t=z^2)$$$$\mathrm{\Delta }=1-4=-3\Rightarrow t_1=\frac{1+i\sqrt{3}}{2}=e^{\frac{i\pi }{3}}and{t}_2=\frac{1-i\sqrt{3}}{2}=e^{-\frac{i\pi }{3}}\Rightarrow$$$$W\left(z\right)=\frac{(z^2+1)e^{-z^2}}{(z^2-e^{\frac{i\pi }{3}})(z^2-e^{-\frac{i\pi }{3}})}=\frac{(z^2+1)e^{-z^2}}{(z-e^{\frac{i\pi }{6}})(z+e^{\frac{i\pi }{6}})(z-e^{-\frac{i\pi }{6}})(z+e^{-\frac{i\pi }{6}})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=2i\pi \{Res(W,e^{\frac{i\pi }{6}})+Res(W,-e^{-\frac{i\pi }{6}})\}$$$$...becontinued...$$$$$$$$$$

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