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Question Number 34421 by abdo mathsup 649 cc last updated on 06/May/18

$$letA={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2-j}withj=e^{i\frac{2\pi }{3}}$$$$extractReAandIm\left(A\right)andcalculsteitsvalues.$$

Commented by abdo mathsup 649 cc last updated on 07/May/18

$$wehaveA={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2-(-\frac{1}{2}+i\frac{\sqrt{3}}{2})}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2+\frac{1}{2}-i\frac{\sqrt{3}}{2}}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2+\frac{1}{2}+i\frac{\sqrt{3}}{2}}{{(x^2+\frac{1}{2})}^2+\frac{3}{4}}dx\Rightarrow$$$$Re\left(A\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2+\frac{1}{2}}{{(x^2+\frac{1}{2})}^2+\frac{3}{4}}dxand$$$$Im\left(A\right)=\frac{\sqrt{3}}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{{(x^2+\frac{1}{2})}^2+\frac{3}{4}}letintroduce$$$$thecomplexfunction\phi \left(z\right)=\frac{1}{z^2-j}$$$$\phi \left(z\right)=\frac{1}{(z-\sqrt{j})(z+\sqrt{j})}=\frac{1}{(z-e^{i\frac{\pi }{3}})(z+{\mathrm{e}}^{\mathrm{i}\frac{\pi }{3}})}$$$$thepolesof\phi are{e}^{i\frac{\pi }{3}},-e^{i\frac{\pi }{3}}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,e^{i\frac{\pi }{3}})$$$$Res(\phi ,e^{i\frac{\pi }{3}})=\frac{1}{2e^{i\frac{\pi }{3}}}=\frac{1}{2}{e}^{-i\frac{\pi }{3}}$$$$=\frac{1}{2}(cos(-\frac{\pi }{3})+isin(-\frac{\pi }{3}))$$$$=\frac{1}{2}(\frac{1}{2}-i\frac{\sqrt{3}}{2})$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{1}{2}(\frac{1}{2}-i\frac{\sqrt{3}}{2})$$$$=i\frac{\pi }{2}+\pi \frac{\sqrt{3}}{2}\Rightarrow Re\left(A\right)=\pi \frac{\sqrt{3}}{2}$$$$andIm\left(A\right)=\frac{\pi }{2}.$$

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